Question

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

Answer 1

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

To prove $cos\theta -sin\theta =\sqrt{2}sin\theta$

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

on squaring both sides we get

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

$si{n}^2\theta +co{s}^2\theta +2sin\theta cos\theta =2co{s}^2\theta$

$si{n}^2\theta -co{s}^2\theta +2sin\theta cos\theta =0$

$-si{n}^2\theta +co{s}^2\theta -2sin\theta cos\theta =0$

or adding $2{\sin }^2\theta$ both sides we get

$si{n}^2\theta +co{s}^2\theta -2sin\theta cos\theta =2si{n}^2\theta$

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

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

Hence, proved