Question

Question 4 (a) Prove the following identity \[ \frac{\cos x-\sin x}{\cos x+\sin x}=\frac{\cos 2 x}{1+\sin 2 x} \]

(b) Find the general solution (in radians) of the following equation \[ \sin 6 x+\cos

Answer 1

See below

Explanation:

`\displaystyle \[ (\cos x-\sin x)/(\cos x+\sin x)\\\\\\=(\cos x-\sin x)/(\cos x+\sin x)\cdot(\cos x+\sin x)/(\cos x+\sin x)\\\\=(\cos^2x-\sin x^2)/(\cos^2x+2\sin x\cos x+\sin^2 x)\\\\=(\cos 2x)/(1+2\sin x\cos x)\\\\=(\cos 2x)/(1+\sin 2x)`

Thus, the identity is proven.

Note the following:

`2\sin x\cos x=\sin 2x` <-- Double-Angle Identity

`\cos^2x+\sin^2x=1` <-- Pythagorean Identity

`\cos^2x-\sin^2x=\cos 2x` <-- Double-Angle Identity