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
Solution
<p>$$\begin{aligned}
& \cos ^{-1} x-\frac{\pi}{2}+\cos ^{-1} y=\alpha \\
& \cos ^{-1} x+\cos ^{-1} y=\frac{\pi}{2}+\alpha \\
& \because \quad \alpha \in\left[\frac{-\pi}{2}, \pi\right] \\
& \text { then } \frac{\pi}{2}+\alpha \in\left(0, \frac{3 \pi}{2}\right) \\
& \cos ^{-1}\left(x y-\sqrt{1-x^2} \sqrt{1-y^2}\right)=\frac{\pi}{2}+\alpha \\
& x y-\sqrt{1-x^2} \sqrt{1-y^2}=-\sin \alpha \\
& x y+\sin \alpha=\sqrt{1-x^2} \sqrt{1-y^2} \\
& x^2 y^2+\sin ^2 \alpha+2 x y \sin \alpha=1-x^2-y^2+x^2 y^2 \\
& \underbrace{x^2+y^2+2 x y \sin \alpha}_E=\cos ^2 \alpha
\end{aligned}$$</p>
<p>Now, minimum value of $E$ is 0.</p>
About this question
Subject: Mathematics · Chapter: Inverse Trigonometric Functions · Topic: Domain and Range
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