1.68: Double-angle identity, equation and integral

4PM1/2/January/2019 — Question 11 · 17 marks

\cos(A+B)=\cos A\cos B-\sin A\sin B

(a) (i) Using the above identity, show that

\cos 2x=1-2\sin^{2}x.

(ii) Hence show that

\frac{13\sin x-2\cos 2x-10}{4\sin x-3}=4+\sin x.

(7)

(b) Hence solve, in radians to 3 significant figures, the equation

10+2\cos\left(2\theta+\frac{\pi}{3}\right)-13\sin\left(\theta+\frac{\pi}{6}\right)=2\sin\left(\theta+\frac{\pi}{6}\right)+8

for

\pi\leqslant\theta\leqslant 2\pi

(5)

\int_{0}^{\pi/2}\frac{13\sin x-2\cos 2x-10+4x\sin x-3x}{4\sin x-3}\,\mathrm{d}x.

(5)