( ) Trigonometric identities and equations, Mixed exercise 10

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1 Trigonometric identities and equations, Mixed exercise 0 a is in the third quadrant, so cos is ve. The angle made with the horizontal is. So cos cos a cos 0 0 b sin sin ( ) sin 4 b is in the fourth quadrant so sin is ve. The angle to the horizontal is 48. So sin sin 48 c 90 is in the third quadrant so tan 90 is +ve. The angle to the horizontal is 0. So tan90 + tan0 c cos80 ( see graph of cosθ ) d tan 40 tan ( ) + tan 60 ( third quadrant) So tan 40 + e tan tan 4 ( second quadrant) So tan Using sin A+ cos A sin A + sin A 4 sin A ± But A is in the second quadrant ( obtuse ), so sin A is + ve. So sin A + sin A Using tan A cos A tan A rationalising the denominator Pearson Education Ltd 0. Coping permitted for purchasing institution onl. This material is not copright free.

2 4 Draw a right-angled triangle with an angle of φ, where φ +. b c θ + θ (use sin cos ) θ θ θ θ θ sin sin cos sin sin θ 4 sin 4 4 cos θ + sin θcos θ + sin θ ( cos θ sin θ) + + (since sin θ cos θ ) a b Use Pthagoras theorem to find the hpotenuse. x So x sinφ As B is reflex and tan B is +ve, B is in the third quadrant. So sin B sinφ From the diagram cos φ. B is in the third quadrant. So cos B cosφ 4 4 a Factorise cos θ sin θ. (This is the difference of two squares. 4 4 cos θ sin θ ( cos θ sin θ)( cos θ sin θ) ( cos θ sin θ) ( as cos θ + sin θ ) So cos θ sin θ cos θ sin θ b Factorise sin θ sin θcos θ. sin θ sin θcos θ ( θ) sin θ cos sin x+ cos x sin x+ cos x 6 a sin x+ 4 cos x sin x+ cos x sin x sin x cos x 4 cos x sin x cos x (divide both sides b cos x) So tan x b sin xcos + cos xsin sin x sin 4 cos x cos sin xcos cos xsin + cos xcos cos xcos sin x sin 4 cos x cos cos xcos cos xcos tan x+ tan tan xtan 4 tan xtan tan 4 + tan x tan tan x 4 + tan x 4 + tan x So tan tan x a b LHS + sin + sin + cos ( θ ) θ θ θ + θ + θ + θ sin since sin cos + sinθ + sin RHS 4 LHS cos θ + sin θ ( cos θ) ( θ) + sin θ ( θ) sin + sin θ 4 sin θ + sin θ + sin θ 4 sin + sin θ 4 cos θ + sin θ RHS (using sin θ + cos θ ) Pearson Education Ltd 0. Coping permitted for purchasing institution onl. This material is not copright free.

3 8 a sinθ has no solutions as sinθ. b sinθ cosθ tanθ Look at the graph of tanθ in the interval 0 θ 60. There are two solutions. c The minimum value of sinθ is. The minimum value of cos θ is. Each minimum value is for a different θ. So the minimum value of sinθ + cos θ is alwas greater than. There are no solutions of sinθ + cosθ as the LHS can never be zero. 0 a As sin ( 90 θ) cos θ, sin ( 90 θ) cos θ So 4 cos θ sin ( 90 θ) 4cosθ cosθ cosθ b a θ ( θ) Using, 4 cos sin 90 is equivalent to cosθ so cosθ Let θ and solvecos in the interval The calculator solution is 48.9 As cos is + ve, is in the first and fourth d Solving tanθ + 0 is equivalent to tanθ solving tan θ, which has no solutions. So there are no solutions. 9 a 4x + 4x ( 4x ) + ( 4x ) ( 4x )( + ) b Using a with x sin θ, cosθ 4sin cos cos θ θ θ + 4sinθ cosθ 0 So 4sinθ cosθ cosθ + 0 So 4sinθ cosθ 0 or cosθ + 0 4sinθ cosθ 0 tanθ 4 The calculator solution is θ 4.0. tanθ is +ve so θ is in the first and third So θ 4.0,94 cosθ + 0 cosθ So θ + 80 from graph Solutions are θ 4.0,80,94 Read off all solutions in the interval ,.8,408.9,6.8, 68.9,0.8 So θ 6.,04,6,4, 6, 44 s.f. a sin θ cos θ sin θ cos θ sin θ tan θ since tan θ cos θ So tan θ 0. b Solve tan θ 0. in the interval 0 θ < 60 or tan 0. where θ, 0 < 0. Pearson Education Ltd 0. Coping permitted for purchasing institution onl. This material is not copright free.

4 b The calculator solution for tan 0. is 6.. As tan is +ve, is in the first and third b sin θ 0. in the interval 0 θ < 60. Solve sin 0., where θ, 0 < 0. The calculator solution is As sin is +ve, is in the first and second Read off solutions for in the interval 0 < 0. 6., 06., 86., 66. θ So θ.,0.,9., 8. d.p. a cos( θ + ) 0. Solve cos 0., where θ +, < 4. Your calculator solution for 60. As cos is +ve, is in the first and fourth Read off all solutions in the interval < 4. 00, 40 θ + 00, 40 So θ, 4 Read off solutions in the interval 0 < ,.6, 404.4, 49.6 θ So θ.,6.8,0.,4.8 d.p. Multipl both sides of the equation b ( cos x), provided cos x. Note: In the interval given, cos x is never equal to. So cos x+ 0. cos x cos x So cos x Solve cos where x, 0 < < 40. The calculator solution is 60. As cos is + ve, is in the first and fourth Pearson Education Ltd 0. Coping permitted for purchasing institution onl. This material is not copright free. 4

5 Read off solutions for in the interval 0 < < , 00, 40 So x 0, 0, 0 4 Using sin θ + cos θ cos θ cosθ cos θ θ θ cos cos 0 ( θ )( θ ) cos + cos 0 So cosθ + 0 or cosθ 0 For cosθ + 0, cosθ The calculator solution is.8. As cos θ is ve, θ is in the second a and third b 0, 0, 90, 0, 0, 80, 0, 0, 0, 690, 90, 00 i.e. x 00, 90, 690, 0, 0, 0, 0, 0, 90, 0, 0, 80 So x 0, 0, 0, 90, 0, 0, 0, 0, 0, 0, 0, 90 6 a b The graphs intersect at two places so there are two solutions to the equation in the given range. c sin x cos x sin x cos x tan x x.,. θ.8, 8. For cosθ, θ 0 (See graph and check the given interval.) So solutions are θ 0,.8, 8.,60 a The student found additional solutions after dividing b three rather than before. The students has not applied the full interval for the solutions. b Let x a Using the cosine rule a + c b cos B ac 6 + cos B cos B 9 cos B b Using Pthagoras theorem 9 40 sin As x, then as 60 x 60 So So the interval for is sin B 40 0 Pearson Education Ltd 0. Coping permitted for purchasing institution onl. This material is not copright free.

6 8 a Using the sine rule sin Q sin P q p sin Q sin sin Q sin Q Challenge tan 4 x tan x + 0 (tan x )(tan x ) 0 tan x or tan x tan x ± or ± x 4,, 4,,, 4., 4., 4.,., 0. So the solutions are x 4, 4.,.,,, 4., 0., b Using Pthagoras theorem or identit cos x+ sin x cosq for the acute angle As Q is obtuse, it is in the second quadrant where cos Q is negative. So cosq 9 a sin x cos x can be written as sin x ( sin x) which reduces to 4sin x b 4sin x sin x 4 sin x ± x 60, 0, 60, 0 So the solutions are x 0.0, 60.0, 60.0, cos x + 4sin x can be written as ( sin x) + 4sin x which can be reduced to sin x + 4sin x 4 0 (sin x )(sin x + ) 0 sin x or sin x sin x has no solutions. So x 4.8, 8.,.8, 8. So the solutions are x 8.,.8, 4.8, 8. Pearson Education Ltd 0. Coping permitted for purchasing institution onl. This material is not copright free. 6