Question

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

Answer 1

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

$\Rightarrow \tan \theta =\frac{1}{\sqrt{2}+1}\times \frac{\sqrt{2}-1}{\sqrt{2}-1}=\frac{sqrt2-1}{1}$

As $\tan \theta =\frac{p}{B},\sin \theta =\frac{p}{H}$

$\Rightarrow \sin \theta =\frac{\sqrt{2}-1}{\sqrt{1^2+(\sqrt{2}-1{)}^2}}=\frac{\sqrt{2}-1}{\sqrt{4-2\sqrt{2}}}$

and $\cos \theta =\frac{B}{H}=\frac{1}{\sqrt{4-2\sqrt{2}}}\Rightarrow \sin \theta +\cos \theta =\frac{\sqrt{2}}{\sqrt{4-2\sqrt{2}}}=\frac{1}{\sqrt{2-\sqrt{2}}}$.