Question

Solve ${\sin }^{-1}\left\{\frac{\sin x+\cos x}{\sqrt{2}}\right\},-\frac{3\pi }{4}<x<\frac{\pi }{4}$

Answer 1

Let ${\sin }^{-1}\left\{\frac{\sin x+\cos x}{\sqrt{2}}\right\}=\theta$

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

$\Rightarrow \sin x\cos \frac{\pi }{4}+\cos x\sin \frac{\pi }{4}=\sin \theta$ where $\frac{1}{\sqrt{2}}=\sin \frac{\pi }{4}$ and $\frac{1}{\sqrt{2}}=\cos \frac{\pi }{4}$

$\Rightarrow \sin (x+\frac{\pi }{4})=\sin \theta$ where $\sin A\cos B+\cos A\sin B=\sin (A+B)$

LHS and RHS are positive.Hence $\sin \theta$ is positive in first quadrant and second quadrant.

$\sin (x+\frac{\pi }{4})$ is positive when $x=\frac{\pi }{4}$

and $x=\pi -\frac{\pi }{4}=\frac{3\pi }{4}$

and $x=-\pi +\frac{\pi }{4}=\frac{-3\pi }{4}$ in clockwise direction.

Hence ${\sin }^{-1}\left\{\frac{\sin x+\cos x}{\sqrt{2}}\right\}={\sin }^{-1}\left\{\sin (x+\frac{\pi }{4})\right\}=x+\frac{\pi }{4}$