Question

If $\cos \theta +\sin \theta =\sqrt{2}\cos \theta$, prove that $\cos \theta -\sin \theta =\sqrt{2}\sin \theta$.

Answer 1

Given: $\cos \theta +\sin \theta =\sqrt{2}\cos \theta$,

Squaring both the sides, we get,

$\Rightarrow (\cos \theta +\sin \theta {)}^2=(\sqrt{2}\cos \theta {)}^2$

$\Rightarrow {\cos }^2\theta +{\sin }^2\theta +2\cos \theta \sin \theta =2{\cos }^2\theta$

$\Rightarrow {\sin }^2\theta -{\cos }^2\theta +2\cos \theta \sin \theta =0$

Subtracting, $2{\sin }^2\theta$ from both sides, we get,

$\Rightarrow -{\sin }^2\theta -{\cos }^2\theta +2\cos \theta \sin \theta =-2{\sin }^2\theta$

$\Rightarrow {\sin }^2\theta +{\cos }^2\theta -2\cos \theta \sin \theta =2{\sin }^2\theta$

$\Rightarrow (\cos \theta -\sin \theta {)}^2=(\sqrt{2}\sin \theta {)}^2$

$\Rightarrow \cos \theta -\sin \theta =\sqrt{2}\sin \theta$

Hence, proved.