1 TRIGONOMETRY. 1.0 Introduction. 1.1 Sum and product formulae. Objectives

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1 TRIGONOMETRY Chapter Trigonometry Objectives After studying this chapter you should be able to handle with confidence a wide range of trigonometric identities; be able to express linear combinations of sine and cosine in any of the forms Rsin( θ ± α) or Rcos( θ ± α); know how to find general solutions of trigonometric equations; be familiar with inverse trigonometric functions and the associated calculus.. Introduction In the first Pure Mathematics book in this series, you will have encountered many of the elementary results concerning the trigonometric functions. These will, by and large, be taken as read in this chapter. However, in the first few sections there is some degree of overlap between the two books: this will be good revision for you.. Sum and product formulae You may recall that sin( A + B)=sin Acos B + cos Asin B sin( A B)=sin Acos B cos Asin B Adding these two equations gives sin( A + B)+sin( A B)=sin Acos B () Let C = A + B and D = A B, then C + D = A and C D = B. Hence A = C + D, B= C D and () can be written as

2 Chapter Trigonometry sin C + sin D = sin C + D cos C D This is more easily remembered as 'sine plus sine = twice sine( half the sum)cos( half the difference)' Activity In a similar way to above, derive the formulae for (a) sin C sin D (b) cosc + cos D (c) cosc cos D By reversing these formulae, write down further formulae for (a) sin E cos F (b) cos E cos F (c) sin E sin F Example Show that cos59 +sin 59 = cos4. Solution Firstly, sin 59 =cos, since sin θ = cos( 9 θ) So LHS = cos 59 +cos = cos cos = cos 45 cos4 = cos4 Example = cos4 = RHS Prove that sin x + sin x + sin x = sin x( + cos x). Solution LHS = sin x + ( sin x + sin x) =sin x + sin x + x cos x x = sin x + sin x cos x = sin x + cos x ( )

3 Chapter Trigonometry Example Write cos 4x cos x sin 6x sin x as a product of terms. Solution Now cos 4x cos x = cos( 4x + x)+cos 4x x = cos5x + cosx and sin 6x sin x = { cos( 6x x) cos 6x + x ( )} = cosx cos9x Thus, LHS = cos5x + cosx cosx + cos9x = (cos5x + cos9x) = cos 5x + 9x = cos 7x cosx The sum formulae are given by 5x 9x cos sin A + sin B = sin A + B cos A B sin A sin B = cos A + B sin A B cos A + cos B = cos A + B cos A B cos A cos B = sin A + B sin A B and the product formulae by sin Acos B = (sin(a + B) + sin(a B)) cos Acos B = (cos(a + B) + cos(a B)) sin Asin B = (cos(a B) cos(a + B))

4 Chapter Trigonometry Exercise A. Write the following expressions as products: (a) cos5x cosx (b) sinx sin 7x (c) cosx + cos9x (e) cos π 5 + cos 4π 5 (f) sin 4 +sin5 +sin6 (g) cos4 +sin 4 (d) sinx + sinx 4π 8π + cos + cos 5 5. Evaluate in rational/surd form sin75 +sin5. Write the following expressions as sums or differences: (a) cos7x cos5x (b) cos x cos 5x 4. Establish the following identities: (a) cosθ cosθ = 4sin θ cosθ (b) sin6x + sin 4x sin x = 4cosxsin x cos x (c) (d) (e) sin 4A + sin6a + sin A sin 4A sin6a sin A = cot A ( ) + sin( A B) ( ) + cos( A B) = tan A sin A + B cos A + B ( ) + cos ( θ + 6 ) cos θ + sin( θ + ) + sin( θ + 6 ) = tanθ + tanθ 5. Write cosx + cos6x + cos4x + cosx as a product of terms. 6. Express cosx cos x cos7x cos5x as a product of terms. (c) sin π 4 θ cos π 4 + θ (d) sin65 cos5. Linear combinations of sin and cos Expressions of the form a cosθ + bsin θ, for constants a and b, involve two trig functions which, on the surface, makes them difficult to handle. After working through the following activity, however, you should be able to see that such expressions (called linear combinations of sin and cos linear since they involve no squared terms or higher powers) can be written as a single trig function. By re-writing them in this way you can deduce many results from the elementary properties of the sine or cosine function, and solve equations, without having to resort to more complicated techniques. For this next activity you will find it very useful to have a graph plotting facility. Remember, you will be working in radians. Activity Sketch the graph of a function of the form y = asin x + b cos x (where a and b are constants) in the range π x π. 4

5 Chapter Trigonometry From the graph, you must identify the amplitude of the function and the x-coordinates of (i) the crossing point on the x-axis nearest to the origin, and (ii) the first maximum of the function as accurately as you can. An example has been done for you; for y = sin x + cos x, you can see that amplitude R. 4 y RR crossing point nearest to the origin O at x = α = π 4 maximum occurs at x = β = π 4 Try these for yourself : π α β π x - (a) (c) (e) y = sin x + 4 cos x (b) y = cos x 5sin x y = 9 cos x +sin x (d) y = 5sin x 8cos x y = sin x + 5cos x (f) y = cos x sin x In each case, make a note of R, the amplitude; α, the crossing point nearest to O; β, the x-coordinate of the maximum. In each example above, you should have noticed that the curve is itself a sine/cosine 'wave'. These can be obtained from the curves of either y = sin x or y = cos x by means of two simple transformations (taken in any order).. A stretch parallel to the y-axis by a factor of R, the amplitude, and. A translation parallel to the x-axis by either α or β (depending on whether you wish to start with sin x or cos x as the original function). Consider, for example y = sin x + cos x. This can be written in the form y = Rsin( x + α),since Rsin(x + α) = R{ sin x cosα + cos x sin α} = Rcos α sin x + Rsin α cos x The R ( > ) and α should be chosen so that this expression is the same as sin x + cos x. 5

6 Chapter Trigonometry Thus Rcosα = and Rsin α = Dividing these terms gives tanα = α = π 4 Squaring and adding the two terms gives Since cos α + sin α =, R cos α + R sin α = + R ( cos α + sin α) = R = R = Thus sin x + cos x = sin x + π 4 Activity Express the function sin x + cos x in the form sin x + cos x = Rcos(x α) Find suitable values for R and α using the method shown above. Another way of obtaining the result in Activity is to note that so that sin θ = cos π θ since cos( θ) = cosθ. sin x + cos x = sin x + π 4 = cos π x + π 4 = cos π 4 x = cos x π 4 6

7 Chapter Trigonometry Example Write 7sin x 4 cos x in the form Rsin( x α) where R > and < α < π. Solution Assuming the form of the result, 7sin x 4cos x = Rsin(x α) = Rsin x cosα Rcos xsinα To satisfy the equation, you need Rcosα = 7 Rsin α = 4 Squaring and adding, as before, gives Thus R = = 65 cosα = 7 65, sin α = 4 or tan α = α =.59 radians, to sig. figs. so 7sin x 4 cos x = 65 sin(x.59) Exercise B Write (in each case, R > and < α < π ). sin x + 4cos x in the form Rsin(x + α ). 4cos x + sin x in the form Rcos ( x α ). 5sin x 8cos x in the form Rsin( x α ) 4. 6cos x sin x in the form Rcos( x + α ) 5. sin x cos x in the form Rsin( x α ) 6. 4cos x + sin x in the form Rcos( x α ) 7. cosx sin x in the form Rcos( x + α ) 8. cos x + 5sin x in the form Rsin( x + α ) 7

8 Chapter Trigonometry. Linear trigonometric equations In this section you will be looking at equations of the form a cos x + bsin x = c for given constants a, b and c. Example Solve cos x + sin x = for x 6. Solution Method Note that cos x and sin x are very simply linked using cos x + sin x = so a 'rearranging and squaring' approach would seem in order. Rearranging: cos x = sin x Squaring: 9 cos x = 4 4sin x + sin x 9 sin ( x)= 4 4sin x + sin x = sin x 4sin x 5 The quadratic formula now gives sin x = 4 ± 6 and giving sin x.9485 or x = 69.,.8 or., 7.7 ( d.p.) Method Write cos x + sin x as Rcos( x α) (or Rsin( x α)) cos x + sin x = Rcos( x α) Firstly, R = + = so cos x + sin x = cos x + sin x ( cos x cosα + sin x sin α) 8

9 Chapter Trigonometry Thus cosα = or sin α = 7 or tan α = => α =8.4 The equation cos x + sin x = can now be written as cos( x 8.4 )= cos( x 8.4 )= and x 8.4 =cos x 8.4 =5.77 or 9. x = or x = 69. or 7.7 ( d.p.) The question now arises as to why one method yields four answers, the other only two. If you check all four answers you will find that the two additional solutions in Method do not fit the equation cos x + sin x =. They have arisen as extra solutions created by the squaring process. (Think of the difference between the equations x = and x = 4: the second one has two solutions.) If Method is used, then the final answers always need to be checked in order to discard the extraneous solutions. Exercise C. By writing 7sin x + 6cos x in the form ( )( R >, <α < 9 ) solve the equation Rsin x + α 7sin x + 6cos x = 9 for values of x between and 6.. Use the 'rearranging and squaring' method to solve (a) 4cosθ + sinθ = (b) sinθ cosθ = for θ 6.. Write cosθ + sinθ as Rcos( θ α ), where R > and < α < π and hence solve cosθ + sinθ = for θ π. 4. Solve (a) 7cos x 6sin x = 4 for 8 x 8 (b) 6sinθ + 8cosθ = 7 for θ 8 (c) 4cos x + sin x = 5 for x 6 (d) sec x + 5tan x + = for x π 9

10 Chapter Trigonometry.4 More demanding equations In this section you will need to keep in mind all of the identities that you have encountered so far including the Addition Formulae, the Sum and Product Formulae and the Multiple Angle Identities in order to solve the given equations. Example Solve cos5θ + cosθ = cosθ for θ 8 Solution Using cos A + cos B = cos A + B cos A B LHS = cosθ cosθ Thus cosθ cosθ = cosθ cosθ ( cosθ )= Then (a) cosθ = or θ = 9, 7, 45 θ =,9,5 (b) cosθ = cosθ = θ = 6, θ =,5 as already found. Solutions are θ = (twice), 9,5 (twice). [Remember, for final solutions in range θ 8, solutions for θ must be in range θ 8 =54.] Exercise D. Solve for θ 8 : (a) cosθ + cosθ = (b) sin 4θ + sinθ = (c) sinθ + sinθ = sin θ.. Find all the values of x satisfying the equation sin x = sin π x for x π.

11 Chapter Trigonometry. By writing θ as θ + θ, show that cosθ = 4cos θ cosθ and find a similar expresion for sinθ in terms of powers of sinθ only. Use these results to solve, for θ 6, (a) cosθ + cosθ = (b) sinθ = sin θ (c) cosθ cosθ = tan θ 4. Show that tan x + cot x = cosecx. Hence solve tan x + cot x = 8cosx for x π. 5. Solve the equation sin x + sinx + sin5x = for x 8 6. Find the solution x in the range x 6 for which sin 4x + cosx =. 7. (a) Given t = tan θ, write down tanθ in terms of t and show that cosθ = t + t Find also a similar expression for sinθ in terms of t. (b) Show that sinθ tanθ = t t 4 ( t ) (c) Hence solve sinθ tanθ = 6cot θ for values of θ in the range <θ <6..5 The inverse trigonometric functions In the strictest sense, for a function f to have an inverse it has to be ('one-to-one'). Now the three trigonometric functions sine, cosine and tangent are each periodic. Thus the equation sin x = k, for k, has infinitely many solutions x. A sketch of the graph of y = sin x is shown opposite. When working on your calculator, if you find sin.5, say, a single answer is given, despite there being infinitely many to choose from. In order to restrict a 'many-to-one' function of this kind into a function, so that the inverse function gives a unique answer, the range of values is restricted. This can be done in a number of ways, but the most sensible way is to choose a range of values x which includes the acute angles. This is shown on the diagram opposite. y y = sin x -π π 4π x - y Thus for k, sin x = k x = sin k x will be assigned a unique value of x in the range π x π : these are the principal values of the inverse-sine function. Activity 4 By drawing the graphs y = cos x and y = tan x, find the ranges of principal values of the inverse-cosine and inverse-tangent functions. (These should include the acute angles of x.)

12 and Chapter Trigonometry Note that the inverse functions are denoted here by sin x, cos, tan x These are not the same as ( sin x) = sin x, etc. and to avoid this confusion, some texts denote the inverse functions as arcsin x, arccos x, arctan x..6 General solutions Up until now you have been asked for solutions of trigonometric equations within certain ranges. For example: Solve sin x = for < x <8 or Find the values of θ for which sin θ + sin θ = with θ π. At the same time, you will have been aware that even the simplest trig equation can have infinitely many solutions: sin θ = ( θ radians) is true when θ =, π,π,π, K also for all negative multiples of π as well. Overall, one could say that the equation sin θ = has general solution θ = n π where n is an integer. Moreover, there are no values of θ which satisfy this equation that do not take this form. Thus, ' θ = n π ' describes all the values of θ satisfying ' sin θ = ' as n is allowed to take any integer value. This is what is meant by a general solution. General solution for the cosine function For 9 x 9 (radians can come later), the graph of y = cos x is shown below. y P. V. y = cos x y = k -7-6 α 6 7 x -

13 Chapter Trigonometry The line y = k (k is chosen as positive here, but as long as k, the actual value is immaterial) is also drawn on the sketch. The principal value of x for which x = cos k (representing the point when y = k and y = cos x intersect) is circled and labelled 'P.V.' Since the cosine function is periodic with period 6, all other solutions to the equation cos x = k corresponding to this principal value are obtained by adding, or subtracting, a multiple of 6 to it. The points of intersection of the two graphs representing these solutions are circled also. Now the cosine curve here is symmetric in the y-axis. So if α is the principal value of x for which cos x = k, then α is also a solution, and this is not obtained by adding, or subtracting, a multiple of 6 to, or from, α. All the remaining solutions of the equation can be obtained by adding or subtracting a multiple of 6 to or from α. The general solution of the equation cos x = k ( k ) is then x = 6n ± α where α = cos k is the principal value of the inverse cosine function and n is an integer. In radians, using 6 π radians, the general solution looks like x = nπ ± α, n an integer. Activity 5 Use the graphs of y = tan x and y = sin x to find the general solutions of the equations (in degrees) of the equations and tan x = k ( < k < ) sin x = k ( k ). In each case, let α be the principal value concerned, let n be an integer, and express the general solutions in terms of radians once the results have been found in terms of degrees.

14 Chapter Trigonometry These results are summarised as follows: In radians In degrees If sin θ = sin α, then If sin θ = sin α, then θ = nπ + ( ) α θ = 8n + ( ) α If cosθ = cosα, then If cosθ = cosα, then θ = nπ ± α θ = 6n ± α If tan θ = tan α, then If tan θ = tan α, then θ = nπ + α θ = 8n + α [In each case, n is an integer.] The AEB's Booklet of Formulae gives only the set of results for θ, α in radians, but you should be able to convert the results into degrees without any difficulty by remembering that 8 π radians, etc. Example Find the general solution, in degrees, of the equation tan θ = Solution tan θ = tan θ = tan 6 θ = 8n + 6 quoting the above result θ = ( 6n + ) [ K; n = θ= 4 ; n = θ = ; n = = 8 ; K] Sometimes, you may have to do some work first. Example Find the general solution, in radians, of the equation 8sin θ +5cosθ = 6. 4

15 Chapter Trigonometry Solution Rewriting the LHS of this equation in the form Rsin( θ + α), for instance, gives R = 8 +5 = 7 and cosθ = 8 7 or sin θ = 5 7 or tan θ = 5 8 so that θ.8 radians. [Check this working through to make sure you can see where it comes from.] The equation can now be written as 7sin( θ +.8)=6 sin( θ +.8)= 6 = sin.67 7 (principal value of sin 6 7 ) θ +.8 = nπ + ( ) n.67 θ = nπ + ( ) n.67.8 One could proceed to make this more appealing to the eye by considering the cases n even and n odd separately, but there is little else to be gained by proceeding in this way. Note on accuracy: although final, non-exact numerical answers are usually required to three significant places, the.67 and.8 in the answer above are really intermediate answers and hence are given to 4 significant figure accuracy. However, unless a specific value of n is to be substituted in order to determine an individual value of θ, you will not be penalised for premature rounding provided your working is clear and the answers correspond appropriately. This final example illustrates the sort of ingenuity you might have to employ in finding a general solution of some equation. Example Find the values of x for which cos x sin 4x =. Solution Using the result sin A = cos π A, the above equation can be written as cos x = cos π 4x whence x = nπ ± π 4x 5

16 Chapter Trigonometry i.e. x = nπ + π 4x or x = nπ π + 4x 5x = nπ + π or x = nπ+ π x = n π 5 + π or x = n π +π 6 Wait a moment! Although the second solution is correct, n is merely an indicator of some integer; positive, negative or zero. It is immaterial then, whether it is denoted as positive or negative: you could write this solution as x = k π + π 6 ( k = n) for some integer k, or alternatively, as x = n π + π 6. In this case, the general solution takes two forms. An alternative approach could have re-written the equation as sin 4x = sin π x 4x = nπ + ( ) n π x When n is odd 4x = nπ π + x x = nπ π x = n π π 6 = ( n )π 6 (n odd) and when n is even 4x = nπ + π x 5x = nπ + π x = n π 5 + π = ( n + ) π (n even) 6

17 Chapter Trigonometry The very first way might be considered preferable since the general solution for cos is less clumsy than that for sin. This example also highlights another important point: two equivalent sets of answers may look very different from each other and yet still both be correct. Exercise E. Find the general solutions, in degrees, of the equations (a) sin x =.766 (b) tan( θ 45 ) = (c) cos x =.7 (d) cot( 6 θ)= (e) 5sin x + cos x = 4 (f) 4cosθ + sinθ = (g) cosθ + cosθ = (h) tan 4x = (i) sin 7x sin x = cos4x. Find the general solutions, in radians, of the equations (a) tan x = (b) cos x + π 6 = (e) 6 sinθ cosθ = (f) cosθ 4sinθ = (g) cos x cosx = (h) tan x + cot x = [Note: cot A = tan π A (i) tan θ + 5secθ + = (j) cos4x + cos6x = cos5x and tan ( A ) = tan A ]. Prove the identity cos4x + 4cosx = 8cos 4 x. Hence find the general solution, in radians, of the equation cos4θ + 8cosθ =. (c) sin x =.5 (d) sec x + π 4 =.7 Calculus of the inverse trigonometric functions The prospect of having to differentiate the function y = sin x may seem rather daunting. However, we can write y = sin x as sin y = x. (Taking the principal range values π y π ensures that this can be done.) Then, using the Chain Rule for differentiation, d ( sin y)= d ( dy sin y) dy dy = cos y so that sin y = x differentiates to give cos y dy = dy = cos y 7

18 Chapter Trigonometry Now, using cos y = sin y, so that cos y =± sin y =± x dy =± x y A quick look at the graph of sin x shows that the gradient of the inverse-sine curve is always positive (technically speaking, infinitely so at x =±) and so y = sin x differentiates to x dy = x Activity 6 Use the above approach to find dy when y = tan x. Find the derivative of cos x also, and decide why it is not necessary to learn this result as well as the result for the derivative of sin x when reversing the process and integrating. The results d ( sin x)= x d ( tan x)= + x and the corresponding integrations = sin x + C x = tan x + C (C constant) + x are special cases of the more general results: 8

19 Chapter Trigonometry = sin x a x a + C = x a + x a tan + C for constant C. a The results, in this form, are given in the AEB's Booklet of Formulae, and may be quoted when needed. Activity 7 (a) (b) Use the substitution x = asin θ to prove the result = sin x a x a + C. Use the substitution x = a tan θ to prove the result a + x = x a tan a + C. Example Evaluate Solution 4 4x Now 4x = 4 4 x = 4 x so that 4 = 4x 4 4 x (and this is now the standard format, with a = ) = x sin 4 = [ sin x 4 ] = sin sin = π 6 = π 9

20 Chapter Trigonometry Example Evaluate + x, giving your answer correct to 4 decimal places. Solution + x = [so a = here] ( ) + x = x tan = tan tan [Important note: work in radians] ( ) =.79 (4 d.p.) Exercise F Evaluate the following integrals, giving your answers to four significant figures. [Remember to work in radians.] x 6. x x x 7 7 x Evaluate the following integrals exactly: 6 + x * x 5 5 x 4 x x + x

21 Chapter Trigonometry.8 The 't = tan A' substitution You will have already encountered the result: tan θ = tan θ tan θ which arises from the identity tan A + tan B tan( A + B)= tan A tan B Setting θ = A then yields the result tan A = t t where t = tan A In the triangle shown opposite, tan A = t, and the t hypotenuse, h, is given by Pythagoras' theorem: h = ( t )+( t) A h t t = t +t 4 +4t =+t +t 4 ( ) = +t So h = + t and sin A = t t and cos A = + t + t This would seem, at first sight, to be merely an exercise in trigonometry manipulation, but these results (which are given in the AEB's Booklet of Formulae) have their uses, particularly in handling some otherwise tricky trigonometric integrations. Incidentally, the above working pre-supposes that angle A is acute, and this is generally the case in practice. The identities are valid, however, for all values of A in the range A π (and hence all values of A). Although, when A = π,π, Ktan A is not defined, the limiting values of sin A, cos A and tan A are still correct.

22 Chapter Trigonometry Example Use the substitution t = tan x to show that the indefinite integral of sec x is ln tan π 4 + x. Solution t = tan x dt = sec x dt = sec x dt = + tan x dt + t = Also, sec x = cos x = + t, using one of the above results. t Then Now sec x= +t t. dt + t = = = t t dt ( )( +t) dt + t + dt by partial fractions t = ln + t ln t = ln + t t π tan π 4 + tan x = 4 + tan x tan π 4 tan = + t x t (since tan π 4 = ) so that sec x=ln tan π 4 + x

23 Chapter Trigonometry Activity 8 Use the above identities for cos x and tan x to prove that + tan sec x + tan x = x tan x The results sec x=ln sec x + tan x = ln tan π 4 + x are given in the AEB's Booklet of Formulae, as is the result cosec x =ln tan x Since this latter result is much easier to establish, it has been set as an exercise below. Example Use the substitution t = tan θ to evaluate exactly Solution Also, π dθ 4 cosθ + sin θ t = tan θ dt dθ = sec θ dt + t = dθ cosθ = t t and sin θ = + t + t Changing the limits: θ = t = and θ = π t = tan π 4 = so Thus θ, π (,) π dθ = 4 cosθ + sin θ 4 4t + t + 6t + t. dt + t

24 Chapter Trigonometry = = dt 4 4t + 6t dt + t t = 5 t + 5 dt + t = 5 ln t + ln + t 5 = 5 ln+ 5 ln 5 ln 5 ln = 5 ln 6 In the following exercise, you may have to use the tan or sin integrals from the previous section. Exercise G. Use the substitution t = tan x to evaluate (c) π dθ + 5sinθ π π + sin, giving your answer to 4 decimal x 4 places.. By writing t = tan x, show that (a) (b) (c) cosec x=ln tan x + C = tan + cos x x + C cos = sin tan x cos x x + C. Use the t = tan θ substitution to evaluate exactly the integrals 4. (a) By using the identity tan A = t t where t = tan A, and setting A = π, show that 6 tan π = (b) Evaluate, to four decimal places, the integral π π 6 dθ + cosθ 5. By setting t = tan x, find the indefinite integral sec x cos x (a) π dθ 5 + 4cosθ (b) π tan θ dθ 5 + 4cosθ 4

25 Chapter Trigonometry.9 Harder integrations In this final section of the chapter, all of the integrations involve the standard results for sin and tan, but you may have to do some work to get them into the appropriate form. Before you start, here are a few reminders of the algebraic techniques which you will need, and also one or two calculus results. To give you a clear idea of how they work out in practice, they are incorporated into the following set of examples. Example By writing x + 7x + + x ( ) x ( ) in terms of partial fractions, show that x + 7x + ln π + x 4 ( ) x ( ) = π Solution x + 7x + + x ( ) x ( ) Ax + B + x + C x Multiplying throughout by the denominator Substituting x = gives x + 7x + ( Ax + B) ( x)+c( + x ) = 5C C = 4 Substituting x = gives = B + 4 B = Comparing x coefficients: = A+4 A = Thus Then x + 7x + + x ( ) x x + 7x + + x ( ) x ( ) x ( ) = + x + 4 x x + x x x 5

26 Chapter Trigonometry The reason for splitting the integration up in this way is to separate the directly integrable bits. Note that ( ) ( ) f x f x = ln f ( x ) + constant i.e. when the 'top' is exactly the differential of the 'bottom', the integral is natural log of the 'bottom'. Now d ( + x )=x and d ( x )=, so the constants in the numerators need jiggling but, apart from this, you should see that the three integrals are log, tan and log respectively: = x + x = ln [ ( x +) ] tan x + x 4 x [ ] 4ln x [ ( ) ] [Strictly speaking, the log integrals should be x + and x, but + x is always positive and x is positive for x between and.] = ln ln ( ) ( tan tan ) 4ln ln ( ) = ln π 4 + 4ln = ln π 4 Example x + Evaluate, giving your answer correct to three x significant figures. Solution x + = x x x + x Now the second integral on the RHS is clearly a sin integral. What about the first one? 6

27 Chapter Trigonometry You might just recognise where the x in the numerator comes from if you think about it long enough. To save time: try calling u = x = ( x ) ; then du = x ( ) 6x ( ) using the Chain Rule = x x So x = x x [This method is referred to as 'integration by recognition'] and x + = x x x + x [ ] = x + x sin = + + sin sin =.59 (to s.f.) Example Integrate exactly the integrals (a) x 6x + 4 (b) + 5x x Solution (a) x = 6x + 4 = x x ( x ) [Now x x + 4 = ( x ) + by completing the square: the factor of was taken out first in order to make this easier to cope with.] [This is a + x with a = and 'x' = x, which is allowed when a single x is involved.] 7

28 Chapter Trigonometry = tan x = tan x [ ([ ] ) ] = ( tan tan ) = = π 9 π (b) = + 5x x + 5 x x Completing the square: + 5 x x = x 5 x+ = x This gives integral = = 9 6 x 5 4 ( 4 ) ( x 4 5 ) ( ) = x 5 sin 4 4 ( ) = 4x 5 sin = sin sin = sin = sin 8

29 Chapter Trigonometry Exercise H. By expressing fractions, evaluate x + ( x) 4 + x ( ) in terms of partial 6. 4 x π Show that = 4 + x ln. 7. Use partial fractions to help evaluate the integral x + x ( ) ( ) 4 + x giving your answer to three significant figures. x +. Show that = 9 + x 4 ln 4 + π 8.. Given that y = 4 x, find an expression for dy and deduce that x = C 4 x 4 x for some constant C. Hence evaluate exactly x 4 x 4. Determine the values of the constants A, B and C such that x + x x A + Bx + x + C + x x + x + 7 Show that + x = + ln + π 5. By 'completing the square' in each of the following cases, evaluate exactly the integrals: (a) x x (b) x x x ( ) + x ( ) 8. Evaluate the following integrals: (a) 4 5x + 4 x + 4 (b) 4 + x x + x 9. Determine the values of the constants A, B and C for which f( x)= x +x 4 x x+4 A+B x x x+4 + C x x+4 Hence evaluate f( x). 4x. (a) Prove that 5 + x = ln. 4 4x + (b) Use the result of (a) to evaluate 5 + x, giving your answer correct to decimal places. * x. Show that = π +6x x. * + x. Show that = x 4 8π + 5 ( ). (c) (d) x x x x [Hint: x = x x x x ] (e) x x (f) x + 6x + 9

30 Chapter Trigonometry. Miscellaneous Exercises. Prove the identity tanθ + cotθ cosecθ. Find, in radians, all the solutions of the equation tan x + cot x = 8cosx in the interval < x < π. (AEB). Find, in radians, the general solution of the equation 6tan θ = 4sin θ +. (AEB). Prove the identity cot θ + tanθ cosecθ. Hence find the values of θ, in the interval <θ <8 for which ( cot θ + tanθ ) = 4. (AEB) 4. Find, in terms of π, the general solution of the equation tan 4 x 4 tan x + =. (AEB) 5. Solve the equation tanθ secθ =,giving all solutions in the interval <θ <6. (AEB) 6. By expanding cos θ 6 ( ), express 7cosθ + 8cos( θ 6 ) in the form sin( θ + α ), where <α < 9, and state the value of α to the nearest.. Hence find the solutions of the equation 7cosθ + 8cos( θ 6 ) = 6.5 in the interval <θ <6, giving your answer to the nearest.. (AEB) 7. Find all solutions in the interval θ 6 of the equation sinθ cosθ = k when (a) k =, and (b) k =. (AEB) 8. Find the general solution of the equation sin x + cos x = for x radians. (AEB) 9. Show that ( ) sin x + sin 4x + sin6x sin 4x + cosx Hence prove the identity sinxsin 4x ( sin x + sin 4x + sin6x)sin x Deduce that sin π = 6 +. (AEB). Prove the identity ( cos A + cos B) + ( sin A + sin B) + cos( A B). Hence solve the equation ( cos4θ + cosθ ) + ( sin 4θ + sinθ ) = sinθ giving the general solution in degrees.. Given that < x, y <, prove that +x. Express f( x)= in partial ( +x )( +x) fractions. Prove that the area of the region enclosed by the curve with equation y = f( x), the coordinate axes and the line x = is π 4 + ln 9. (AEB). Express 5cosθ + sinθ in the form Rsin( θ + α ), where R > and <α < 9. The function f is defined by f θ ( ) = 6 5cosθ sinθ for θ 6. State the greatest and least values of f and the values of θ, correct to the nearest., at which these occur. 4. Show that x +x integration by parts, evaluate 6 x 5. Express x Hence show that ( ) ( ) + x 6 x x ( ) ( ) + x = ln. Hence using tan x. (AEB) in partial fractions. = ln 6 + 5π 6 (AEB) 6. Express sinθ sinθ in terms of cos θ. Hence show that if sinθ = λ sin θ, where λ is a constant, then either sinθ = or 4cos θ λ cosθ =. Determine the general solution, in degrees, of the equation sinθ = sin θ. (AEB) 7. Express cos x 4sin x in the form Acos( x + α ), where A > and α is acute, stating the value of α to the nearest.. 4 (a) Given that f( x)= cos x 4sin x + 7 : (i) Write down the greatest and least values of f(x) and the values of x to the nearest. in the interval 8 < x <8 at which these occur; (ii) Find the general solution, in degrees, of the tan x + tan y = tan x + y xy Deduce the value of tan + tan 5 + tan 8 (AEB) equation f( x)= 6. (b) Solve the equation cos x 4sin x = 5cosx, giving your answers to the nearest. in the interval < x <8. (Oxford)

31 Chapter Trigonometry 8. Given that cosθ + 4sinθ Rcos( θ α ), where R > and α π, state the value of R and the value of tanα. (a) For each of the following equations, solve for θ in the interval θ π and give your answers in radians correct to one decimal place: (i) cosθ + 4sinθ = (ii) cosθ + 4sin θ = 5cosθ. (b) The curve with equation y =, between x = π and cos x + 4sin x + 7 x = π, cuts the y-axis at A, has a maximum point at B and a minimum point at C. Find the coordinates of A, B and C. (AEB) 9. Given that f ( x )=9sin x + π 6 + 5cos x + π, use the formulae for sin( A + B) and cos( A + B) to express f(x) in the form C cos x + D sin x, where C and D are integers. Hence show that f(x) can be written in the form 6 cos( x α ) giving a value for α in radians to three significant figures. (Oxford) + tanθ. Prove the identity + sin θ + tan θ. By using the substitution t = tanθ, or otherwise, find the general solution, in radians, of the equation ( tanθ )( + sin θ ) =. (AEB). (a) Starting from the identity cos A + B prove the identity ( ) cos Acos B sin Asin B, cosθ cos θ. (b) Find the general solution of the equation sinθ + tanθ cosθ =, giving your answer in radians in terms of π. (c) Prove the identity cos θ cos θ cosθ cos4θ. (d) By substituting θ = π 5 prove that cos π 5 cos π 5 in the identity in (c), =. (e) Hence find the value of cos π 5 in the form a + b 5, stating the values of a and b. (AEB). Use identities for cos( C + D) and cos( C D) to prove that cos A + cos B = cos A + B cos A B. Hence find, in terms of π, the general solution of the equation cos5θ + cosθ = cosθ. Using both the identity for cos A + cos B, and the corresponding identity for sin A sin B, show that ( ). cos5α sinα = sinα cos4α + cosα The triangle PQR has angle QPR = α( ), angle PQR = 5α and RP = RQ. Show that sin5α = sinα and deduce that cos4α + cosα =. By solving a quadratic equation in cosα, or otherwise, find the value of α, giving your answer to the nearest.. (AEB) 7x x. Given that ( x) ( x +) A x + Bx + C x +, determine the values of A, B and C. A curve has 7x x equation y =. Prove that the area ( x) ( x +) of the region enclosed by the curve, the x-axis and the line x = is 7 ln π 4. (AEB) 4. Use the substitution x = + tanθ to evaluate the x integral x giving your answer x + 5 correct to two decimal places. (AEB) 5. By expressing cosθ sin θ and other similar expressions as the difference of two sines, prove the identity ( cosθ + cosθ + cosθ +)sin θ sin 7θ. Express cosθ and cosθ in terms of cosθ and deduce the identity ( 8cos θ + 4cos θ 4cosθ )sin θ sin 7θ. Hence, or otherwise, show that cos π 7, cos 4π 7 and cos 6π are the roots of the equation 7 8x + 4x 4x =. (AEB) 6. Assuming the identities sinθ sinθ 4sin θ and cosθ 4cos θ cosθ prove that cos5θ 5cosθ cos θ +6cos 5 θ.

32 Chapter Trigonometry (a) Find the set of values of θ in the interval < θ < π for which cos5θ > 6cos 5 θ. (b) Find the general solution, in radians, of the equation cos x + cosx + cos5x =. (AEB) 7. Express f ( θ ) = 4cosθ + sinθ in the form Rcos( θ α ) where R > and < α < π. (a) A rectangle OABC is formed from the origin, the point A( 4cosθ,), the point B, and the point C, sinθ and express the perimeter of the rectangle in terms of f ( θ ). Hence find the greatest perimeter of the rectangle as θ varies in the ( ). State the coordinates of B range θ π and state the coordinates of B for which this greatest perimeter occurs. (b) A curve has the equation y = x π. Show that ( 4cos x + sin x) the region enclosed by the curve, the x-axis, and the lines x = and x = π has area. (AEB)