Considering the principal values of the inverse trigonometric functions, ^-1(√32 x+(1)/(2)…

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

Accepted Answer

$\begin{gathered} {\sin }^{-1}(\frac{\sqrt{3}}{2}x+\frac{1}{2}\sqrt{1-{x}^{2}}),-\frac{1}{2}0 \end{gathered}$

$\begin{gathered} {\sin }^{-1}(\frac{\sqrt{3}}{2}\cos 	heta +\frac{1}{2}\sin 	heta )={\sin }^{-1}( \sin (\frac{\pi }{3}+	heta )) \ \frac{\pi }{3}+	heta \in (\frac{7\pi }{2},\pi ) \end{gathered}$

$\begin{gathered} ={\sin }^{-1}(\sin (\pi -(\frac{\pi }{3}+	heta ))) \ ={\sin }^{-1}(\sin (\frac{2\pi }{3}-	heta )) \ =\frac{2\pi }{3}-	heta \ =\frac{2\pi }{3}-{\cos }^{-1}x \ =\frac{2\pi }{3}-(\frac{\pi }{2}-{\sin }^{-1}x) \ =\frac{\pi }{6}+{\sin }^{-1}x \end{gathered}$

Solutions · 1