Question

Prove that $\frac{{\cos }^{3}x-{\sin }^{3}x}{\cos x-\sin x}=\frac{1}{2}(2+\sin 2x)$

Answer 1

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

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

$=1+\sin x\cos x$ (as ${\cos }^{2}x+{\sin }^{2}x=1)$

$=\frac{1}{2}(2+2\sin x\cos x)$

$=\frac{1}{2}(2+\sin 2x)$