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 ...

Accepted Answer

$\begin{gathered} {\cos }^{-1}x-\frac{\pi }{2}+{\cos }^{-1}y=\alpha \ {\cos }^{-1}x+{\cos }^{-1}y=\frac{\pi }{2}+\alpha \ \because \, \alpha \in [\frac{-\pi }{2},\pi ] \ \frac{\pi }{2}+\alpha \in (0,\frac{3\pi }{2}) \ {\cos }^{-1}(xy-\sqrt{1-{x}^{2}}\sqrt{1-{y}^{2}})=\frac{\pi }{2}+\alpha \ xy-\sqrt{1-{x}^{2}}\sqrt{1-{y}^{2}}=-\sin \alpha \ xy+\sin \alpha =\sqrt{1-{x}^{2}}\sqrt{1-{y}^{2}} \ {x}^{2}{y}^{2}+{\sin }^{2}\alpha +2xy\sin \alpha =1-{x}^{2}-{y}^{2}+{x}^{2}{y}^{2} \ \underset{E}{\underset{}{{x}^{2}+{y}^{2}+2xy\sin \alpha }}={\cos }^{2}\alpha \end{gathered}$

Now, minimum value of $E$ is 0.

Solutions · 1