Target Mathematics

1.95: Trigonometric identities and equations

4PM1/1/June/2015 — Question 8 · 17 marks

Using the identities
cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A+B)=\cos A\cos B-\sin A\sin B
sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A+B)=\sin A\cos B+\cos A\sin B
(a) (i) show that cos⁡2A=1−2sin⁡2A\cos2A=1-2\sin^2A, (3)
(ii) express sin⁡2A\sin2A in terms of sin⁡A\sin A and cos⁡A\cos A, simplifying your answer. (1)
(b) Hence show that sin⁡3A=3sin⁡A−4sin⁡3A\sin3A=3\sin A-4\sin^3A. (4)
(c) Solve, for −90∘≤A≤90∘-90^\circ\le A\le90^\circ, the equation 8sin⁡3A−6sin⁡A=18\sin^3A-6\sin A=1. (4)
(d) (i) Find ∫sin⁡3θ dθ\displaystyle\int\sin^3\theta\,d\theta.
(ii) Evaluate ∫0π/4sin⁡3θ dθ\displaystyle\int_0^{\pi/4}\sin^3\theta\,d\theta, giving your answer in the form a−b2c\dfrac{a-b\sqrt2}{c}, where a,b,ca,b,c are integers. (5)