Question

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

Answer 1

$\because \cos \theta +\sin \theta =\sqrt{2}\cos \theta$ ... (i)

Let $\cos \theta -\sin \theta =k$ ...(ii)

squaring and adding the two equations , we get

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

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

$\Rightarrow 2-2{\cos }^2\theta =k^2$

$\Rightarrow k^2=2{\sin }^2\theta$

$\Rightarrow k=\sqrt{2}\sin \theta$

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