Question

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

Answer 1

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

we have to prove that $\cos \theta +\sin \theta =\sqrt{2}\cos \theta$

Now,

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

squaring both sides,

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

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

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

Adding a ${\cos }^2\theta$ on both sides of the equation we get

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

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

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

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

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

Hence, proved.