Considering the principal values of the inverse trigonometric functions, sin^(-1) (√3/2x+1/2√(1-x^2)), -1/2<x<1/√2, is equal to

Question

Considering the principal values of the inverse trigonometric functions, \(\sin ^{-1}\left(\frac{\sqrt{3}}{2} x+\frac{1}{2} \sqrt{1-x^{2}}\right),-\frac{1}{2}<x<\frac{1}{\sqrt{2}},\) is equal to

(1) \(\frac{\pi}{4}+\sin ^{-1} x\)

(2) \(\frac{\pi}{6}+\sin ^{-1} x\)

(3) \(\frac{-5 \pi}{6}-\sin ^{-1} x\)

(4) \(\frac{5 \pi}{6}-\sin ^{-1} x\)

Answer 1

Correct option is: (2) \(\frac{\pi}{6}+\sin ^{-1} x\)  

\( \sin ^{-1}\left(\frac{\sqrt{3}}{2} x+\frac{1}{2} \sqrt{1-x^{2}}\right), \frac{-1}{2}<x<\frac{1}{\sqrt{2}}\)

\(\Rightarrow \text{Let}\ \sin ^{-1}(\mathrm{x})=\theta \quad \frac{-\pi}{6}<\theta<\frac{\pi}{4}\)

\(\Rightarrow \mathrm{x}=\sin \theta,\) then

\(\Rightarrow \sin ^{-1}\left(\frac{\sqrt{3}}{2} \sin \theta+\frac{1}{2} \cos \theta\right)\)

\(\Rightarrow \sin ^{-1}\left(\sin \left(\theta+\frac{\pi}{6}\right)\right)=\theta+\frac{\pi}{6}\)

\(\Rightarrow \sin ^{-1}(\mathrm{x})+\frac{\pi}{6}\)