Question

If cos x + sin x = $\sqrt{2}cosx$ then show that cosx-sinx =$\sqrt{2}sinx$

Answer 1

$\cos x+\sin x=\sqrt{2}\cos x$

Squaring both sides, we get

${\cos }^{2}x+{\sin }^{2}x+2\cos x\sin x=2{\cos }^{2}x$

$2\cos x\sin x={\cos }^{2}x-{\sin }^{2}x$

$2\sin x\cos x=(\cos x+\sin x)(\cos x-\sin x)$

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

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

$\Rightarrow \frac{2\sqrt{2}\sin x}{\cos x-\sin x}=\sqrt{2}\times \sqrt{2}$

$\Rightarrow \frac{2\sqrt{2}\sin x}{\cos x-\sin x}=2$

$\therefore \cos x-\sin x=\sqrt{2}\sin x$