Given that the inverse trigonometric function assumes principal values only.

Question

Given that the inverse trigonometric function assumes principal values only. Let $x, y$ be any two real numbers in $$[-1,1]$$ such that $$\cos ^{-1} x-\sin ^{-1} y=\alpha, \frac{-\pi}{2} \leq \alpha \leq \pi$$.
Then, the minimum value of $$x^2+y^2+2 x y \sin \alpha$$ is

Accepted Answer

$$\begin{aligned}
& \cos ^{-1} x-\frac{\pi}{2}+\cos ^{-1} y=\alpha \
& \cos ^{-1} x+\cos ^{-1} y=\frac{\pi}{2}+\alpha \
& \because \quad \alpha \in\left[\frac{-\pi}{2}, \pi\right] \
& \text { then } \frac{\pi}{2}+\alpha \in\left(0, \frac{3 \pi}{2}\right) \
& \cos ^{-1}\left(x y-\sqrt{1-x^2} \sqrt{1-y^2}\right)=\frac{\pi}{2}+\alpha \
& x y-\sqrt{1-x^2} \sqrt{1-y^2}=-\sin \alpha \
& x y+\sin \alpha=\sqrt{1-x^2} \sqrt{1-y^2} \
& x^2 y^2+\sin ^2 \alpha+2 x y \sin \alpha=1-x^2-y^2+x^2 y^2 \
& \underbrace{x^2+y^2+2 x y \sin \alpha}_E=\cos ^2 \alpha
\end{aligned}$$

Now, minimum value of $E$ is 0.

Solutions · 1