Know the following (or better yet, learn a couple and be able to derive the rest, quickly, from your knowledge of the trigonometric functions):

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1 OCR Core Module Revision Sheet The C exam is hour 0 minutes long. You are allowed a graphics calculator. Before you go into the exam make sureyou are fully aware of the contents of theformula booklet you receive. Also be sure not to panic; it is not uncommon to get stuck on a question (I ve been there!). Just continue with what you can do and return at the end to the question(s) you have found hard. If you have time check all your work, especially the first question you attempted...always an area prone to error. Trigonometry J.M.S. We define tanθ sinθ cosθ. This identity is very useful in solving equations like sinθ cosθ = 0 which yields tanθ =. The solutions of this in the range 0 θ 60 are θ = 6.4 and θ = 4.4 to one decimal place. Know the following (or better yet, learn a couple and be able to derive the rest, quickly, from your knowledge of the trigonometric functions): θ sinθ cosθ tanθ undefined Be able to sketch sinθ, cosθ and tanθ in both degrees and radians. By considering a right angled triangle (or a point on the unit circle) we can derive the important result sin θ+cos θ. This is useful in solving certain trigonometric equations. Worked example; solve = cos θ +sinθ for 0 θ 60. = cos θ+sinθ = ( sin θ)+sinθ get rid of cos θ, 0 = sin θ +sinθ quadratic in sinθ, 0 = sin θ sinθ factorise as normal, 0 = (sinθ+)(sinθ ). So we just solve sinθ = and sinθ =. Therefore θ = 0 or θ = 0 or θ = 90. The above relation is also useful in converting between the different trigonometric functions. For example if cosθ = 6 7 then, to find sinθ, do not use cos on your calculator J.M.Stone

2 and then sin the answer. Instead sin θ +cos θ =, sin θ =, sinθ = ± 49 = ± 7. Without further information you must keep both the positive and negative solution. If a question tells you that the angle is acute, obtuse or reflex then you must visualise the appropriate graph and interpret. For example given that sinθ = and that θ is obtuse, find the value of cosθ. By the argument above you will find that 8 cosθ = ± = ±. However, given an obtuse angle (90 < θ < 80 ) the cosine graph is negative, so the final answer should be cosθ =. You must be careful when you see things like tanxsinx = tanx. It isso tempting to divide both sides by tanx to yield sinx =. But you must bring everything to one side and factorise; tanxsinx tanx = 0 tanx(sinx ) = 0. The full set of solutions can then be by solving tanx = 0 and sinx = 0. [It is completely analogous to x = x. If we divide by x we find x =, but we know this has missed the solution x = 0. However when we factorise we find x(x ) = 0 and both solutions are found.] Given a trigonometric equation it is always best first to isolate the trigonometric function on its own; for example 9cos(...)+ = 7 cos(...) = 9. For complicated trigonometric equations where you are not just cos ing, sin ing or tan ing a single variable (x, θ, t or the like), it is often easiest to make a substitution. For example to solve cos(x+0) = 4 in the range 0 x 60 the desired substitution is clearly u = x + 0, but you must remember to also convert the range also (many students forget this) so: cos(x+0) = 4 0 x 60, cosu = 4 0 u 70, u =...,...,...,.... However, we don t want solutions in u, so we need to use x = u 0 on each u solution to get x =...,...,..., J.M.Stone

3 Sequences You must be comfortable with -notation. It works as follows; you put in the number at the bottom of the and then keep summing until you reach the top number. For example: 8 (i+) = = 7, i=4 n (i +i) = (+)+(4+)+(9+)+ +(n +n). i= A sequence is a list of numbers in a specific order. A series is a sum of the terms of a sequence. Sequences are sometimes defined recursively. For example the sequence a n+ = a n + with a = 0 defines the sequence 0,,6,9... We know that this is an arithmetic sequence which can also be defined deductively by a n = 0+(n ). An arithmetic sequence increases or decreases by a constant amount. The letter a always denotes the first term and d is the difference between the terms (negative for a decreasing sequence!). The nth term is denoted a n and satisfies the important relationship a n = a+(n )d. For example if told the third term of a sequence is 0 and the seventh term is 4 then we can use the above equation to find the a and d. 0 = a+( )d 4 = a+(7 )d 4d = 4 d = 6 a =. The sum of the n terms of an arithmetic sequence is given by S = n (First+Last) = n (a+(n )d). For example the sum of the first 0 terms of a sequence is 0 and the first term is 4. What is the difference? Binomial Theorem S = n (a+(n )d) 0 = 0 (8+(0 )d) d =. Binomial expansion allows us to expand (a + b) n for any integer n. Best explained by means of an example; expand (x y).. Begin by considering prototype expansion of (a+b).. So (a+b) = ( ) 0 a + ( ) a 4 b+ ( ) a b + ( ) a b + ( ) 4 ab 4 + ( ) b.. Calculate binomial coefficients either on calculator or by drawing a mini Pascal s Triangle to give (a+b) = a +a 4 b+0a b +0a b +ab 4 +b. If you want to multiply instead then use -notation. For example n i = 4 n n! i= J.M.Stone

4 4. Next notice that in our case a = x and b = y and substitute in to get (x y) = (x) +(x) 4 ( y)+0(x) ( y) +0(x) ( y) +(x)( y) 4 +( y).. Tidying up we get (x y) = x 80x 4 y +80x y 40x y +0xy 4 y. It is worth noting that when the expansion is of the form (something another thing) n, then the signs will alternate. Also of note is the way each individual component is constructed. For example; find the ( x coefficient in the expansion of ( x) 7. The component with x is given by 7 ) () ( x) = 04x, so the coefficient is 04. ( ) n = n C r = r n!. For example r!(n r)! Know that (+x) n expands thus: ( n (+x) n = + ) x+ ( ) =!!! = 4 ( ) ( ) = 0. ( ) n x + = +nx+ n(n ) (This is particularly useful in Core 4.) Sine & Cosine Rules The sine rule states for any triangle sina a ( ) n x + +x n x + n(n )(n ) x + +x n.! = sinb b = sinc. c The cosine rule states that a = b +c bccosa. Practice both sine and cosine rules on page 9. By considering half of a general parallelogram we can show that the area of any triangle is given by A = absinc. You must be good at bearing problems which result in triangles. Remember to draw lots of North lines and remember also that they are all parallel; therefore you can use Corresponding, Alternate and Allied angle theorems... revise your GCSE notes! Bearings are measured clockwise from North and must contain three digits. For example Integration. 0.. Calculus is the combined study of differentiation and integration (and their relationship). A good description is that calculus is the study of change in the same way that geometry is the study of shapes. Integration is the reverse of differentiation. That is if dy dx = f(x) then y = f(x)dx. For example if dy dx = x then y = x dx = 4 x4 +c. The general rule is therefore ax n dx = axn+ n+ +c. ydx is an indefinite integral because there are no limits on the integral sign. When evaluating these integrals never forget an arbitrary constant added on at the end. For example 6x dx = x +c. 4 J.M.Stone

5 b a ydx is a definite integral and is the area between the curve and the x-axis from x = a to x = b. Areas under the x-axis are negative. (For areas between the curve and the y-axis switch the x and the y and use q p xdy between y = p and y = q.) To find the area between two curves between x = a and x = b evaluate b a (top bottom)dx. For example, given that y = x + x, find the area under the curve from x = and x =. x+ x dx = x +x dx [ = = ] x +x ( + = 0 8. ) ( ) + To calculate integrals where one of the limits is infinite ( or ), proceed as normal until you input the or into the integral. Then you must not write such things as, +,,, and the like. You must think what these things would equal and just write down the number; in the previous four cases you would get 0, 0, and. (If, in the C exam, you think that when you put in you get an infinite answer, chances are you ve made a mistake somewhere.) For example: 8 x dx = 8 x dx [ = 4 ] x = (0) ( 4 ) =. Geometric Sequences A geometric sequence is one where the terms are multiplied by a constant amount. For example,,4,8,6,...,[ n ] is a geometric sequence with a = and r =. The n th term is given by a n = ar n. So for the above example the 0th term is a 0 = 9 = 488. The sum of n terms of a geometric sequence is given by ( r n ) S = a or (equivalently) by r ( ) r n S = a. r For example sum the first 0 terms of 4,,,,...,[4 ( n ]. ) This is given by ) 0 ) S = 4(( = J.M.Stone

6 If the ratio (r) lies between and (i.e. < r < ) then there exists a sum to infinity given by S = a r. Therefore S for the above example is S = 4 terms is very close to S. Exponentials & Logarithms = 8. We can see that the sum to 0 (In these notes if I write logx I mean log 0 x. If I mean a different base, I will write it explicitly as log a x. When we see log a x we say log to the base a of x.) We define a logarithm to be the solution (in x) to the equation a x = b. It is written x = log a b. The fundamental relationship is therefore a x = b x = log a b. From weseethat log a bmeans: Thenumberahastoberaisedto, tomakeb. Therefore some simple logarithms can be calculated without a calculator: log 8 = the number has to be raised to, to make 8 =, log = the number 0 has to be raised to, to make 0000 = 4, log 9 = the number 9 has to be raised to, to make =, ( = 9 = 9 ) log a a = the number a has to be raised to, to make a =. We see from that logarithms pluck out powers from equations. Therefore if you ever see an equation with the unknown in the power, then that is the clue that you will need to use logarithms. For example to solve 7 x = we discover 7 x =, x = log 7, x = log 7+. However we need to build on because not all equations are this simple (e.g. x = 7 x+ ) and not all calculators can calculate log 7. You can also use to eliminate logarithms from an equation. Given an equation of the form log a ( ) = b, you can eliminate the logarithm instantly to get ( ) = a b. A good way to remember this is Girvan s Bullying Base. So if we have log x = 8, then the bullying base knocks the log out of the way and moves to the other side and squeezes up the 8 to put it in its place; therefore x = 8. Logarithms and exponentials (powers) are the inverse functions of each other (as can be seen from if one puts one into the other). Therefore So if loga =.4 then a = 0.4. From a colleague I respect; Mr Girvan log 0 0 x = x and 0 log 0 x = x. 6 J.M.Stone

7 There are some rules that can be derived from that must be learnt. They are (for all bases): log(ab) = loga+logb log = 0 log( a b ) = loga logb log aa = log(a n ) = nloga log( a ) = loga log a b = log cb log c a. Whenweneedtosolveanequationwheretheunknownisintheexponentsuchas x = 8 take log 0 of both sides and simplify: Factors and Remainders x = 8 log 0 ( x ) = log 0 8 (x )log 0 = log 0 8 x = log 08 log 0 x = ( log0 8 log 0 + x =. (sf). Need to know how to divide any polynomial by a linear factor of the form ax b. For example divide x +x +x 6 by x. (Always devote a column to each power of x.) x +4x + x +x +x +x 6 +x x +4x +x +4x 8x +x 6 +x +6 ) So the remainder is +6. Therefore x +x +x 6 = (x )(x +4x+)+6. If the remainder is zero, then the divisor is said to be a factor of the original polynomial. The Factor Theorem states: (x a) is a factor of f(x) f(a) = 0. [More generally (but used less often in exams) is: (ax b) is a factor of f(x) f( b a ) = 0.] For example if x +ax +8x 4 has (x ) as a factor, find a. From factor theorem we know f() = 0, so we discover +a +8 4 = 0, and therefore a =. 7 J.M.Stone

8 The Remainder Theorem states: When f(x) is divided by (x a) the remainder is f(a). [More generally (but used less often in exams) is: When f(x) is divided by (ax b) the remainder is f( b a ).] Notice that the factor theorem is a subsetof theremainder theorem. In the factor theorem all remainders are zero, by definition. For example if told that when f(x) = x +x x 7 is divided by x the remainder is, we know f() =. Worked example: f(x) = x + x + kx. The remainder when f(x) is divided by (x ) is four times the remainder when f(x) is divided by (x+). Find k. We know Radians f() = 4 f( ) + +k = 4[ ( ) + ( ) k ] k =. There are (by definition) π radians in a circle. So 60 = π. To convert from degrees π to radians we use the conversion factor of 80. For example to convert 4 to radians we calculate 4 π 80 = π 80 4 rad. From radians to degrees we use its reciprocal π. When using radians the formulae for arc length and area of a sector of a circle become simpler. They are s = rθ and A = r θ. The Trapezium Rule The area under any curve can be approximated by the Trapezium Rule. The governing formula is given by (and contained in the formula booklet you will have in the exam) b a ydx h[y 0 +y n +(y +y + +y n )], where h is the width of each trapezium, y 0 and y n are the end heights and y +y + + y n are the internal heights. By considering the shape of the graph in the interval over which you are approximating it should be clear whether your estimate of the area is an over or under-estimate of the true area For example if you were to estimate π 0 sinxdx (above, left) using the trapezium rule, due to the shape of the curve, the trapezia would all fall below the curve, so we would obtain an under-estimate. However, with 0 x dx (above, right) we would obtain an over-estimate. 8 J.M.Stone

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