Solve the following
Question:

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

Solution:

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

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

$=\cos ^{-1}\left[\sin \frac{\pi}{4} \sin x+\cos \frac{\pi}{4} \cdot \cos x\right]=\cos ^{-1}\left[\cos \left(\frac{\pi}{4}-x\right)\right]$

$y=\frac{\pi}{4}-x$ $\left[\because-\frac{\pi}{4}<x<\frac{\pi}{4}\right]$

Differentiating both sides w.r.t. $x$

Thus, $\quad \frac{d y}{d x}=-1$

Administrator

Leave a comment

Please enter comment.
Please enter your name.