Question

If $\sin \theta +\cos \theta =\sqrt{2}\cos \theta$, then the general value of $\theta$ is

Answer 1

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

sguasing both sides, we get

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

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

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

$-(co{s}^2\theta -si{n}^2\theta )+2sincos\theta =0$

$\Rightarrow -cos2\theta +sin2\theta =0$ [$\therefore$ $co{s}^2\theta -si{n}^2\theta =cos2\theta 2sin\theta cos\theta =sin2\theta ]$

$\Rightarrow sin20=cos20$

$\Rightarrow sin20=sin(\frac{\pi }{2}-20)$

$\therefore 20=\frac{\pi }{2}-20$

$\Rightarrow 20+20=\frac{\pi }{2}$

$\Rightarrow 40=\frac{\pi }{2}$

$\Rightarrow 0=\frac{\pi }{8}$