Question

Differentiate \[\sin^{- 1} \left\{ \frac{\sin x + \cos x}{\sqrt{2}} \right\}, - \frac{3 \pi}{4} < x < \frac{\pi}{4}\] ?

Answer 1

\[\text{ Let, y } = \sin^{- 1} \left\{ \frac{\sin x + \cos x}{\sqrt{2}} \right\}\]

\[ \Rightarrow y = \sin^{- 1} \left\{ \sin x\left( \frac{1}{\sqrt{2}} \right) + \cos \left( \frac{1}{\sqrt{2}} \right) \right\}\]

\[ \Rightarrow y = \sin^{- 1} \left\{ \sin x \cos\frac{\pi}{4} + \cos x \sin\frac{\pi}{4} \right\}\]

\[ \Rightarrow y = \sin^{- 1} \left\{ \sin\left( x + \frac{\pi}{4} \right) \right\} . . . \left( i \right)\]

\[\text{ Here, } \frac{- 3\pi}{4} < x < \frac{\pi}{4}\]

\[ \Rightarrow \frac{- 3\pi}{4} + \frac{\pi}{4} < x + \frac{\pi}{4} < \frac{\pi}{4} + \frac{\pi}{4}\]

\[ \Rightarrow \frac{- \pi}{2} < x + \frac{\pi}{4} < \frac{\pi}{2}\]

\[\text{ From } \left( i \right) \text{ we get }, \]

\[ \Rightarrow y = x + \frac{\pi}{4} \left[ Since, \sin^{- 1} \left( \sin\theta \right) = \theta, \text{ if }\theta \in \left[ \frac{- \pi}{2}, \frac{\pi}{2} \right] \right]\]

\[\text{ Differentiating it with respect to x }, \]

\[\frac{d y}{d x} = 1 + 0\]

\[ \therefore \frac{d y}{d x} = 1\]