Considering the principal values of the inverse trigonometric functions, ^-1(√32x+12√1-x^2…

Question

Considering the principal values of the inverse trigonometric functions, $\sin^{-1}\left(\dfrac{\sqrt3}{2}x+\dfrac12\sqrt{1-x^2}\right)$, $-\dfrac12<x<\dfrac{1}{\sqrt2}$, is equal to

Accepted Answer

Let $x=\cos\theta$, $\theta\in\left(\dfrac{\pi}{4},\dfrac{2\pi}{3}\right)$, so $\sqrt{1-x^2}=\sin\theta$ (positive). Then

$$\sin^{-1}\left(\frac{\sqrt3}{2}\cos\theta+\frac12\sin\theta\right)=\sin^{-1}\left(\sin\left(\theta+\frac{\pi}{3}\right)\right).$$

Since $\theta+\dfrac{\pi}{3}\in\left(\dfrac{7\pi}{12},\pi\right)$ (outside $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$), use $\sin^{-1}(\sin\phi)=\pi-\phi$ there:

$$=\sin^{-1}\left(\sin\left(\pi-\theta-\frac{\pi}{3}\right)\right)=\pi-\theta-\frac{\pi}{3}=\frac{2\pi}{3}-\theta=\frac{2\pi}{3}-\cos^{-1}x=\frac{2\pi}{3}-\left(\frac{\pi}{2}-\sin^{-1}x\right)=\frac{\pi}{6}+\sin^{-1}x.$$

Solutions · 1