Question

Find $n$: $\frac{{\cos }^{3}x-{\sin }^{3}x}{\cos x-\sin x}=\frac{1}{n}(n+\sin nx)$

Answer 1

$\because a^3-b^3=(a-b)(a^2+b^2+ab)$

LHS- $\frac{{\cos }^{3}x-{\sin }^{3}x}{\cos x-\sin x}$

$\Rightarrow \frac{(\cos x-\sin x)({\cos }^{2}x+{\sin }^{2}x+\cos x\sin x)}{\cos x-\sin x}$

$\because {\cos }^2\theta +{\sin }^2\theta =1$

$\Rightarrow 1+\cos x\sin x$

$2\cos \theta \sin \theta =\sin 2\theta$

$\Rightarrow 1+\frac{2\cos x\sin x}{2}$

$\Rightarrow 1+\frac{\sin 2x}{2}$

$\Rightarrow \frac{2+\sin 2x}{2}$

$\Rightarrow n=2$