GCE Further Mathematics (6360) Further Pure Unit 2 (MFP2) Textbook. Version: 1.4

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1 GCE Further Mathematics (660) Further Pure Uit (MFP) Tetbook Versio: 4

2 MFP Tetbook A-level Further Mathematics 660 Further Pure : Cotets Chapter : Comple umbers 4 Itroductio 5 The geeral comple umber 5 The modulus ad argumet of a comple umber 6 4 The polar form of a comple umber 8 5 Additio, subtractio ad multiplicatio of comple umbers of the form + iy 9 6 The cojugate of a comple umber ad the divisio of comple umbers of the form +iy 0 7 Products ad quotiets of comple umbers i their polar form 8 Equatig real ad imagiary parts 9 Further cosideratio of z z ad arg(z z ) 4 0 Loci o Argad diagrams 5 Chapter : Roots of polyomial equatios Itroductio Quadratic equatios Cubic equatios 4 4 Relatioship betwee the roots of a cubic equatio ad its coefficiets 7 5 Cubic equatios with related roots 8 6 A importat result 7 Polyomial equatios of degree 8 Comple roots of polyomial equatios with real coefficiets Chapter : Summatio of fiite series 8 Itroductio 9 Summatio of series by the method of differeces 40 Summatio of series by the method of iductio 45 4 Proof by iductio eteded to other areas of mathematics 48 Chapter 4: De Moivre s theorem ad its applicatios 5 4 De Moivre s theorem 54 4 Usig de Moivre s theorem to evaluate powers of comple umbers 56 4 Applicatio of de Moivre s theorem i establishig trigoometric idetities Epoetial form of a comple umber The cube roots of uity 7 46 The th roots of uity The roots of z, where is a o-real umber 77 cotiued overleaf

3 MFP Tetbook A-level Further Mathematics 660 Further Pure : Cotets (cotiued) Chapter 5: Iverse trigoometrical fuctios 85 5 Itroductio ad revisio 86 5 The derivatives of stadard iverse trigoometrical fuctios 89 5 Applicatio to more comple differetiatio 9 54 Stadard itegrals itegratig to iverse trigoometrical fuctios Applicatios to more comple itegrals 96 Chapter 6: Hyperbolic fuctios 0 6 Defiitios of hyperbolic fuctios 0 6 Numerical values of hyperbolic fuctios 04 6 Graphs of hyperbolic fuctios Hyperbolic idetities Osbore s rule 0 66 Differetiatio of hyperbolic fuctios 67 Itegratio of hyperbolic fuctios 4 68 Iverse hyperbolic fuctios 5 69 Logarithmic form of iverse hyperbolic fuctios 6 60 Derivatives of iverse hyperbolic fuctios 9 6 Itegrals which itegrate to iverse hyperbolic fuctios 6 Solvig equatios 5 Chapter 7: Arc legth ad area of surface of revolutio 7 Itroductio 7 Arc legth 7 Area of surface of revolutio 7 Aswers to the eercises i Further Pure 4

4 MFP Tetbook A-level Further Mathematics 660 Chapter : Comple Numbers Itroductio The geeral comple umber The modulus ad argumet of a comple umber 4 The polar form of a comple umber 5 Additio, subtractio ad multiplicatio of comple umbers of the form iy 6 The cojugate of a comple umber ad the divisio of comple umbers of the form iy 7 Products ad quotiets of comple umbers i their polar form 8 Equatig real ad imagiary parts 9 Further cosideratio of z z ad arg( z z) 0 Loci o Argad diagrams This chapter itroduces the idea of a comple umber Whe you have completed it, you will: kow what is meat by a comple umber; kow what is meat by the modulus ad argumet of a comple umber; kow how to add, subtract, multiply ad divide comple umbers; kow how to solve equatios usig real ad imagiary parts; uderstad what a Argad diagram is; kow how to sketch loci o Argad diagrams 4

5 MFP Tetbook A-level Further Mathematics 660 Itroductio You will have discovered by ow that some problems caot be solved i terms of real umbers For eample, if you use a calculator to evaluate 64 you get a error message This is because squarig every real umber gives a positive value; both ( 8) ad ( 8) are equal to 64 As caot be evaluated, a symbol is used to deote it the symbol used is i i i It follows that i The geeral comple umber The most geeral umber that ca be writte dow has the form i y, where ad y are real umbers The term iy is a comple umber with beig the real part ad y the imagiary part So, both i ad 4i are comple umbers The set of real umbers, (with which you are familiar), is really a subset of the set of comple umbers, This is because real umbers are actually umbers of the form 0i 5

6 MFP Tetbook A-level Further Mathematics 660 The modulus ad argumet of a comple umber Just as real umbers ca be represeted by poits o a umber lie, comple umbers ca be represeted by poits i a plae The poit P(, y) i the plae of coordiates with aes O ad Oy represets the comple umber iy ad the umber is uiquely represeted by that poit The diagram of poits i Cartesia coordiates represetig comple umbers is called a Argad diagram y r P(, y) O θ If the comple umber iy is deoted by z, ad hece z i, y z ( mod zed ) is defied as the distace from the origi O to the poit P represetig z Thus z OP r The modulus of a comple umber z is give by z y The argumet of z, arg z, is defied as the agle betwee the lie OP ad the positive -ais usually i the rage (π, π) The argumet of a comple umber z is give y by arg z, where ta You must be careful whe or y, or both, are egative 6

7 MFP Tetbook A-level Further Mathematics 660 Eample Fid the modulus ad argumet of the comple umber i Solutio The poit P represetig this umber, z, is show o the diagram y P, z ( ) ad ta θ Therefore, arg z π O Note that whe ta, θ could equal π or π However, the sketch clearly shows that θ lies i the secod quadrat This is why you eed to be careful whe evaluatig the argumet of a comple umber Eercise A Fid the modulus ad argumet of each of the followig comple umbers: (a) i, (b) i, (c) 4, (d) i Give your aswers for arg z i radias to two decimal places Fid the modulus ad argumet of each of the followig comple umbers: (a) i, (b) 4i, (c) 7i Give your aswers for arg z i radias to two decimal places 7

8 MFP Tetbook A-level Further Mathematics The polar form of a comple umber y I the diagram alogside, y rsi rcos ad P(, y) If P is the poit represetig the comple umber z i, y it follows that z may be writte i the form rcos irsi This is called the polar, or modulus argumet, form of a comple umber O θ r A comple umber may be writte i the form z r(cos isi ), where z r ad arg z For brevity, r(cos isi ) ca be writte as (r, θ) Eercise B Write the comple umbers give i Eercise A i polar coordiate form Fid, i the form i y, the comple umbers give i polar coordiate form by: (a) z cosπ π isi, (b) 4 4 π π 4cos isi 8

9 MFP Tetbook A-level Further Mathematics Additio, subtractio ad multiplicatio of comple umbers of the form + iy Comple umbers ca be subjected to arithmetic operatios Cosider the eample below Eample 5 Give that z 4i ad z i, fid (a) z z, (b) z z ad (c) zz Solutio (a) zz ( 4i) ( i) 4i (b) z z ( 4i) ( i) 6i (c) zz ( 4i)( i) 4i6i8i i 8 (sice i ) i I geeral, if z ai b ad z a i b, z z ( a a ) i( b b ) z z ( a a ) i( b b ) zz aa bb i( abab) Eercise C Fid z z ad z z whe: (a) z i ad z i, (b) z 6i ad z i 9

10 MFP Tetbook A-level Further Mathematics The cojugate of a comple umber ad the divisio of comple umbers of the form + iy The cojugate of a comple umber z iy (usually deoted by z * or z ) is the comple umber z* i y Thus, the cojugate of i is i, ad that of i is i O a Argad diagram, the poit represetig the comple umber z * is the reflectio of the poit represetig z i the -ais The most importat property of z * is that the product zz * is real sice zz* ( i y)( i y) iy iy i y y * zz z Divisio of two comple umbers demads a little more care tha their additio or multiplicatio ad usually requires the use of the comple cojugate Eample 6 z Simplify z, where z 4i ad z i Solutio 4i (4i)( i) i (i)(i) 4i6i8i i i 4i 50i 5 i z multiply the umerator ad deomiator of z by z *, ie ( i) so that the product of the deomiator becomes a real umber Eercise D z For the sets of comple umbers z ad z, fid z where (a) z 4 i ad z i, (b) z 6i ad z i 0

11 MFP Tetbook A-level Further Mathematics Products ad quotiets of comple umbers i their polar form If two comple umbers are give i polar form they ca be multiplied ad divided without havig to rewrite them i the form i y Eample 7 Fid z Solutio z if z π π cos isi ad z π π 6 6 π π π π zz cos isi cos isi π π π π 6 6 π π 6 6 cos isi π π π π π π π π 6 cos cos isi cos isi cos i si si cos πcos π si πsi π i si πcos π cos πsi π cos isi 6cos isi Notig that arg z is π, it follows that the modulus of zz is the product of the modulus of 6 z ad the modulus of z, ad the argumet of zz is the sum of the argumets of z ad z Usig the idetities: cos( A B) cos Acos Bsi Asi B si( A B) si Acos Bcos Asi B Eercise E z (a) Fid cos isi z z (b) What ca you say about the modulus ad argumet of z? if z cosπ π isi ad z π π 6 6

12 MFP Tetbook A-level Further Mathematics 660 Eample 7 If z r, ad z r,, show that zz rr Solutio zz r(cos isi ) r(cos isi ) cos( ) isi( ) rr coscos i(sicos cossi ) i sisi rr (cos cos si si ) i(si cos cos si ) rr cos( ) isi( ) If z ( r, ) ad z ( r, ) the zz ( rr, ) with the proviso that π may have to be added to, or subtracted from, if is outside the permitted rage for There is a correspodig result for divisio you could try to prove it for yourself z r (, ) ad (, ) the z, r same proviso regardig the size of the agle If z r z r with the

13 MFP Tetbook A-level Further Mathematics Equatig real ad imagiary parts z Goig back to Eample 6, z ca be simplified by aother method 4i Suppose we let a ib The, i ( i)( ai b) 4i abi( b a) 4i Now, a ad b are real ad the comple umber o the left had side of the equatio is equal to the comple umber o the right had side, so the real parts ca be equated ad the imagiary parts ca also be equated: ab ad ba 4 Thus b ad a, givig i as the aswer to a ib as i Eample 6 While this method is ot as straightforward as the method used earlier, it is still a valid method It also illustrates the cocept of equatig real ad imagiary parts If aibc i d, where a, b, c ad d are real, the a c ad b d Eample 8 Fid the comple umber z satisfyig the equatio Solutio ( 4i) z( i) z* i Let z ( a i b), the z* ( a i b) Thus, ( 4i)( ai b) ( i)( ai b) i Multiplyig out, a4iaib4i baiaibi b i Simplifyig, abi( 5a4 b) i Equatig real ad imagiary parts, ab, 5a4b So, a ad b Hece, z i Eercise F If z π z π 6, ad,, fid, i polar form, the comple umbers (a) zz, (b) z z, (c) z, (d) z, (e) z z Fid the comple umber satisfyig each of these equatios: (a) ( i) z i, (b) ( z i)(i) 7i, (c) z iz*

14 MFP Tetbook A-level Further Mathematics Further cosideratio of z z ad arg( z z) Sectio 5 cosidered simple cases of the sums ad differeces of comple umbers Cosider ow the comple umber z z z, where z iy ad z i y The poits A ad B represet z ad z, respectively, o a Argad diagram y A, ) B, ) ( y ( y O C The z z z ( ) i( y y) ad is represeted by the poit C (, y y) This makes OABC a parallelogram From this it follows that z z OC ( ) ( y y), that is to say z z is the legth AB i the Argad diagram Similarly arg( z z) is the agle betwee OC ad the positive directio of the -ais This i tur is the agle betwee AB ad the positive directio If the comple umber z is represeted by the poit A, ad the comple umber z is represeted by the poit B i a Argad diagram, the z z AB, ad arg( z z) is the agle betwee AB directio of the -ais ad the positive Eercise G Fid z z ad arg( z z) i (a) z i, z 7 5i, (b) z i, z 4 i, (c) z i, z 4 5i 4

15 MFP Tetbook A-level Further Mathematics Loci o Argad diagrams A locus is a path traced out by a poit subjected to certai restrictios Paths ca be traced out by poits represetig variable comple umbers o a Argad diagram just as they ca i other coordiate systems Cosider the simplest case first, whe the poit P represets the comple umber z such that z k This meas that the distace of P from the origi O is costat ad so P will trace out a circle z k represets a circle with cetre O ad radius k If istead zz k, where z is a fied comple umber represeted by the poit A o a Argad diagram, the (from Sectio 9) z z represets the distace AP ad is costat It follows that P must lie o a circle with cetre A ad radius k zz k represets a circle with cetre z ad radius k Note that if zz k, the the poit P represetig z ca ot oly lie o the circumferece of the circle, but also aywhere iside the circle The locus of P is therefore the regio o ad withi the circle with cetre A ad radius k Now cosider the locus of a poit P represeted by the comple umber z subject to the coditios zz z z, where z ad z are fied comple umbers represeted by the poits A ad B o a Argad diagram Agai, usig the result of Sectio 9, it follows that AP BP because z z is the distace AP ad z z is the distace BP Hece, the locus of P is a straight lie zz z z represets a straight lie the perpedicular bisector of the lie joiig the poits z ad z Note also that if zz z z the locus of z is ot oly the perpedicular bisector of AB, but also the whole half plae, i which A lies, bouded by this bisector All the loci cosidered so far have bee related to distaces there are also simple loci i Argad diagrams ivolvig agles The simplest case is the locus of P subject to the coditio that arg z, where is a fied agle y O α P 5

16 MFP Tetbook A-level Further Mathematics 660 This coditio implies that the agle betwee OP ad O is fied ( ) so that the locus of P is a straight lie arg z represets the half lie through O iclied at a agle to the positive directio of O Note that the locus of P is oly a half lie the other half lie, show dotted i the diagram above, would have the equatio arg z π, possibly π if π falls outside the specified rage for arg z I eactly the same way as before, the locus of a poit P satisfyig arg( zz), where z is a fied comple umber represeted by the poit A, is a lie through A arg( z z ) iclied at a agle to the positive directio of O represets the half lie through the poit z y P A α O Note agai that this locus is oly a half lie the other half lie would have the equatio arg( zz ) π, possibly π Fially, cosider the locus of ay poit P satisfyig arg( zz) This idicates that the agle betwee AP ad the positive -ais lies betwee ad, so that P ca lie o or withi the two half lies as show shaded i the diagram below y A β α O 6

17 MFP Tetbook A-level Further Mathematics 660 Eercise H Sketch o Argad diagrams the locus of poits satisfyig: (a) z, (b) arg( z ) π, (c) z i 5 4 Sketch o Argad diagrams the regios where: (a) z i, (b) π arg( z 4 i) 5π 6 Sketch o a Argad diagram the regio satisfyig both z i ad 0arg z π 4 4 Sketch o a Argad diagram the locus of poits satisfyig both z i z i ad z i 4 7

18 MFP Tetbook A-level Further Mathematics 660 Miscellaeous eercises Fid the comple umber which satisfies the equatio ziz* 4 i, where z * deotes the comple cojugate of z [AQA Jue 00] The comple umber z satisfies the equatio ziz (a) Fid z i the form a i b, where a ad b are real (b) Mark ad label o a Argad diagram the poits represetig z ad its cojugate, z * (c) Fid the values of z ad z z* [NEAB March 998] The comple umber z satisfies the equatio zz* z z* i, where z * deotes the comple cojugate of z Fid the two possible values of z, givig your aswers i the form a i b [AQA March 000] 4 By puttig z i y, fid the comple umber z which satisfies the equatio z z* i, i where z * deotes the comple cojugate of z [AQA Specime] 5 (a) Sketch o a Argad diagram the circle C whose equatio is z i (b) Mark the poit P o C at which z is a miimum Fid this miimum value (c) Mark the poit Q o C at which arg z is a maimum Fid this maimum value [NEAB Jue 998] 8

19 MFP Tetbook A-level Further Mathematics (a) Sketch o a commo Argad diagram (i) the locus of poits for which z i, (ii) the locus of poits for which arg z π 4 (b) Idicate, by shadig, the regio for which z i ad arg z π 4 [AQA Jue 00] 7 The comple umber z is defied by z i i (a) (i) Epress z i the form a i b (ii) Fid the modulus ad argumet of z, givig your aswer for the argumet i the form p π where p (b) The comple umber z has modulus ad argumet 7π The comple umber z is defied by z zz (i) Show that z π 4 ad arg z 6 (ii) Mark o a Argad diagram the poits P ad P which represet z ad z, respectively (iii) Fid, i surd form, the distace betwee P ad P [AQA Jue 000] 9

20 MFP Tetbook A-level Further Mathematics (a) Idicate o a Argad diagram the regio of the comple plae i which 0arg π z (b) The comple umber z is such that 0arg π z ad π arg z π 6 (i) Sketch aother Argad diagram showig the regio R i which z must lie (ii) Mark o this diagram the poit A belogig to R at which z has its least possible value (c) At the poit A defied i part (b)(ii), z z A (i) Calculate the value of z A (ii) Epress z A i the form a i b [AQA March 999] 9 (a) The comple umbers z ad w are such that z 4i i ad w 4 i i Epress each of z ad w i the form a i b, where a ad b are real (b) (i) Write dow the modulus ad argumet of each of the comple umbers 4 i ad i Give each modulus i a eact surd form ad each argumet i radias betwee π ad π (ii) The poits O, P ad Q i the comple plae represet the comple umbers 0 0i, 4 i ad i, respectively Fid the eact legth of PQ ad hece, or otherwise, show that the triagle OPQ is right-agled [AEB Jue 997] 0

21 MFP Tetbook A-level Further Mathematics 660 Chapter : Roots of Polyomial Equatios Itroductio Quadratic equatios Cubic equatios 4 Relatioship betwee the roots of a cubic equatio ad its coefficiets 5 Cubic equatios with related roots 6 A importat result 7 Polyomial equatios of degree 8 Comple roots of polyomial equatios with real coefficiets This chapter revises work already covered o roots of equatios ad eteds those ideas Whe you have completed it, you will: kow how to solve ay quadratic equatio; kow that there is a relatioship betwee the umber of real roots ad form of a polyomial equatio, ad be able to sketch graphs; kow the relatioship betwee the roots of a cubic equatio ad its coefficiets; be able to form cubic equatios with related roots; kow how to eted these results to polyomials of higher degree; kow that comple cojugates are roots of polyomials with real coefficiets

22 MFP Tetbook A-level Further Mathematics 660 Itroductio You should have already met the idea of a polyomial equatio A polyomial equatio of degree, oe with as the highest power of, is called a quadratic equatio Similarly, a polyomial equatio of degree has as the highest power of ad is called a cubic equatio; oe with 4 as the highest power of is called a quartic equatio I this chapter you are goig to study the properties of the roots of these equatios ad ivestigate methods of solvig them Quadratic equatios You should be familiar with quadratic equatios ad their properties from your earlier studies of pure mathematics However, eve if this sectio is familiar to you it provides a suitable base from which to move o to equatios of higher degree You will kow, for eample, that quadratic equatios of the type you have met have two roots (which may be coicidet) There are ormally two ways of solvig a quadratic equatio by factorizig ad, i cases where this is impossible, by the quadratic formula Graphically, the roots of the equatio a b c 0 are the poits of itersectio of the curve y a b c ad the lie y 0 (ie the -ais) For eample, a sketch of part of y 8 is show below y ( 4, 0) (, 0) (0, 8) The roots of this quadratic equatio are those of ( )( 4) 0, which are ad 4

23 MFP Tetbook A-level Further Mathematics 660 A sketch of part of the curve y 4 4 is show below y (0, 4) O (, 0) I this case, the curve touches the -ais The equatio ( ) 0 ad, a repeated root 44 0 may be writte as Not all quadratic equatios are as straightforward as the oes cosidered so far A sketch of part of the curve y 4 5 is show below y (0, 5) O (, ) This curve does ot touch the -ais so the equatio 45 0 caot have real roots Certaily, 4 5 will ot factorize so the quadratic formula b b 4ac a has to be used to solve this equatio This leads to 4 60 ad, usig ideas from Chapter, this becomes 4 i or i It follows that the equatio 45 0 does have two roots, but they are both comple umbers I fact the two roots are comple cojugates You may also have observed that whether a quadratic equatio has real or comple roots depeds o the value of the discrimiat b 4 ac The quadratic equatio a b c 0, where a, b ad c are real, has comple roots if b 4ac 0

24 MFP Tetbook A-level Further Mathematics 660 Eercise A Solve the equatios (a) 60 0, (b) 06 0 Cubic equatios As metioed i the itroductio to this chapter, equatios of the form a b c d 0 are called cubic equatios All cubic equatios have at least oe real root ad this real root is ot always easy to locate The reaso for this is that cubic curves are cotiuous they do ot have asymptotes or ay other form of discotiuity Also, as, the term a becomes the domiat part of the epressio ad a (if a 0), whilst a whe Hece the curve must cross the lie y 0 at least oce If a 0, the as, ad a as ad this does ot affect the result a 4

25 MFP Tetbook A-level Further Mathematics 660 A typical cubic equatio, below y y a b c d with a 0, ca look like ay of the sketches The equatio of this curve has three real roots because the curve crosses the lie y 0 at three poits O y y O O I each of the two sketch graphs above, the curve crosses the lie y 0 just oce, idicatig just oe real root I both cases, the cubic equatio will have two comple roots as well as the sigle real root Eample (a) Fid the roots of the cubic equatio 0 (b) Sketch a graph of y Solutio y 4 (a) If f( ), the f() 0 Therefore is a factor of f() f( ) ( )( 4) ( )( )( ) Hece the roots of f() = 0 are, ad (b)

26 MFP Tetbook A-level Further Mathematics 660 Eample Fid the roots of the cubic equatio Solutio Let f ( ) 4 6 The f () Therefore is a factor of f(), ad f ( ) ( )( 6 ) The quadratic i this epressio has o simple roots, so usig the quadratic formula o 6 0, 4 a i i b b ac Hece the roots of f ( ) 0 are ad i Eercise B Solve the equatios (a) 5 0, (b) (c) 4 0, 0 0 6

27 MFP Tetbook A-level Further Mathematics Relatioship betwee the roots of a cubic equatio ad its coefficiets As a cubic equatio has three roots, which may be real or comple, it follows that if the geeral cubic equatio a b c d 0 has roots, ad, it may be writte as a ( )( )( ) 0 Note that the factor a is required to esure that the coefficiets of are the same, so makig the equatios idetical Thus, o epadig the right had side of the idetity, a b c d a( )( )( ) a a( ) a( ) a The two sides are idetical so the coefficiets of ad ca be compared, ad also the umber terms, ba( ) ca( ) d a If the cubic equatio a b c d 0 has roots, ad, the b, a c, a d a Note that meas the sum of all the roots, ad that meas the sum of all the possible products of roots take two at a time Eercise C Fid, ad for the followig cubic equatios: (a) 7 5 0, (b) The roots of a cubic equatio are, ad If, the cubic equatio 7 ad 5, state 7

28 MFP Tetbook A-level Further Mathematics Cubic equatios with related roots The eample below shows how you ca fid equatios whose roots are related to the roots of a give equatio without havig to fid the actual roots Two methods are give Eample 5 The cubic equatio 4 0 has roots, ad Fid the cubic equatios with: (a) roots, ad, (b) roots, ad, (c) roots, ad Solutio: method From the give equatio, 0 4 (a) Hece From which the equatio of the cubic must be or 6 0 (b) 4 ( ) 66 ( )( ) (4) 00 ( )( )( ) Hece the equatio of the cubic must be 0 8

29 MFP Tetbook A-level Further Mathematics 660 (c) So that the cubic equatio with roots, ad is or 4 0 9

30 MFP Tetbook A-level Further Mathematics 660 The secod method of fidig the cubic equatios i Eample 5 is show below It is ot always possible to use this secod method, but whe you ca it is much quicker tha the first Solutio: method (a) As the roots are to be, ad, it follows that, if X, the a cubic equatio i X must have roots which are twice the roots of the cubic equatio i As the equatio i is 4 0, if you substitute X the equatio i X becomes as before or X X 4 0, X 6X 0 (b) I this case, if you put X i 4 0, the ay root of a equatio i X must be less tha the correspodig root of the cubic i Now, X gives X ad substitutig ito 4 0 gives ( X ) ( X ) 4 0 which reduces to X X 0 (c) I this case you use the substitutio X or X For 4 0 X X O multiplyig by X, this gives X 4X 0 or 4X X 0 as before 4 0 this gives Eercise D The cubic equatio 47 0 has roots, ad Usig the first method described above, fid the cubic equatios whose roots are (a), ad, (b), ad, (c), ad Repeat Questio above usig the secod method described above Repeat Questios ad above for the cubic equatio 6 0 0

31 MFP Tetbook A-level Further Mathematics A importat result If you square you get ( ) ( )( ) So,, or for three umbers, ad This result is well worth rememberig it is frequetly eeded i questios ivolvig the symmetric properties of roots of a cubic equatio Eample 6 The cubic equatio has roots, ad Fid the cubic equatios with (a) roots, ad, (b) roots, ad [Note that the direct approach illustrated below is the most straightforward way of solvig this type of problem] Solutio (a) ( ) Hece the cubic equatio is (b) 5 6 usig the same result but replacig with with, ad with Thus 6 (5) 46 ( ) Hece the cubic equatio is 46 0

32 MFP Tetbook A-level Further Mathematics Polyomial equatios of degree The ideas covered so far o quadratic ad cubic equatios ca be eteded to equatios of ay degree A equatio of degree has two roots, oe of degree has three roots so a equatio of degree has roots Suppose the equatio a b c d k 0 has roots,,, the b, a c, a d, a ( ) k util, fially, the product of the roots a Remember that is the sum of the products of all possible pairs of roots, is the sum of the products of all possible combiatios of roots take three at a time, ad so o I practice, you are ulikely to meet equatios of degree higher tha 4 so this sectio cocludes with a eample usig a quartic equatio Eample 7 4 The quartic equatio has roots,, ad Write dow (a), (b) (c) Hece fid Solutio (a) (b) (c) 4 6 ( Now ( This shows that the importat result i Sectio 6 ca be eteded to ay umber of letters Hece ( ) ( ) 0

33 MFP Tetbook A-level Further Mathematics 660 Eercise E 4 The quartic equatio 58 0 has roots,, ad (a) Fid the equatio with roots,, ad (b) Fid 8 Comple roots of polyomial equatios with real coefficiets Cosider the polyomial equatio f( ) a b c k Usig the ideas from Chapter, if p ad q are real, f( pi q) a( pi q) b( pi q) k ui v, where u ad v are real Now, f( i ) ( i ) ( i ) p q a p q b p q k uiv sice i raised to a eve power is real ad is the same as i raised to a eve power, makig the real part of f ( p i q) the same as the real part of f ( p i q) But i raised to a odd power is the same as i raised to a odd power multiplied by, ad odd powers of i comprise the imagiary part of f ( p i q) Thus, the imagiary part of f ( p i q) is times the imagiary part of f ( p i q) Now if p iq is a root of f ( ) 0, it follows that u iv 0 ad so u 0 ad v 0 Hece, uiv 0 makig f ( p i q) 0 ad p iq a root of f ( ) 0 If a polyomial equatio has real coefficiets ad if p i q, where p ad q are real, is a root of the polyomial, the its comple cojugate, p i q, is also a root of the equatio It is very importat to ote that the coefficiets i f ( ) 0 must be real If f ( ) 0 has comple coefficiets, this result does ot apply

34 MFP Tetbook A-level Further Mathematics 660 Eample 8 The cubic equatio k 0, where k is real, has oe root equal to i Fid the other two roots ad the value of k Solutio As the coefficiets of the cubic equatio are real, it follows that i is also a root Cosiderig the sum of the roots of the equatio, if is the third root, ( i) ( i), To fid k, k ( i)( i)( ) 5, k 5 Eample 8 The quartic equatio roots has oe root equal to i Fid the other three Solutio As the coefficiets of the quartic are real, it follows that i is also a root Hece (i) ( i) is a quadratic factor of the quartic Now, (i) (i) (i) (i) (i)(i) 5 4 Hece 5 is a factor of Therefore 45 ( 5)( a b) Comparig the coefficiets of, a a 4 Cosiderig the umber terms, 5b 5 b Hece the quartic equatio may be writte as ( 5)( 4) 0 ( 5)( )( ) 0, ad the four roots are i,i, ad 4

35 MFP Tetbook A-level Further Mathematics 660 Eercise F A cubic equatio has real coefficiets Oe root is ad aother is i Fid the cubic equatio i the form a bc 0 The cubic equatio roots has oe root equal to i Fid the other two The quartic equatio three roots has oe root equal to i Fid the other 5

36 MFP Tetbook A-level Further Mathematics 660 Miscellaeous eercises The equatio 4 0, p where p is a costat, has roots, ad, where 0 (a) Fid the values of ad (b) Fid the value of p [NEAB Jue 998] The umbers, ad satisfy the equatios ad (a) Show that 0 (b) The umbers, ad are also the roots of the equatio 0, p q r where p, q ad r are real (i) Give that 4i ad that is real, obtai ad (ii) Calculate the product of the three roots (iii) Write dow, or determie, the values of p, q ad r [AQA Jue 000] The roots of the cubic equatio are, ad 4 0 (a) Write dow the values of, ad (b) Fid the cubic equatio, with iteger coefficiets, havig roots, ad [AQA March 000] 4 The roots of the equatio are, ad (a) Write dow the value of (b) Give that i is a root of the equatio, fid the other two roots [AQA Specime] 6

37 MFP Tetbook A-level Further Mathematics The roots of the cubic equatio 0, p q r where p, q ad r are real, are, ad (a) Give that, write dow the value of p (b) Give also that 5, (i) fid the value of q, (ii) eplai why the equatio must have two o-real roots ad oe real root (c) Oe of the two o-real roots of the cubic equatio is 4i (i) Fid the real root (ii) Fid the value of r [AQA March 999] 6 (a) Prove that whe a polyomial (b) The polyomial g is defied by 5 f is divided by a, g 6 p q, where p ad q are real costats Whe remaider is (i) Fid the values of p ad q [AQA Jue 999] (ii) Show that whe the remaider is a f g is divided by i, where i, the g is divided by i, the remaider is 6i 7

38 MFP Tetbook A-level Further Mathematics 660 Chapter : Summatio of Fiite Series Itroductio Summatio of series by the method of differeces Summatio of series by the method of iductio 4 Proof by iductio eteded to other areas of mathematics This chapter eteds the idea of summatio of simple series, with which you are familiar from earlier studies, to other kids of series Whe you have completed it, you will: kow ew methods of summig series; kow which method is appropriate for the summatio of a particular series; uderstad a importat method kow as the method of iductio; be able to apply the method of iductio i circumstaces other tha i the summatio of series 8

39 MFP Tetbook A-level Further Mathematics 660 Itroductio You should already be familiar with the idea of a series a series is the sum of the terms of a sequece That is, the sum of a umber of terms where the terms follow a defiite patter For istace, the sum of a arithmetic progressio is a series I this case each term is bigger tha the preceedig term by a costat umber this costat umber is usually called the commo differece Thus, 58 4 is a series of 5 terms, i arithmetic progressio, with commo differece The sum of a geometric progressio is also a series Istead of addig a fied umber to fid the et cosecutive umber i the series, you multiply by a fied umber (called the commo ratio) Thus, is a series of 6 terms, i geometric progressio, with commo ratio A fiite series is a series with a fiite umber of terms The two series above are eamples of fiite series 9

40 MFP Tetbook A-level Further Mathematics 660 Summatio of series by the method of differeces Some problems require you to fid the sum of a give series, for eample sum the series 4 I others you have to show that the sum of a series is a specific umber or a give epressio A eample of this kid of problem is show that 4 The method of differeces is usually used whe the sum of the series is ot give Suppose you wat to fid the sum, ur, of a series r u u u u where the terms follow a certai patter The aim i the method of differeces is to epress th the r term, which will be a fuctio of r (just as th is the r term of the first series rr above), as the differece of two epressios i r of the same form I other words, u r is epressed as f rfr, or possibly f r f r, where f r is some fuctio of r If you ca epress u r i this way, it ca be see that settig r ad the r gives uu f f f f f f If this idea is eteded to the whole series, the r u f f r u f f r u f f 4 Now, addig these terms gives r u f f r u f f f f +f f uu uuu f f +f f +f f 4 +f 4 The left had side of this epressio is the required sum of the series, had side, early all the terms cacel out: f,f,f 4,,f f f as the sum of the series ur O the right r all cacel leavig just 40

41 MFP Tetbook A-level Further Mathematics 660 Eample Fid the sum of the series Solutio 4 Clearly, this is ot a familiar stadard series, such as a arithmetic or geometric series Nor is the aswer give So it seems that the method of differeces ca be applied th As above, the r term, u r, is give by We eed to try to split up u r The oly rr sesible way to do this is to epress i partial fractios Suppose rr A B rr r r The, A( r) Br Comparig the coefficiets of r, A B 0 Comparig the costat terms, A Hece B ad u r r r r r Hece, i this case the f r metioed previously would be, o Now, writig dow the series term by term, r r r 4 4 r ( ) ( ) r ( ) r with f r, r ad so Addig the colums, the left had side becomes Because the,, etc terms r rr ( ) cacel, the right had side becomes, amely the first left had side term ad the last right had side term Hece, r rr ( ) ( ) 4

42 MFP Tetbook A-level Further Mathematics 660 Eample Show that Solutio r r r r 4 r Hece fid r r The left had side of the idetity has a commo factor, r r r r r r r r r r rr r r r r r r r 4r Now, if r r r so that f( ), the 4r is of the form 4 r r r, f( r) r r f r f( r) Listig the terms i colums, as i Eample, r 4 0 r 4 r r r Addig the colums, it ca be see that the left had side is Summig the right had side, all the terms cacel out ecept those shaded i the scheme above, so the sum is Hece, r r 4 0 Hece, r 4 r 0, as required r 4

43 MFP Tetbook A-level Further Mathematics 660 I both Eamples ad, oe term o each lie cacelled out with a term o the et lie whe the additio was doe Some series may be such that a term i oe lie cacels with a term o a lie two rows below it Eample Sum the series Solutio As i Eample, the way forward is to epress Let rr r r Multiplyig both sides by rr, A r B r A B 4 i partial fractios Comparig the coefficiets of r, A B 0, so A B Comparig the costat terms, A B Hece A ad B Thus, 4 4 r r 4 r 4 r Now substitute r,,, r r [ote that othig will cacel at this stage] r [ote that will cacel o the first row ad the third row] 4 5 r r r r

44 MFP Tetbook A-level Further Mathematics 660 There will be two terms left at the begiig of the series whe the colums are added, ad Likewise, there will be two terms left at the ed of the series the right 4 4 had part of the r ad r rows Therefore, additio gives Eercise A (a) Simplify r r r r (b) Use your result to obtai r r (a) Show that rr ( )( r) ( r)( r)( r) rr ( )( r)( r) (b) Hece sum the series ( )( )( ) r rr r r (a) Show that (b) Deduce r r r 6r r 44

45 MFP Tetbook A-level Further Mathematics 660 Summatio of a series by the method of iductio The method of iductio is a method of summig a series of, say, terms whe the sum is give i terms of Suppose you have to show that the sum of terms of a series is S() If you assume that the summatio is true for oe particular iteger, say k, where k, the you are assumig that the sum of the first k terms is S(k) You may thik that this rather begs the questio but it must be uderstood that the result is assumed to be true for oly oe value of, amely k You the use this assumptio to prove that the sum of the series to k terms is Sk that is to say that by addig oe etra term, the et term i the series, the sum has eactly the same form as S() but with replaced by k Fially, it is demostrated that the result is true for To summarise: Assume that the result of the summatio is true for k ad prove that it is true for k Prove that the result is true for Statemet shows that, by puttig k (which is kow to be true from Statemet ), the result must be true for ; ad Statemet shows that by puttig k the result must be true for ; ad so o By buildig up the result, it ca be said that the summatio result is true for all positive itegers There is a formal way of writig out the method of iductio which is show i the eamples below For coveiece, ad compariso, the eamples worked i Sectio are used agai here Eample Show that r r r Solutio Assume that the result is true for k; that is to say 4 k kk k Addig the et term to both sides, 4 k kk kk k kk The k k r r k k k r kk kk k k kk 45

46 MFP Tetbook A-level Further Mathematics 660 kk kk k k k, k which is of the same form but with k replacig k Hece, if the result is true for k, it is true for k But it is true for because the left had side is, ad the right had side is Therefore the result is true for all positive itegers by iductio Eample Show that r r 4 Solutio Assume that the result is true for k, that is to say k r The, addig the et term to both sides r k k 4 k r r k k k 4 k k 4k 4 k k 4k 4 4 k k 4 k k, 4 which is of the same form but with k replacig k Hece, if the result is true for k, it is true for k But it is true for k because the left had side is, ad the right had side is Therefore the result is true for all positive itegers by iductio 4 46

47 MFP Tetbook A-level Further Mathematics 660 Eercise B Prove the followig results by the method of iductio: 4 (a) (b) 6 (c) rr r (d) r r r 7 6!! 47

48 MFP Tetbook A-level Further Mathematics Proof by iductio eteded to other areas of mathematics The method of iductio is certaily useful i the summatio of series but it is ot cofied to this area of mathematics This chapter cocludes with a look at its use i three other coectios sequeces, divisibility ad de Moivre s theorem for positive itegers Eample 4 applicatio to sequeces A sequece u, u, u, is defied by u u u Prove by iductio that for all, u Solutio Assume that the result is true for k, that is to say k u k k The, usig the relatioship give, u k uk k k k k k k k k k k k k k k k k which is of the same form as uk but with k replacig k Hece, if the result is true for k, it is true for k But whe k, u as give Therefore the result is true for all positive itegers by iductio 48

49 MFP Tetbook A-level Further Mathematics 660 Eample 4 applicatio to divisibility Prove by iductio that if is a positive iteger, Solutio The best approach is a little differet to that used so far 7 is divisible by 8 Assume that the result is true for k, i other words that 7 is divisible by 8 k k Whe k the epressio is 7 Cosider 7 k 7, the differece betwee the values whe k ad k This epressio is equal to k k k k k k 8 ( k) k Thus, if k 7 is divisible by 8, ad clearly 8 k is divisible by 8, it follows that k 7 is also divisible by 8 I other words, if the result is true for k, it is true for k But for, 7 6 ad is divisible by 8 Hece, for all positive itegers by iductio 7 is divisible by 8 Eample 4 applicatio to de Moivre s theorem for positive itegers Prove by iductio that for itegers, Solutio Assume that the result is true for k, that is to say k cos isi cos k isi k Multiplyig both sides by cos isi, cos isi cos isi k cos isi cos isicos k isi kcos isi k cos isi cos k cos isi k cos isi cos k i si ksi k k k k cosk isi k, cos cos si si i si cos cos si i which is of the same form but with k replacig k Hece, if the result is true for k it is true for k But whe k, cos i si cos i si Therefore the result is true for all positive itegers by iductio 49

50 MFP Tetbook A-level Further Mathematics 660 Eercise C Prove the followig results by the method of iductio i all eamples is a positive iteger: (a) (b) is divisible by 6 is divisible by 7 [Hit: cosider f 5f 5 f 5 ] d d (c) where [Hit: use the formula for differetiatig a product] (d) is divisible by 50

51 MFP Tetbook A-level Further Mathematics 660 Miscellaeous eercises Use the idetity to show that [AQA Jue 999] r r r r r r r (a) Use the idetity 4 r r r r r r to show that r 4r (b) Hece fid r r, givig your aswer as a product of three factors i terms of [AQA Jue 000] Prove by iductio that [AQA March 999] r r r Prove by iductio, or otherwise, that [NEAB Jue 998] r r r!! 5 Prove by iductio that for all itegers 0, 7 is divisible by [AQA Specime] 6 Use mathematical iductio to prove that for all positive itegers [AEB Jue 997] r rr 5

52 MFP Tetbook A-level Further Mathematics A sequece u, u, u, is defied by u, u, u Prove by iductio that for all, u [AQA Jue 999] 8 Verify the idetity r r r r r r r r Hece, usig the method of differeces, prove that [AEB Jauary 998] r rr 9 The fuctio f is defied for all o-egative itegers r by f r r r (a) Verify that f f r r Ar for some iteger A, statig the value of A (b) Hece, usig the method of differeces, prove that [AEB Jauary 000] r r 0 For some value of the costat A, r A r r r r (a) By settig, or otherwise, determie the value of A (b) Use mathematical iductio to prove the result for all positive itegers (c) Deduce the sum of the ifiite series [AEB Jue 000] 5

53 MFP Tetbook A-level Further Mathematics 660 Chapter 4: De Moivre s Theorem ad its Applicatios 4 De Moivre s theorem 4 Usig de Moivre s theorem to evaluate powers of comple umbers 4 Applicatio of de Moivre s theorem i establishig trigoometric idetities 44 Epoetial form of a comple umber 45 The cube roots of uity 46 The th roots of uity 47 The roots of z, where is a o-real umber This chapter itroduces de Moivre s theorem ad may of its applicatios Whe you have completed it, you will: kow the basic theorem; be able to fid shorter ways of workig out powers of comple umbers; discover alterative methods for establishig some trigoometric idetities; kow a ew way of epressig comple umbers; kow how to work out the th roots of uity ad, i particular, the cube roots; be able to solve certai types of polyomial equatios 5

54 MFP Tetbook A-level Further Mathematics De Moivre s theorem I Chapter (sectio 4), you saw a very importat result kow as de Moivre s theorem It was proved by iductio that, if is a positive iteger, the cos isi cos isi De Moivre s theorem holds ot oly whe is a positive iteger, but also whe it is egative ad eve whe it is fractioal Let be a egative iteger ad suppose k The k is a positive iteger ad k cos isi cos isi k cos isi cos k isi k Some of the results obtaied i Chapter ca ow be put to use I order to remove i from the deomiator of the epressio above, the umerator ad deomiator are multiplied by the comple cojugate of the deomiator, i this case cos k isi k Thus, cosk isik cos k i si k cos k i si k cos k i si k cos k isi k cos k i si k cos k i si k cos k i si k cos k isi k cos k si k cos k isi k coskisi k cos isi, as required If is a fractio, say p q where p ad q are itegers, the q p p pθ pθ cos θisi θ cos qisi q q q q q cos pθ isi pθ But p is also a iteger ad so p cos p isi p cos isi th Takig the q root of both sides, p p p q cos isi cos isi q q [ q is a iteger] 54

55 MFP Tetbook A-level Further Mathematics 660 p p It is importat to poit out at this stage that cos isi is just oe value of q q p q cos isi A simple eample will illustrate this If π, p ad q, the cos π isiπ cos π isi π i But cos π isiπ cosπ ad si π 0 ad i So i is oly oe value of cos π isiπ There are, i fact, q differet values of cos π show i sectio 46 cos isi cos isi for positive ad egative itegers, ad fractioal values of isiπ p q ad this will be 55

56 MFP Tetbook A-level Further Mathematics Usig de Moivre s theorem to evaluate powers of comple umbers Oe very importat applicatio of de Moivre s theorem is i the additio of comple umbers of the form a i b The method for doig this will be illustrated through eamples Eample 4 Simplify π π cos isi 6 6 Solutio It would, of course, be possible to multiply cos π isi π by itself three times, but this would 6 6 be laborious ad time cosumig eve more so had the power bee greater tha Istead, cos π isi π cos π isi π cos π isi π 0i i 56

57 MFP Tetbook A-level Further Mathematics 660 Eample 4 Fid i 0 Solutio i the form a i b Clearly it would ot be practical to multiply i by itself te times De Moivre s theorem could provide a alterative method but it ca be used oly for comple umbers i the form cos isi, ad i is ot i this form A techique itroduced i Chapter (sectio 4) ca be used to epress it i polar form# y O a Argad diagram, i is represeted by the i poit whose Cartesia coordiates are, Now, r ad ta so that i cos π isi π 6 6 Thus, 0 π 6 0 ad i cos isi ote that is raised to the power 0 as well π 0π cos isi π π i 5 i O θ r 57

58 MFP Tetbook A-level Further Mathematics 660 Eample 4 Simplify π π Solutio cos isi 6 6 De Moivre s theorem applies oly to epressios i the form cos isi ad ot cos isi, so the epressio to be simplified must be writte i the form cos π π isi 6 6 cos π isi π cos π isi π cos π π isi 6 6 cos π π isi cos π isi π i Note that it is apparet from this eample that cos isi cos isi It is very importat to realise that this is a deductio from de Moivre s theorem ad it must ot be quoted as the theorem 58

59 MFP Tetbook A-level Further Mathematics 660 Eample 44 Fid i Solutio i the form a i b The comple umber i is represeted by the poit whose Cartesia coordiates are, o the Argad diagram show here, r α y O θ Hece, r 6 4, ad i i 4cos π ta ta so that π Thus isi π π π 4 cos isi cos π isi π Eercise 4A Prove that cos isi cos isi Epress each of the followig i the form a i b: (a) cos isi 5 cos π isi π 5 5 (b) 0 (c) π π cos isi (d) i 6 (e) i 4 (g) i i (f) 5 59

60 MFP Tetbook A-level Further Mathematics Applicatio of de Moivre s theorem i establishig trigoometric idetities Oe way of showig how these idetities ca be derived is to use eamples The same priciples are used whichever idetity is required Eample 4 Show that cos 4 cos cos Solutio There are several ways of establishig this result The epasio of cos A B ca be used to epress cos i terms of cos settig A ad B Similarly, the epasio of cos ca be used to give cos i terms of cos Usig de Moivre s theorem gives a straightforward alterative method cos isi cos isi usig the biomial epasio of p q cos i cos si cos si i si usig i cos cos i si cos i si i si Now cos is the real part of the left-had side of the equatio, ad the real parts of both sides ca be equated, cos cos cos si cos coscos sice cos si 4cos cos Note that this equatio will also give si by equatig the imagiary parts of both sides of the equatio 60

61 MFP Tetbook A-level Further Mathematics 660 Eample 4 Epress ta 4 i terms of ta Solutio ta 4 si 4 so epressios for si 4 ad cos 4 i terms of si ad cos must be cos 4 established to start with Usig de Moivre s theorem, 4 4 cos 4cos isi6cos isi 4cosisi isi usig the biomial epasio cos4 isi4 cos isi cos 4i cos si 6 cos si 4i cos si si usig i Equatig the real parts o both sides of the equatio, 4 4 cos 4 cos 6 cos si si, ad equatig the imagiary parts, si 4 4cos si 4cossi Now, ta 4 si 4 cos 4 4cos si 4cossi 4 4 cos 6 cos si si 4 Dividig every term by cos gives 4 si 4 si cos ta 4 cos 4 6 si si cos cos 4 But ta si so cos ta 4 4ta 4ta 4 6ta ta 6

62 MFP Tetbook A-level Further Mathematics 660 Eercise 4B Epress si i terms of si Epress ta i terms of ta Epress si 5 i terms of si 4 Show that 6 4 cos 6 cos 48cos 8cos 6

63 MFP Tetbook A-level Further Mathematics 660 So far si, cos ad ta have bee epressed i terms of si, cos ad ta De Moivre s theorem ca be used to epress powers of si, cos ad ta i terms of sies, cosies ad tagets of multiple agles First some importat results must be established Suppose z cos si i The isi isi z cos z cos cos isi So, z cos isi Addig, ad subtractig, cos isi z z cos, z z i si z Also, z cos isi Combiig z ad z z cos isi cos isi z If z cos isi z cos z z isi z isi cos cos isi as before, z z z cos, z i si If z cos isi, z z cos z z isi A commo mistake is to omit the i i isi, so make a poit of rememberig this result carefully 6

64 MFP Tetbook A-level Further Mathematics 660 Eample 4 5 Show that Solutio cos cos 5 5cos 0 cos 6 Suppose z cos i si The z cos z 5 5 ad cos z z 5 4 z 5z z0z z 0z z 5zz z 5 5 z 5z 0z0 z5 z z So z z z cos z 5z 5 0z0 5 z z z z z z Usig the results established earlier, 5 z cos5, 5 z z cos, z z cos z Hece cos5 cos5 5cos 0cos 5 cos cos 5 5cos 0 cos, as required 6 Oe very useful applicatio of the eample above would be i itegratig 5 5 cos cos cos 5 5cos 0 cos 6 si5 5si 0si c, where c is a arbitrary costat 6 5 Eample 44 (a) Show that cos si si si 6 (b) Evaluate π 0 cos si d 64

65 MFP Tetbook A-level Further Mathematics 660 Solutio (a) Multiplyig these, Now z cos z z z si z z z z z z z 8cos 8i si 64i cos si z z z z 6 z 6 z z z z z z 6 z z 6 z z 6 z z 6 z z isi 6 ad isi z z 64i cos si isi 6 isi isi 6 6isi Thus, Dividig both sides by 64i, cos si si 6 si si si 6, as required (b) π π 0 cos si si si 6 d 0 cos cos π 0 This sectio cocludes with a eample which uses the ideas itroduced here ad eteds ito other areas of mathematics 65

66 MFP Tetbook A-level Further Mathematics 660 Eample 45 (a) Show that 4 cos5 cos 6cos 0cos 5 (b) Show that the roots of the equatio are (c) Deduce that cos π cos π Solutio 4 cos rπ 0 for r =,, 7 ad 9 (a) Usig the ideas itroduced at the begiig of this sectio, cos5 isi 5 cos isi 5 Usig the biomial theorem for epasio, the right-had side of this equatio becomes cos 5cos isi 0cos isi 0cos isi 5cos isi isi Not every term of this epressio has to be simplified As cos5 is the real part of the left-had side of the equatio, it equates to the real part of the right-had side The real part of the right-had side of the equatio comprises those terms with eve powers if i i them, sice i ad is real 5 4 cos 5 cos 0 cos i si 5cos i si Thus, 5 4 cos 0 cos si 5cos si 5 cos 0cos cos 5cos cos cos 0 cos 0 cos 5cos 0 cos 5cos 5 6 cos 0 cos 5cos 4 cos 6cos 0cos 5 usig cos si 66

67 MFP Tetbook A-level Further Mathematics 660 (b) Now whe cos 5 0, either cos 0 or cos 0 cos 5 0 So, puttig cos, the roots of are the values of cos for which cos5 0, provided cos 0 But if cos5 0, 5 π, π, 5π, 7π, 9π, π, π, so that π, π, 5π, 7π, 9π, π, π, Also, cos π is the same as cos 9π, 0 0 ad cos π is the same as cos 7π, so that, 0 0 although there is a ifiite umber of values of, there are oly five distict values of cos ad these are cos π, cos π, cos 5π, cos 7π ad cos 9π Now cos 5π cos π 0 ad π is, of course, a root of cos 0, so that the roots of the 0 4 equatio are cos π, cos π, cos 7π ad cos 9π The roots may be writte i a slightly differet way as cos 7π cos π π 0 0 cos π, 0 ad cos 9π cos 9π π 0 0 cos π 0 Thus the four roots of the quartic equatio ca be writte as ad cos π 0 4 cos π 0 (c) From the ideas set out i Chapter (sectio 7), the product of the roots of the quartic 4 equatio is 5 so that 6 cos π cos π cos π cos π Ad hece, cos π cos π

68 MFP Tetbook A-level Further Mathematics 660 Eercise 4C If z cos i si write, i terms of z: (a) cos 4 (b) cos 7 (c) si 6 (d) si Prove the followig results: 4 (a) cos4 8cos 8cos (b) 5 si 5 6si 0si 5si (c) si 6 si cos 5 cos 6 cos (d) ta ta ta ta 68

69 MFP Tetbook A-level Further Mathematics Epoetial form of a comple umber Both cos ad si ca be epressed as a ifiite series i powers of, provided that is measured i radias These are give by 4 6 cos ( )! 4! 6!! ad 5 7 si ( )! 5! 7!! There is also a series for e give by If i is substituted for i the series for e, 4 e!! 4!! 4 i i i i i e i!! 4!! 4 i i!! 4! Regroupig, 4 i e i,! 4!! ad, usig the previous results for si ad cos, i e cos isi It is also importat to ote that if z cos isi, the z cos isi cos isi i e, ad if z r cos isi, the i re z ad z i r e If z r cos isi, i r i the z e ad z r e 69

70 MFP Tetbook A-level Further Mathematics 660 i r The form e is kow as the epoetial form of a comple umber ad is clearly liked to the polar form very closely Aother result ca be derived from the epoetial form of a comple umber: i e cos isi i e cos isi So, cos isi Addig these or i i e e cos isi cos isi cos, i i cos e e Subtractig gives i i e e cos isicos isi or isi, i i si e e i cos e si e i i e e i i i Eample 44 Epress i i the form i re Solutio The comple umber i is represeted by the poit with the coordiates, o a Argad diagram y Hece, ad so that r 8, ta π, 4 πi 4 i= 8 e O r θ, Eercise 4D i Epress the followig i the form re : (a) i b) i (c) i (d) i 70

71 MFP Tetbook A-level Further Mathematics The cube roots of uity The cube roots of are umbers such that whe they are cubed their value is They must, therefore, satisfy the equatio z 0 Clearly, oe root of z is z so that z must be a factor of z Factorisig, z z z z 0 Now z 0 is a cubic equatio ad so has three roots, oe of which is z The other two come from the quadratic equatio z z 0 If oe of these is deoted by w, the w satisfies z z 0 so that w w 0 It ca also be show that if w is a root of z, the w is also a root i fact, the other root Substitutig w ito the left-had side of 6 z gives w w w, as Thus the three cube roots of are, w ad ca be epressed i the form a ib by solvig z w sice w is a solutio of w, where w ad 4 z i It does t matter whether w is labelled as i or as of the other I other words, if w i the If w i, the w i i i 4 i 4 i 4 i w i The cube roots of uity are, w ad w ww 0 ad the o-real roots are z w are o-real Of course, w z 0 usig the quadratic formula: i w, where i because each is the square 7

72 MFP Tetbook A-level Further Mathematics 660 Both w ad w ca be epressed i epoetial form Take w i ; w ca be represeted by the poit whose Cartesia coordiates are, o a Argad diagram, α r y θ O From the diagram, π, π ad πi e r, ad π, where ta Thus, πi w e The other root is w πi 4πi e e ad ca also be writte as y Plottig the three cube roots of uity o a Argad diagram shows three poits equally spaced (at itervals of π ) roud a circle of radius as show i the diagram alogside π π,0 Eample 45 Simplify w 7 8, w where w is a comple cube root of Solutio w 7 6 w w w w w ww 8 6 w w w w w w w 7 8 because w w 0 because, w w ww 7

73 MFP Tetbook A-level Further Mathematics 660 Eample 45 Show that 0 w w ww Solutio ww 0 so w w, ad so o So the deomiators of the left-had side of the equatio ca be replaced to simplify to w w Multiplyig the first term of this epressio by w i the umerator ad deomiator, ad the secod term by w similarly gives w w w w w w w 0 as w w 0 as Eercise 4E If w is a comple cube root of, fid the value of 0 (a) w w (b) w w (c) w w 7

74 MFP Tetbook A-level Further Mathematics The th roots of uity The equatio z clearly has at least oe root, amely z, but it actually has may more, most of which (if ot all) are comple I fact, if is odd z is the oly real root, but if is eve z is also a real root because raised to a eve power is To fid the remaiig roots, the right-had side of the equatio z has to be eamied I 0 0 πi epoetial form, e because e cos0isi0i0 But also, e because πi πi e cosπ isiπ i0 Ideed e k where k is ay iteger Substitutig the k right-had side of the equatio z by this term gives z e πi Takig the th root of both sides gives show below ad so o util Thus, kπi kπi z e Differet iteger values of k will give rise to differet roots, as k 0 gives e, 0 πi k gives e cos π isi π, 4πi k gives e cos 4π isi 4π, πi π π k gives e cos isi z e 0,,,, gives the distict roots of the equatio z There are o more roots because if k is set equal to, e e cosπ isiπ, which is the same root as that give by k 0 πi πi πi πi πi πi πi Similarly, if k is set equal to, e e e e e e e which is the same root as that give by k, ad so o The roots of z ca be illustrated o a Argad diagram All the roots lie o the circle z because the modulus of every root is Also, the amplitudes of the comple umbers represetig the roots are π 4π 6π π,,,, I other words, the roots are represeted by poits equally spaced aroud the uit circle at agles of π startig at, 0 the poit represetig the real root z πi The equatio z has roots kπi z e k 0,,,, πi 6πi e 4πi e πi e π π π 0 e π e πi 74

75 MFP Tetbook A-level Further Mathematics 660 Eample 46 Fid, i the form a i b, the roots of the equatio Argad diagram 6 z ad illustrate these roots o a Solutio Therefore 6 πi z e k z e kπi 6 kπi e k 0,,,, 4, 5 Hece the roots are k 0, z πi k, z e cos π i si π i πi k, z e cos π isi π i πi k, z e cos π isiπ 4πi k 4, z e cos 4π isi 4π i 5πi k 5, z e cos 5π i si 5π i To summarise, the si roots are z, z i ad these are illustrated o the Argad diagram alogside πi e y πi e 4πi e 5πi e Two further poits are worth otig Firstly, you may eed to give the argumets of the roots betwee π ad π istead of betwee 0 ad π I eample 46, the roots would be give as kπi z e for k 0,,, Secodly, a give equatio may ot ivolve uity for eample, if eample 46 had cocered z z kπi e kπi 6 6 z 64, the solutio would have bee writte z e k 0,,,, 4, 5 ad the oly differece would be that the modulus of each root would be istead of, with 6 the cosequece that the si roots of z 64 would lie o the circle z istead of z 75

76 MFP Tetbook A-level Further Mathematics 660 Of course, there are variatios o the above results For eample, you may eed to fid the 4 5 roots of the equatio zz z z z 0 This looks dautig but if you ca recogise the left-had side as a geometric progressio with commo ratio z, it becomes more straightforward Summig the left-had side of the equatio, z zz z z z 0, z so that the five roots of zz z z z 0 are five of the roots of z 0 The 6 root to be ecluded is the root z because z is idetermiate whe z So the roots z 4 5 of zz z z z 0 are z i ad, whe writte i the form a i b Eercise 4F Write, i the form a i b, the roots of: (a) 4 z (b) 5 z (c) 0 z I each case, show the roots o a Argad diagram Solve the equatio 4 z z z z 0 Solve the equatio z4z 8z 0 4 By cosiderig the roots of 5 z, show that cos π cos 4π cos 6π cos 8π

77 MFP Tetbook A-level Further Mathematics The roots of z =α where α is a o-real umber Every comple umber of the form a ib ca be writte i the form e, where r is real ad lies i a iterval of π (usually from 0 to π or from π to π) Suppose that i re iπi i πi pq p q Now e e e usig e e e i πi e because e ikπi i kπi Similarly e e e i e also So, z i kπi re ad, takig the th root of both sides, z r e i kπi i r i kπ r e k 0,,,,, π r e These roots ca be illustrated o a Argad diagram as before All lie o the circle z r ad are equally spaced aroud the circle at i itervals of π Whe k 0, z r e ad this could be take as the startig poit for the itervals of π π π iπ r e i The equatio z, where re, i kπ has roots z re k 0,,,, 77

78 MFP Tetbook A-level Further Mathematics 660 Eample 47 Fid the three roots of the equatio z i Solutio First, i must be epressed i epoetial form From the diagram alogside, r 8, ta, ad π 4 So, i πi 8 e 4 πi kπi Hece, z 8e 4 y r θ, Takig the cube root of each side, z e 8 k πi e k 0,, So the roots are πi 4 kπi k 0, z e k, z e πi 9πi 7πi 7πi k, z e or e cos π isi π whe k 0, ad so o The roots ca also be writte This chapter closes with oe further eample of the use of the priciples discussed Eample 47 Solve the equatio 5 5 z z Solutio givig your aswers i the form a i b At first sight, it is temptig to use the biomial epasio o z 5 but this geerates a 5 quartic equatio (the terms i z cacel) which would be difficult to solve Istead, because kπi e, the equatio ca be writte as 5 kπi 5 z z e 78

79 MFP Tetbook A-level Further Mathematics 660 Takig the fifth root of each side, kπi 5 ze z k,,,4 Note that k 0 is ecluded because this would give z z, ad i ay case as the equatio is really a quartic equatio it will have oly four roots Solvig the equatio for z, kπi z e 5 k,,,4 or z kπi 5 e The et step is ew to this sectio ad is well worth rememberig The term e 5 ca be writte as cos kπ isi kπ makig the deomiator have the form p i q The umerator 5 5 ad deomiator of the right-had side of the equatio ca the be multiplied by p iq to remove i from the deomiator As p would the equal cos kπ ad q would equal 5 si kπ, this would be a rather cumbersome method Istead, the umerator ad 5 deomiator of the right-had side of the equatio are multiplied by e will be apparet later) Thus, z kπi, e 5 kπi 5 kπi (for reasos which So But e i e i i si so that z e e kπi 5 kπi kπi kπi e e e e kπi 5 kπi kπi 5 5 e kπi kπi 5 5 kπ e e isi ad so, 5 cos kπi 5 z e isi kπ 5 kπ π isi k 5 5 isi kπ 5 cot k π i 5 icot k π 5 k,,, 4 as required 79

80 MFP Tetbook A-level Further Mathematics 660 Eercise 4G Solve the followig equatios: 4 (a) z 6i (b) z i (c) 8 z i (d) z (e) z 8i (f) 5 5 z z 80

81 MFP Tetbook A-level Further Mathematics 660 Miscellaeous eercises 4 (a) Write dow the modulus ad argumet of the comple umber 64 4 (b) Hece solve the equatio z 64 0 givig your aswers i the form r cos isi, where r 0 ad π π (c) Epress each of these four roots i the form a ib ad show, with the aid of a diagram, that the poits i the comple plae which represet them form the vertices of a square [AEB Jue 996] (a) Epress each of the comple umbers i ad i i the form r cos isi, where r 0 ad π π (b) Usig your aswers to part (a), (i) show that i i 0 5 i, (ii) solve the equatio z i i givig your aswers i the form a i b, where a ad b are real umbers to be determied to two decimal places [AQA Jue 00] (a) By cosiderig z cos i si ad usig de Moivre s theorem, show that 4 si 5 si 6si 0si 5 (b) Fid the eact values of the solutios of the equatio (c) Deduce the eact values of aswers [AQA Jauary 00] si π ad si π, 5 5 eplaiig clearly the reasos for your 8

82 MFP Tetbook A-level Further Mathematics (a) Show that the o-real cube roots of uity satisfy the equatio z z 0 (b) The real umber a satisfies the equatio, a a where is oe of the o-real cube roots of uity Fid the possible values of a [AQA Jue 000] 5 (a) Verify that πi z 5 e is a root of the equatio z 5 (b) Fid the other four roots of the equatio (c) Mark o a Argad diagram the poits correspodig to the five roots of the equatio Show that these roots lie o a circle, ad state the cetre ad radius of the circle (d) By cosiderig the Argad diagram, fid (i) arg z i terms of π, (ii) z i the form π [AQA Specime] a cos, b where a ad b are itegers to be determied 6 (a) (i) Show that πi 5 w e is oe of the fifth roots of uity (ii) Show that the other fifth roots of uity are, πi w, w ad w (b) Let p ww ad q w w, where w e (i) Show that pq ad pq (ii) Write dow the quadratic equatio, with iteger coefficiets, whose roots are p ad q (iii) Epress p ad q as iteger multiples of cos π ad cos 4π, respectively, 5 5 (iv) Hece obtai the values of cos π ad cos 4π i surd form 5 5 [NEAB Jue 998] 8

83 MFP Tetbook A-level Further Mathematics (a) (i) Use de Moivre s theorem to show that if z cos isi, the z cos z (ii) Write dow the correspodig result for z z (b) (i) Show that z 6 z A z 6 B z, z z z z where A ad B are umbers to be determied (ii) By substitutig z cos isi i the above idetity, deduce that [AQA Jue 000] cos si si si 6 8 (a) (i) Epress e i i e i terms of si (ii) Hece, or otherwise, show that i i i cot, e e (b) Derive epressios, i the form the equatio 6 z i e where π π, for the four o-real roots of w w has oe real root ad four o-real roots (i) Eplai why the equatio has oly five roots i all (ii) Fid the real root (iii) Show that the o-real roots are,,,, z z z z (c) The equatio * 6 4 where z, z, z ad z 4 are the o-real roots of the equatio 6 z (iv) Deduce that the poits i a Argad diagram that represets the roots of equatio (*) lie o a straight lie [AQA March 000] 8

84 MFP Tetbook A-level Further Mathematics (a) Epress the comple umber i i the form i re, where r 0 ad π π (b) Show that oe of the roots of the equatio z i πi is e, ad fid the other two roots givig your aswers i the form e i, is a surd ad π π r where r (c) Idicate o a Argad diagram poits A, B ad C correspodig to the three roots foud i part (b) (d) Fid the area of the triagle ABC, givig your aswer i surd form (e) The poit P lies o the circle through A, B ad C Deotig by w,, ad the comple umbers represeted by P, A, B ad C, respectively, show that w w w 6 [AQA Jue 999] 84

85 MFP Tetbook A-level Further Mathematics 660 Chapter 5: Iverse Trigoometrical Fuctios 5 Itroductio ad revisio 5 The derivatives of stadard iverse trigoometrical fuctios 5 Applicatios to more comple differetiatio 54 Stadard itegrals itegratig to iverse trigoometrical fuctios 55 Applicatios to more comple itegrals This chapter revises ad eteds work o iverse trigoometrical fuctios Whe you have studied it, you will: be able to recogise the derivatives of stadard iverse trigoometrical fuctios; be able to eted techiques already familiar to you to differetiate more complicated epressios; be able to recogise algebraic epressios which itegrate to stadard itegrals; be able to rewrite more complicated epressios i a form that ca be reduced to stadard itegrals 85

86 MFP Tetbook A-level Further Mathematics Itroductio ad revisio You should have already met the iverse trigoometrical fuctios whe you were studyig the A specificatio module Core However, i order to preset a clear picture, ad for the sake of completeess some revisio is icluded i this sectio If y si, we write si y (or arcsi y ) Note that si ormally be writte as y (si ) whe epressed i terms of sie y is ot cosec y which would The use of the superscript - is merely the covetio we use to deote a iverse i the same way as we say that f is the iverse of the fuctio f The sketch of y si will be familiar to you ad is show below For ay give value of there is oly oe correspodig value of y, but for ay give value of y there are ifiitely may values of The graph of y si beig the iverse, is the reflectio of y si i the lie y ad a sketch of it is as show As it stads, for a give value of, y si has ifiitely may values, but if we wish to describe si as a fuctio, we must make sure that the fuctio has precisely oe value I order to overcome this obstacle, we restrict the rage of y to π y π so that the sketch of y si becomes the sketch show 86

87 MFP Tetbook A-level Further Mathematics 660 By doig this, we esure that for ay give value of there is a uique value of y for which y si This value is usually called the pricipal value si is the agle betwee π ad π iclusive whose sie is Notice that the gradiet of y si is always greater tha zero We ca defie cos i a similar way but with a importat differece The sketches of y cos ad y cos are show below 87

88 MFP Tetbook A-level Further Mathematics 660 I this case, it would ot be sesible to restrict y to values betwee π ad π sice for every value of 0 there would be two values of y ad for values of 0 there would be o value of y Istead we choose the rage 0 π ad the sketch is as show cos is the agle betwee 0 ad π iclusive whose cosie is Whe it comes to ta we ca restrict the rage to π ad π ta is the agle betwee π ad π eclusive whose taget is The sketch of y ta is show below Eercise 5A Epress i terms of π the values of: (d) (a) ta cos 0 (e) (b) ta cos (c) si (f) cos ( ) 88

89 MFP Tetbook A-level Further Mathematics The derivatives of stadard iverse trigoometrical fuctios Suppose y si the si y ad, differetiatig implicitly, dy cos y d thus dy d cosy si y usig cos y si y Note that we choose the positive square root This is due to the fact that the gradiet of the graph of y si is always greater tha zero as was show earlier If y si dy d dy For y cos usig similar workig we would arrive at d, this time choosig the egative sig of the square root as the graph of y cos always has a gradiet less tha zero If y cos dy d 89

90 MFP Tetbook A-level Further Mathematics 660 If y ta the we write ta y, ad, differetiatig implicitly, dy sec y d dy or d sec y ta y dy d usig sec y ta y If y ta dy d Eercise 5B Prove that if y cos the dy d 90

91 MFP Tetbook A-level Further Mathematics Applicatios to more comple differetiatio Some methods of differetiatio you should already be familiar with These would iclude the fuctio of a fuctio rule, ad the product ad quotiet rules We will complete this sectio by usig the rules with fuctios ivolvig iverse trigoometrical fuctios Eample 5 If y si ( ), fid d y d Solutio Set u the y si u du d ad dy du u Thus, dy dy du d du d u ( ) 4 4 (usig the fuctio of a fuctio rule) Eample 5 Differetiate si e ad usig the fuctio of a fuctio rule, Set u e ad let y si so that y si u du e d dy du u e dy dy d u e d du d u e e 9

92 MFP Tetbook A-level Further Mathematics 660 Eample 5 If y ta, fid d y d This time we eed to use the product rule ad the fuctio of a fuctio rule For d ta d, set u y ta dy d ta ta d d d d du d du d the ta ta u u 4 Thus dy ta d 4 Eample 54 Differetiate cos If y cos The, usig the quotiet rule, cos dy d cos 9

93 MFP Tetbook A-level Further Mathematics 660 Eercise 5C Differetiate the followig: (a) ta (b) cos (c) si (a) ta (b) e cos (c) si (a) si (b) ta 4 (a) si a b (b) ta a b where a ad b are positive umbers 9

94 MFP Tetbook A-level Further Mathematics Stadard itegrals itegratig to iverse trigoometrical fuctios Geerally speakig, as you have bee taught, formulae from the Formulae ad Statistical Tables Booklet supplied for each AS ad A module apart from MPC ca be quoted without proof However, this does ot preclude a questio requirig a proof of a result from this booklet beig set There are two stadard results, the proofs of which are give here ad the methods for these proofs should be committed to memory The first oe is d a This itegral requires a substitutio Let ata so that d asec d The d asec d a a a ta a sec d a sec d a c a ta c a a d a ta a c a The secod itegral is a d This iterval also requires a substitutio The Let asi d acos d d = a cos d a a a si a cos d a cos c si c a a d si a c 94

95 MFP Tetbook A-level Further Mathematics 660 We ow give two eamples of defiite itegrals Eample 54 Evaluate 0 d 4 We have d 0 ta 4 ta ta 0 π 0 4 π 8 0 Eample 54 Evaluate d 0 9 We have d si 9 si si 0 si si 0 π 0 6 π Eercise 5D Itegrate the followig, leavig your aswers i terms of π d d 4 d 5 4 d 5 0 d 95

96 MFP Tetbook A-level Further Mathematics Applicatios to more comple itegrals I this sectio we will show you by meas of eamples how ufamiliar itegrals ca ofte be reduced to oe or perhaps two, stadard itegrals Most will ivolve completig the square of a quadratic epressio, a method you will o doubt have used may times before i other cotets We will begi by usig eamples of itegrals which i whole or part reduce to d a Eample 55 Fid d 48 Now o completig the square so that d d 48 4 The substitutio u gives du d ad becomes du a stadard form The result is u 4 therefore ta u c or epressig it i terms of, ta c Eample 55 Fid d so that we have d The substitutio u gives du d ad the itegral becomes d u u ta u c or substitutig back ta c We write 96

97 MFP Tetbook A-level Further Mathematics 660 Eample 55 Fid d 4 9 Here the substitutio u trasforms the give itegral ito stadard form for ifu d u du d ad we have u 9 ta u c ta c 6 Fially we give a slightly harder eample of a itegral which uses d i its solutio a Eample 554 Fid 5 d 6 Itegrals of this type where the umerator is a liear epressio i ad the deomiator is a quadratic i usually itegrate to l p( ) ta q( ) where p( ) ad q ( ) are fuctios of ' f d I order to tackle this itegral you eed to remember that itegrals of the form f itegrate to l f c You should have bee taught this result whe studyig the module Core So to start evaluatig this itegral we have to ote that the derivative of 6is 6 ad we rewrite the umerator of the itegral as 6 so that the itegral becomes 6 d 6 ad separatig it ito two halves we write it agai as 6 d d

98 MFP Tetbook A-level Further Mathematics 660 The first itegral itegrates to l 6 i the deomiator ad write it as whilst i the secod we complete the square d The substitutio u leads to du u ta u ta So that 5 d l 6 ta c 6 The fial part of this sectio will show you how itegrals ca sometimes be reduced to d This will be doe by meas of eamples a Eample 555 Fid d 4 As i previous eamples, we eed to complete the square o 4, ad we write 4 4 So that the iterval becomes d 4 98

99 MFP Tetbook A-level Further Mathematics 660 The substitutio u simplifies the result to the stadard form du 4 u si u c Eample 556 si c Fid d 6 I order to complete the square i the deomiator, we write Thus, The substitutio of u 6 4 d d 6 4 reduces the itegral to du 4 u si u c si c Oe fial eample shows how more complicated epressios may be itegrated usig methods show here ad other results which you should have met studyig earlier modules I this particular cotet the result you will eed is that ' f d f c f [This result ca be easily verified usig the substitutio u f ad the itegral becomes du ] u ', sice the, du f d 99

100 MFP Tetbook A-level Further Mathematics 660 Eample 557 Fid d 76 Now the derivative of becomes 76 is 6 or separatig the itegral ito two parts The first itegral is of the form so we write as 6 d 76 6 d ' 6 ad the itegral f d apart from a scaler multiplier, ad so itegrates to f 76, whilst completig the square o the deomiator of the secod itegral, we obtai d 6 which itegrates to si usig the substitutio u 4 Hece, d 76 si c

101 MFP Tetbook A-level Further Mathematics 660 Eercise 5E Itegrate (a) 45 Itegrate (a) Fid (b) 4 5 (b) (c) (a) d (b) 76 d (c) d 4 Fid (a) d (b) d (c) d 0

102 MFP Tetbook A-level Further Mathematics 660 Chapter 6: Hyperbolic Fuctios 6 Defiitios of hyperbolic fuctios 6 Numerical values of hyperbolic fuctios 6 Graphs of hyperbolic fuctios 64 Hyperbolic idetities 65 Osbore s rule 66 Differetiatio of hyperbolic fuctios 67 Itegratio of hyperbolic fuctios 68 Iverse hyperbolic fuctios 69 Logarithmic form of iverse hyperbolic fuctios 60 Derivatives of iverse hyperbolic fuctios 6 Itegrals which itegrate to iverse hyperbolic fuctios 6 Solvig equatios This chapter itroduces you to a wholly ew cocept Whe you have completed it, you will: kow what hyperbolic fuctios are; be able to sketch them; be able to differetiate ad itegrate them; have leared some hyperbolic idetities; uderstad what iverse hyperbolic fuctios are ad how they ca be epressed i alterative forms; be able to sketch iverse hyperbolic fuctios; be able to differetiate iverse hyperbolic fuctios ad recogise itegrals which itegrate to them; be able to solve equatios ivolvig hyperbolic fuctios 0

103 MFP Tetbook A-level Further Mathematics Defiitios of hyperbolic fuctios It was show i Chapter 4 that i i si e e ad i i cos e +e Hyperbolic i fuctios are defied i a very similar way The defiitios of sih ad cosh (ofte called hyperbolic sie ad hyperbolic cosie proouced shie ad cosh ) are: sih e e cosh e e There are four other hyperbolic fuctios derived from these, just as there are four trigoometric fuctios They are: tah sih cosh cosech sih sech cosh coth tah tha cosheck sheck coth I terms of epoetial fuctios, tah sih cosh e e e e e e, e e or, o multiplyig the umerator ad deomiator by e, tah e e Agai, cosech sih e e e e Epoetial forms for sech ad coth ca be foud i a similar way 0

104 MFP Tetbook A-level Further Mathematics 660 Eercise 6A Epress, i terms of epoetials: (a) sech, (b) coth, (c) tah, (d) cosech 6 Numerical values of hyperbolic fuctios Whe fidig the value of trigoometric fuctios, for eample si, the agle must be give i degrees (or radias) There is o uit for whe evaluatig, for eample, sih It is quite i order to speak about sih or cosh: sih e e 6 (to two decimal places); cosh e e 97 (to two decimal places) You ca work out these values o a calculator usig the e butto However, for coveiece most scietific calculators have a hyp butto ad sih ca be obtaied directly by pressig the, hyp ad si buttos i the appropriate order It is worth rememberig that 0 0 sih 0 e e 0; 0 0 cosh 0 e e Eercise 6B Use a calculator to evaluate, to two decimal places: (a) sih 06, (b) tah, (c) sech, (d) tah ( 06), (e) cosh ( 0), (f) coth 4 04

105 MFP Tetbook A-level Further Mathematics Graphs of hyperbolic fuctios The graphs of hyperbolic fuctios ca be sketched easily by plottig poits Some sketches are give below but it would be a good eercise to make a table of values ad cofirm the geeral shapes for yourself It would also be worthwhile committig the geeral shapes of y sih, y cosh ad y tah to memory y y y sih 0 y cosh 0 The sketch of y tah requires a little more cosideratio I Sectio 5, it was show that tah could be writte as tah e e e e Now, as e 0 for all values of, it follows that the umerator i the bracketed epressio above is less tha its deomiator, so that tah Also, as, e 0 ad tah So the graph of y tah has a asymptote at y Now, if the umerator ad deomiator of tah are divided by e, tah e e As e 0 for all values of, it follows that the umerator of this fractio is less tha its deomiator, from which it ca be deduced that tah It ca also be deduced that as e 0 as, so tah as So the graph of y tah has a asymptote at Hece the curve y tah lies betwee y ad y, ad has y as asymptotes y tah y 0 05

106 MFP Tetbook A-level Further Mathematics 660 Eercise 6C Sketch the graphs of (a) y sech, (b) y cosech, (c) y coth 64 Hyperbolic idetities Just as there are trigoometric idetities such as cos si ad there are similar hyperbolic idetities For eample, cosh e e e e, 4 ad sih e e e e from which, by subtractio, 4 cosh sih e e 4 4 e e cos cos, cosh sih Dividig both sides of this equatio by cosh sih cosh cosh cosh tah sech cosh, it follows that sech tah Or agai, dividig both sides by sih istead, cosh sih sih sih sih coth cosech cosech coth 06

107 MFP Tetbook A-level Further Mathematics 660 Now cosider sih cosh y cosh sih y y y e e e e e e y y e e ee y e e y ee y y y y e e e e e e ee 4 e e y e e y 4 ee y e e y e y y e [usig the laws of idices] sih y y e e y I eactly the same way, epressios for sih y, cosh y ad cosh y worked out sih y sih cosh ycosh sih y cosh y cosh cosh ysih sih y ca be Eercise 6D Show that sih y sih cosh y cosh sih y, (a) (b) cosh y cosh cosh y sih sih y You will probably remember that the basic trigoometric formulae for si y cos y ca be used to fid epressios for si, cos, ad ad so o The hyperbolic formulae give above help to fid correspodig results for hyperbolic fuctios 07

108 MFP Tetbook A-level Further Mathematics 660 For eample, because puttig y, sih y sih cosh y cosh sih y, sih sih cosh cosh sih, or sih sih cosh Usig puttig y, or cosh y cosh cosh y sih sih y, cosh cosh cosh sih sih, cosh cosh sih Usig or cosh sih, cosh sih sih sih cosh cosh sih sih cosh cosh cosh sih cosh sih Some eamples will illustrate etesios of these results 08

109 MFP Tetbook A-level Further Mathematics 660 Eample 64 Show that tah tah tah Solutio tah sih cosh sih cosh cosh sih Dividig the umerator ad deomiator by cosh, sih cosh tah cosh cosh sih cosh sih cosh sih cosh tah tah Eample 64 Show that cosh 4 cosh cosh Solutio cosh cosh cosh cosh sih sih cosh cosh sih cosh sih cosh cosh cosh cosh cosh cosh cosh cosh 4cosh cosh [usig cosh sih ] Eercise 6E Usig the epasio of sih, show that sih sih 4sih Epress cosh 4 i terms of cosh 09

110 MFP Tetbook A-level Further Mathematics Osbore s rule It should be clear that the results ad idetities for hyperbolic fuctios bear a remarkable similarity to the correspodig oes for trigoometric fuctios I fact the oly differeces are those of sig for eample, whereas cos si, the correspodig hyperbolic idetity is cosh sih There is a rule for obtaiig the idetities of hyperbolic fuctios from those for trigoometric fuctios it is called Osbore s rule To chage a trigoometric fuctio ito its correspodig hyperbolic fuctio, where a product of two sies appears chage the sig of the correspodig hyperbolic term For eample, because the cos y cos cos y si si y cosh y cosh cosh y sih sih y Note also that because the cos si cosh sih, because si is a product of two sies However, care must be eercised i usig this rule, as the et eample shows It is kow that sec ta but sech tah The reaso that the sig has to be chaged here is that a product of sies is implied because ta si cos It should be oted that Osbore s rule is oly a aid to memory It must ot be used i a proof for eample, that cosh sih The method show i Sectio 64 must be used for that 0

111 MFP Tetbook A-level Further Mathematics Differetiatio of hyperbolic fuctios k You will already have met the derivative of e d e k e k k d Just to remid you, As hyperbolic fuctios ca be epressed i terms of e, it follows that their differetiatio is straightforward For eample, y sih e e dy Therefore, e +e Ad if d cosh y sih k e k e k, dy the k k ke + ke d e k +e k k kcosh k

112 MFP Tetbook A-level Further Mathematics 660 You ca differetiate cosh ad cosh k i eactly the same way; tah ca be differetiated by treatig it as the derivative of the quotiet sih The followig results cosh should be committed to memory: y sih, dy cosh d y cosh, dy sih d y tah, dy sech d Geerally: y sih k, dy k cosh k d y cosh k, dy k sih k d y tah k, dy k sech k d Note that the derivatives are very similar to the derivatives of trigoometric fuctios, ecept that whereas d cos si, d cosh sih d d Eample 66 Differetiate Solutio sih y sih dy cosh d cosh

113 MFP Tetbook A-level Further Mathematics 660 Eample 66 Differetiate cosh cosh 4 Solutio y cosh cosh cosh cosh 4 Usig the product ad chai rules for differetiatio, dy cosh sih 4cosh sih d cosh sih sih cosh 4 Eercise 6F Differetiate the followig epressios: (a) cosh, (d) cosh, (g) cosech (b) cosh, (e) tah, (c) cosh, (f) sech [hit: write sech as cosh ],

114 MFP Tetbook A-level Further Mathematics Itegratio of hyperbolic fuctios k You will already have met the itegral of e Just to remid you, k e d k e k As hyperbolic fuctios ca be epressed i terms of e, ad as it is the reverse of differetiatio, it follows that their itegratio is straightforward sih d cosh c cosh d sih c sech d tah c Of course, geerally sih k cosh k c k For the itegratio of tah, a substitutio is eeded as follows: tah d sih d cosh Puttig u cosh, du sih d So that tah d du u l uc l cosh c tah d l cosh c coth dl sih c Eercise 6G Evaluate the followig itegrals: (a) cosh d, (b) cosh d, (c) sih d, (d) tah d 4

115 MFP Tetbook A-level Further Mathematics Iverse hyperbolic fuctios Just as there are iverse trigoometric fuctios si, cos, etc, so there are iverse hyperbolic fuctios They are defied i a similar way to iverse trigoometric fuctios so, if sih y, the y sih ; ad likewise for the other five hyperbolic fuctios To fid the value of, say, sih usig a calculator, you use it i the same way as you would if it was a trigoometric fuctio (pressig the appropriate buttos for hyperbolic fuctios) The sketches of y sih ad y tah are the reflectios of y sih ad y tah, respectively, i the lie y These sketches are show below Note that the curve y tah has asymptotes at y sih y y tah y 0 0 Note also that y cosh does ot have a iverse This is because the mappig f: cosh is ot a oe-to-oe mappig If you look at the graph of y cosh (i Sectio 5) you will see that for every value of y there are two values of However, if the domai of y cosh is restricted to 0 there will be a oe-to-oe mappig, ad hece a iverse, ad the rage for the iverse will be y 0 5

116 MFP Tetbook A-level Further Mathematics Logarithmic form of iverse hyperbolic fuctios The iverse hyperbolic fuctios cosh, sih logarithms For eample, if y cosh, the cosh y Multiplyig by e, e e e e y y e e y y y y y y ad tah ca be epressed as 0e e This is a quadratic equatio i e y ad ca be solved usig the quadratic formula: Takig the logarithm of each side, y e 4 4 y l Now, Thus y, ad l l so that y l, l 6

117 MFP Tetbook A-level Further Mathematics 660 However, from Sectio 58, if 0 y the y cosh l A similar result for y sih y i terms of e y sih ca be obtaied by writig sih y ad the epressig This gives y l, but as 0 the egative sig has to be rejected because the logarithm of a egative umber is o-real Thus sih l It is straightforward to obtai the logarithmic form of tah y, tah y is writte as sih y cosh y l cosh l sih l tah y tah if, after writig 7

118 MFP Tetbook A-level Further Mathematics 660 Eample 69 Fid, i logarithmic form, tah Solutio tah l l l l Eample 69 Fid, i logarithmic form, sih 4 Solutio sih l l 4 6 l l l Eercise 6H Show that sih l Show that tah l Epress the followig i logarithmic form: (a) cosh, (b) tah, (c) 5 sih 8

119 MFP Tetbook A-level Further Mathematics Derivatives of iverse hyperbolic fuctios As already see, if y sih, the sih y Differetiatig with respect to, dy cosh y d dy So that d cosh y Agai, if y sih, a sih y, a ad dy cosh y d a Thus dy d acosh y The derivatives of cosh i eactly the same way a a a ad usig cosh ysih y cosh, ad also tah a ad tah a are obtaied Eample 60 d tah d Fid the derivative of Solutio y tah tah y dy sech y d dy d sech y usig sech y tah y 9

120 MFP Tetbook A-level Further Mathematics 660 It is suggested that you work through the remaiig results yourself dy y sih : d dy y cosh : d dy y tah : d Geerally, y y y sih : cosh : tah : dy a d a dy a d a dy a a d a Eample 60 Differetiate cosh Solutio y cosh dy d 9 0

121 MFP Tetbook A-level Further Mathematics 660 Eample 60 If y sih, fid d y d whe, givig your aswer i the form a b l c Solutio Differetiate So, whe Eercise 6I sih y as a product sih dy sih d dy 4sih 4 d 4l 4 8 4l l l 8 4l Differetiate the followig: (a) tah, (b) sih, (d) e sih, (e) cosh (c) cosh, 4 If y cosh, fid d y whe, givig your aswer i the form l, d a b where a ad b are irratioal umbers

122 MFP Tetbook A-level Further Mathematics Itegrals which itegrate to iverse hyperbolic fuctios Itegratio ca be regarded as the reverse of differetiatio so it follows, from the results i Sectio 60, that: a d sih c a d cosh c a a d tah c a a a The itegrals of ad, i particular, help to wide the ability to itegrate a a I fact, these results ca be used to itegrate ay epressio of the form, or p qr eve s t, where p 0 The eamples which follow show how this ca be doe p qr Eample 6 Fid d 6 Solutio d d 6 4 sih c 4

123 MFP Tetbook A-level Further Mathematics 660 Eample 6 Fid d 6 Solutio d d 6 d sih c Eample 6 Fid d Solutio I order to evaluate this itegral, you must complete the square i the deomiator Hece, 4 4 d d 4 Substitutig z, for which dz d, gives d dz z 4 cosh z c cosh c

124 MFP Tetbook A-level Further Mathematics 660 Eample 64 Fid 5 60 d Solutio I this case, the itegral is split ito two by writig oe itegral with a umerator which is the derivative of 6 0 Now, d 60 6 d But 56 ad 5 d 6 d d I I, say Because the derivative of 6 0 is 6, the for I the substitutio z 6 0 gives d z 6 ad cosequetly d I dz z z dz z 60 For I, completig the square i the deomiator, So that I d The substitutio u will give d u, d ad I du u sih u sih Therefore, the complete itegral is II 60 sih c 4

125 MFP Tetbook A-level Further Mathematics 660 Eercise 6J Evaluate the followig itegrals: (a) d, (b) d, (c) 9 6 (d) (g) d, (e) 9 49 d, (h) 45 d, (f) 4 d d, 4 5 d, 6 6 Solvig equatios You are likely to meet two types of equatios ivolvig hyperbolic fuctios The methods for solvig them are quite differet The first type has the form acosh bsih c, or a similar liear combiatio of hyperbolic fuctios The correct method for solvig this type of equatio is to use the defiitios of sih ad cosh to tur the equatio ito oe ivolvig e (frequetly a quadratic equatio) Eample 6 Solve the equatio 7sih 5cosh Solutio Usig the defiitios sih e e ad cosh e e, 7 e e 5 e e 7e 7e 5e 5e e 6e Multiplyig throughout by e, e 6 e, or e e 6 0 This is a quadratic equatio i e ad factorizes to e e 0 Hece, e or e The oly solutio possible is l because e sice e 0 for all values of 5

126 MFP Tetbook A-level Further Mathematics 660 Eample 6 Solve the equatio cosh 4sih 6 Solutio The idetity cosh sih is used here The reaso for this ca be see o substitutio istead of havig a equatio ivolvig cosh ad sih, the origial equatio is reduced to oe ivolvig sih oly sih 4sih6 sih 4sih 5 0 This is a quadratic equatio i sih which factorizes to sih 5 sih 0 Hece, sih 5 or sih ad sih 5 or sih, ad or 088 (to two decimal places) The aswers ca also be epressed i terms of logarithms, usig the results from Sectio 69: sih 5 l 5 5 l 5 6 ; sih l l Note that it is ot advisable to use the defiitios of sih ad cosh whe attemptig to solve the equatio i Eample 6 this would geerate a quartic equatio i e which would be difficult to solve Eamples 6K Solve the equatios: (a) 4sih e 9, (b) sih 4cosh 4, (c) (e) cosh sih 5, (d) cosh cosh 4, tah 7sech 6

127 MFP Tetbook A-level Further Mathematics 660 Miscellaeous eercises 6 (a) Epress cosh sih i terms of e (b) Hece evaluate [AQA March 999] 0 d cosh sih (a) Usig the defiitios prove that ad cosh e e sih cosh sih (b) Hece, or otherwise, solve the equatio 8sih sech, sih e e, leavig your aswer i terms of atural logarithms [AEB Jauary 997] (a) By cosiderig sih y or otherwise, prove that (b) Solve the equatio sih l cosh 5sih 8 0, leavig your aswers i terms of atural logarithms [AEB Jauary 996] 4 (a) Show that the equatio ca be epressed as 4sih 0cosh 5 e 5e 0 (b) Hece solve the equatio 4sih 0cosh 5, givig your aswer as a atural logarithm [AQA Jue 00] 7

128 MFP Tetbook A-level Further Mathematics 660 t 5 (a) Startig from the defiitio of cosh t i terms of e, 4cosh tcosht cosh t show that (b) Hece show that the substitutio cosh t trasforms the equatio 6 5 ito cosh t 5 4 (c) The above equatio i has oly oe real root Obtai this root, givig your aswer i p q the form, where p ad q are ratioal umbers to be foud [AQA March 999] 6 (a) Usig the defiitios of sih ad cosh i terms of epoetials, show that tah e e (b) Hece prove that tah l, where [AEB Jue 999] 7 (a) By epressig tah i terms of sih ad cosh, show that (i) (ii) tah sech, d tah sech d (b) (i) Fid tah tah d (ii) Hece, or otherwise, fid [AQA March 000] tah d 8

129 MFP Tetbook A-level Further Mathematics (a) Eplai, by meas of a sketch, why the umbers of distict values of satisfyig the equatio cosh k i the cases k, k ad k are 0, ad, respectively (b) Give that cosh 7 ad sih y 4, 8 (i) epress y i the form l, where is a iteger, (ii) show that oe of the possible values of y is l ad fid the other possible value i the form l a, where a is to be determied [AQA March 000] 9 (a) State the values of for which cosh is defied (b) A curve C is defied for these values of by the equatio y cosh (i) Show that C has just oe statioary poit (ii) Evaluate y at the statioary poit, givig your aswer i the form p l q, where p ad q are umbers to be determied [NEAB March 998] 0 (a) Prove that d tah sech d (b) Hece, or otherwise, prove that d tah d (c) By epressig i partial fractios ad itegratig, show that tah l (d) Show that 0 tah d a l b, where a ad b are umbers to be determied [AQA Jue 990] 9

130 MFP Tetbook A-level Further Mathematics 660 The diagram shows a regio R i the y plae bouded by the curve y sih, the -ais ad the lie AB which is perpedicular to the -ais y A R (a) Give that AB 4, show that OB l cosh l k k (b) (i) Show that k (ii) Show that the area of the regio R is O B l (c) (i) Show that 0 sih d sih l 9 l 9 4 (ii) Hece fid, correct to three sigificat figures, the volume swept out whe the regio R is rotated through a agle of π radias about the -ais [NEAB Jue 998] 0

131 MFP Tetbook A-level Further Mathematics 660 Chapter 7: Arc Legth ad Area of Surface of Revolutio 7 Itroductio 7 Arc legth 7 Area of surface of revolutio This chapter itroduces formulae which allow calculatios cocerig curves Whe you have completed it, you will: kow a formula which ca be used to evaluate the legth of a arc whe the equatio of the curve is give i Cartesia form; kow a formula which ca be used to evaluate the legth of a arc whe the equatio of the curve is give i parametric form; kow methods of evaluatig a curved surface area of revolutio whe the equatio of the curve is give i Cartesia or i parametric form

132 MFP Tetbook A-level Further Mathematics Itroductio You will probably already be familiar with some formulae to do with the arc legth of a curve ad the area of surface of revolutio For eample, the area uder the curve y f above the -ais ad betwee the lies a b ad b is give by A yd a You will also be familiar with the formula V πy d which gives the volume of the solid of revolutio whe that part of the curve betwee the lies a ad b is rotated about the -ais The formulae to be itroduced i this chapter should be committed to memory You should also realise that, as with may problems, the skills eeded to solve them do ot cocer the formulae themselves but ivolve itegratio, differetiatio ad maipulatio of algebraic, trigoometric ad hyperbolic fuctios may of which have bee itroduced i earlier chapters a b y O a b

133 MFP Tetbook A-level Further Mathematics Arc legth The arc legth of a curve is the actual distace you would cover if you travelled alog it I the diagram alogside, s is the arc legth betwee the poits P ad Q o the curve y f, y, Q has coordiates If P has the coordiates y, y ad PN is parallel to the -ais so that agle PNQ is 90º, it follows that PN ad QN y y O y P a δs δ Q δy N y b Now, if P ad Q are fairly close to each other the the arc legth s must be quite short ad PQN be approimately a right-agled triagle Usig Pythagorus theorem, Dividig by, I the limit as 0, s s y ds d y dy d ds dy d d Thus, b dy s d a d If y f, the legth of the arc of curve from the poit where a to the poit where b is give by b dy s d a d

134 MFP Tetbook A-level Further Mathematics 660 A correspodig formula for curves give i terms of a parameter ca also be derived Suppose that ad y are both fuctios of a parameter t As before, Dividig by t, As 0, 0 s y s t ad ds d ds d t s d y t t t dy dt dt dt dy dt dt dt dy d, t t dt dt where t ad t are the values of the parameter at each ed of the arc legth beig cosidered The legth of arc of a curve i terms of a parameter t is give by t t d dy s dt dt dt, where t ad t are the values of the parameter at each ed of the arc The use of these formulae will be demostrated through some worked eamples 4

135 MFP Tetbook A-level Further Mathematics 660 Eample 7 Fid the legth of the curve y cosh betwee the poits where 0 ad Solutio Therefore, Now, y cosh dy sih d dy sih d cosh dy cosh d 0 0 dy s d 0 d sih cosh d sih sih 0 sih 5

136 MFP Tetbook A-level Further Mathematics 660 Eample 7 Show that the legth of the curve (called a cycloid) give by the equatios a y a cos betwee 0 ad π is 8a Solutio si, Therefore, Now, d a cos d dy asi d d dy a d cos a si d a cos cos si a cos a cos a cos a si d dy 4a si d d asi π π d dy s 0 d d 0 asi d a cos π 4acosπ 4acos0 8 a 0 d (usig si cos ) (usig si cos ) 6

137 MFP Tetbook A-level Further Mathematics Area of surface of revolutio If a arc of a curve is rotated about a ais, it forms a surface The area of this surface is kow as the curved surface area or area of surface of revolutio Suppose two closely spaced poits, P ad Q, are take o the curve y f This arc is rotated about the -ais by π radias The coordiates of P ad Q are, y ad y, y, respectively, ad the legth of arc PQ is δs y y You ca see from the diagram that the curved surface geerated by the rotatio is larger tha that of the cylider of width δs obtaied by rotatig the poit P about the -ais, but smaller tha the area of the cylider width δs obtaied by rotatig the poit Q about the same ais Usig the formula S πrh for the area of the curved surface of a cylider, the area of the former is πy s ad that of the latter is π y y s If the actual area geerated by the rotatio of arc PQ about the -ais is deoted by δa, it follows that πy saπy ys or, dividig by δs, πy A π y y s O a b Now as 0, y0 ad s 0 so that the right-had side of the iequality teds to π y Therefore, da πy ds A a b b πyds dy πy d a d (from sectio 7) y P δs δ Q δy The area of surface of revolutio obtaied by rotatig a arc of the curve y f through π radias about the -ais betwee the poits where a ad b is give by b dy A πy d a d 7

138 MFP Tetbook A-level Further Mathematics 660 Eample 7 Fid the area of surface of revolutio whe the curve y cosh betwee the poits where 0 ad is rotated through π radias about the -ais Solutio This was the curve used i eample 7 from there dy cosh d Hece, dy A πy d 0 d Eample 7 0 π cosh cosh d π cosh d 0 π coshd 0 π sih 0 π sih4 π 0 sih0 π sih4 Show that the area of surface of revolutio whe the cycloid curve give by the equatios a si, y a cos betwee 0 ad π is rotated through π radias about the -ais is 64 π a 8

139 MFP Tetbook A-level Further Mathematics 660 Solutio This was the curve used i eample 7 from there Now, d dy asi d d π 0 d π d y A y d 0 d d π = πa cosasi d 0 π 4πa cos si d 0 π 4πa si cos si d 0 π 8πa si cos si d (usig cos cos ) Now, cosider cos si d The substitutio u cos du si d ad the itegral becomes u du u Itegratig for A, 64 π cos A8πa cos cos gives 8π cosπ cos π cos0 cos 0 a 8πa a π 0 9

140 MFP Tetbook A-level Further Mathematics 660 Miscellaeous eercises 7 (a) Show that d tah sech, d (i) d sech sech tah d (ii) (b) A curve C is give parametrically by tah, y sech, 0 (i) Show that d d dy d tah (ii) The legth of arc C measured from the poit 0, to a geeral poit with parameter is s Fid s i terms of ad deduce that, for ay poit o s curve, y e [AQA Specime] (a) Usig oly the defiitios of cosh ad sih i terms of epoetials, (i) determie the eact values of cosh ad sih, where l 9, 4 (ii) establish the idetities cosh cosh sih sih cosh (b) The arc of the curve with equatio y cosh, betwee the poits where 0 ad l 9, is rotated through oe full tur about the -ais to form a surface of 4 revolutio with area S Show that S π 9 l p 4 for some ratioal umber p whose value you should state [AEB Jue 000] 40

141 MFP Tetbook A-level Further Mathematics 660 (a) (i) Usig oly the defiitios cosh e e ad sih e e, prove the idetity cosh sih (ii) Deduce a relatioship betwee sech ad tah (b) A curve C has parametric represetatio sech, y tah (i) Show that d dy sech d d (ii) The arc of the curve betwee the poits where 0 ad l 7 is rotated through oe full tur about the -ais Show that the area of the surface geerated is 6 π square uits 5 [AEB Jue 997] 4 A curve has parametric represetatio si, y cos, 0 π (a) Prove that d dy 4cos d d (b) The arc of this curve, betwee the poits whe 0 ad π is rotated about the -ais through π radias The area of the surface geerated is deoted by S Determie the value of the costat k for which S k ad hece evaluate S eactly [AEB Jue 996] π 0 si cos d, 4

142 MFP Tetbook A-level Further Mathematics The curve C is defied parametrically by the equatios t t, y t, where t is a parameter (a) Show that d dy t dt dt (b) The arc of C betwee the poits where t 0 ad t is deoted by L Determie (i) the legth of L, (ii) the area of the surface geerated whe L is rotated through π radias about the -ais [AEB Jauary 998] 6 (a) Give that a is a positive costat ad that y a sih a, a use differetiatio to determie the value of the costat k for which dy k a d (b) A curve has the equatio y sih The legth of the arc of curve betwee the poits where 0 ad is deoted by L Show that L 5l5 8 [AEB Jauary 000] 4

143 MFP Tetbook A-level Further Mathematics 660 Aswers to Eercises Further Pure Chapter Eercise A (a), π (b) 4 π, (c) 4, π (d), 5π 6 (a) 6, 8 (b) 5, 09 (c) 707, 7 Eercise B (a) cos π isi π (c) 4cosπ (b) cosπ π isi 4 4 cos 5π isi 5π 6 6 isiπ (d) (a) i (b) i Eercise C (a) i, 4 i (b) 8i, 4 i Eercise D 6 8i (b) i 5 (a) Eercise E (a) cos π isi π (b) z z z z ad arg arg z arg z z z 4

144 MFP Tetbook A-level Further Mathematics 660 Eercise F (a) π 6, (b) 5π, 6 (c) 4 π 9, (d) 7, 0 (e) π, 9 (a) 5 05i (b) 4 i (c) i Eercise G (a) 59, 08 (b) 5, 09 (c) 76, 98 Eercise H (a) y (b) y (c) y O O,0 π 4 O, y (a) y (b) 0, 4, O y O y P 0, 4, O, 0, Q The locus is the lie PQ 44

145 MFP Tetbook A-level Further Mathematics 660 Miscellaeous eercises i y 4 z (a) 4 i, 5 5 (b) (c), 5 O 4 i, 4 i, z 5 5 * i (a) y (b) (c) π Q O P,0, 6 y, arg z π 4 z i O, 0 7 (a)(i) i (ii), π 4 (b)(ii) y P 4, π 6 (iii) 0 O, 7π P 8 (a) (b) y y A, 0, 0, 0 (c)(i) (c)(ii) i (a) 4 i,+i (b)(i) 0, 046 ad 0, 0 (ii) 0 45

146 MFP Tetbook A-level Further Mathematics 660 Chapter Eercise A (a) i (b) 5 i Eercise B (a),, (b), i (c), i Eercise C (a) 7,, 5 (b) 4, 7, Eercise D & (a) (b) (c) (a) (b) 9 0 (c) 68 0 Eercise E 4 (a) (b) Eercise F , i i, i 46

147 MFP Tetbook A-level Further Mathematics 660 Miscellaeous eercises (a), 5 (b) (b) (i) 4i, 6 (ii) 50 (iii) 0,, 50 (a) (b) (a) 8 7 (b) i, (a) p (b) (i) q 7 (ii) (c) (i) (ii) (b) (i) p4, q 4 47

148 MFP Tetbook A-level Further Mathematics 660 Chapter Eercise A (a) r (b) (b) 8 6 (b) Miscellaeous eercises (b) 9 (a) 0 (a) (c) 48

149 MFP Tetbook A-level Further Mathematics 660 Chapter 4 Eercise 4A (a) cos5 isi5 (b) (c) i (d) 8i (e) 64 (f) i 64 (g) 447 Eercise 4B si 4si ta ta ta 5 6si 0si 5si Eercise 4C (a) z z 4 4 (b) z z 7 7 (c) i 6 z 6 z (d) i z z Eercise 4D (a) πi 4 e (b) πi e 6 (c) πi 6 e (d) 5πi 6 4e Eercise 4E (a) (b) 7 (c) 8 Eercise 4F (b) kπ kπ (a), i cos isi, k 0, 5 5 (c) cos kπ i si kπ, k 0,,, 4, cos kπ i si kπ, k, 5 5, i 49

150 MFP Tetbook A-level Further Mathematics 660 Eercise 4G (a) (c) (e) 4k πi 8k πi e 8 k 0,,, (b) 6 e k,, 6kπi 8 e 4 k,,,,8 (d) i 4kπi e 6 k 0,, (f) π icot k 5 Miscellaeous eercises 4 (a) 64, +π (b) cos π i si π 4 4, cos π i siπ 4 4, π π cos i si, cos π i siπ 4 4 i, i (c) (a) cos π isi π π π,cos isi (b)(ii) 40i, 086i, 060 8i 4 4 (b) 5 5 (c) 8 4 (b), 5 5, y 5 (b) z e πi+k 5 k, (c) cetre z, radius (d)(i) π (ii) cos π (b)(ii) 0 (iii) cos π, cos 4π 5 5 (iv) 5, (a)(ii) isi (b)(i) A, B 8 (a)(i) si (c)(i) coefficiets of (b) kπi e k, 6 w cacel (ii) 50

151 MFP Tetbook A-level Further Mathematics (a) πi 4 e (b) πi 7πi 4 e, e (c) y (d) B A C Chapter 5 Eercise 5A (a) π 4 (b) π 6 (c) π (d) π 6 (e) π 6 (f) π Eercise 5C (a) 9 (b) 6 9 (c) 4 (a) ta (b) e cos e 4 (c) si 8 4 (a) si 4 9 (b) 6 ta 4 (a) a a b (b) a a b 5

152 MFP Tetbook A-level Further Mathematics 660 Eercise 5D π 6 π π [ote that 4 si si π by drawig a, 4, 5 right-agled triagle] π 4 5 π Eercise 5E 6 (a) ta + c (b) ta + c (c) ta + c 7 7 l ta + c (a) l ta + c (b) (a) si + c (b) si 4 (a) 4 si + c (b) si + c + c (c) si 4 + c (c) si 5 + c 5

153 MFP Tetbook A-level Further Mathematics 660 Chapter 6 Eercise 6A (a) e e (b) e e (c) e e (d) e e Eercise 6B (a) 064 (b) 086 (c) 04 (d) 054 (e) 05 (f) 00 Eercise 6C (a) y (b) y O (c) y O O Eercise 6E 4 8cosh 8cosh Eercise 6F (a) sih (b) 6sih cosh (c) cosh sih (d) sih cosh (e) tah sech (f) sech tah (g) cosech coth 5

154 MFP Tetbook A-level Further Mathematics 660 Eercise 6G (a) sih c sih c (c) cosh sih c (d) tah 4 (b) Eercise 6H (a) l 5 (b) l (c) l Eercise 6I (a) 9 (d) e sih e (b) 9 (c) (e) 6 4 cosh l Eercise 6J (a) sih c (b) cosh c (c) sih c 4 5 (d) sih c (e) sih c (f) cosh c 7 4 (g) sih c (h) cosh c Eercise 6K (a) l (b) 0, l 7 (d) l 5 (c) l, l 4 7 (e) No solutios 54

155 MFP Tetbook A-level Further Mathematics 660 Miscellaeous eercises 6 (a) e (b) l (b) l, l 5 4 (b) l 4 5 (c) 4 7 (b)(i) tah + c (ii) l cosh tah + c y 8(a) (b)(i) l (ii) l 4 two roots oe root o roots O 9 (a) (b)(ii) l 0 (d) l 8 (c)(ii) 76 55

156 MFP Tetbook A-level Further Mathematics 660 Chapter 7 Miscellaeous eercises 7 (b)(ii) s l cosh (a)(i) 97 65, 7 7 (b) (a)(ii) tah sech 4 (b) k 8π, 0 π 5 (b)(i) (ii) 576 π 5 6 (a) 56

157 MFP Tetbook A-level Further Mathematics 660 Copyright 04 AQA ad its licesors All rights reserved AQA retais the copyright o all its publicatios However, registered schools/colleges for AQA are permitted to copy material from this booklet for their ow iteral use, with the followig importat eceptio: AQA caot give permissio to schools/colleges to photocopy ay material that is ackowledged to a third party eve for iteral use withi the cetre 57