Question

If $\cos \theta -\sin \theta =\sqrt{2}\sin \theta$ then show that $\cos \theta +\sin \theta =\sqrt{2}\cos \theta$

Answer 1

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

Squaring on both sides

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

$1-2\cos \sin \theta =2{\sin }^2\theta$

$2\cos \theta \sin \theta =1-2{\sin }^2\theta$

$(\cos \theta -\sin \theta {)}^2=(\cos \theta +\sin \theta {)}^2-4\cos \theta \sin \theta$

$(\sqrt{2}\sin \theta {)}^2=(\cos \theta +\sin \theta {)}^2-2(1-2{\sin }^2\theta )$

$(\cos \theta +\sin \theta {)}^2=2{\sin }^2\theta +2-4{\sin }^2\theta$

$(\cos \theta +\sin \theta {)}^2=2-{\sin }^2\theta =2(1-{\sin }^2\theta )$

$=2{\cos }^2\theta$

$\cos \theta +\sin \theta =\sqrt{2}\cos \theta$.