Trigonometry & Inverse Trigonometry
Maximum of (sin⁻¹x)² + (cos⁻¹x)²
nta_pyq_2026_jan
Grade 12
Question:
Let the maximum value of $(\sin^{-1}x)^2+(\cos^{-1}x)^2$ for $x\in\left[-\dfrac{\sqrt{3}}{2},\dfrac{1}{\sqrt{2}}\right]$ be $\dfrac{m}{n}\pi^2$, where $\gcd(m,n)=1$. Then $m+n$ is equal to _____.
Step-by-Step Solution
Key Concept: Let $\alpha=\sin^{-1}x$. $(\sin^{-1}x)^2+(\cos^{-1}x)^2=\alpha^2+(\tfrac{\pi}{2}-\alpha)^2=2\alpha^2-\pi\alpha+\tfrac{\pi^2}{4}=2(\alpha-\tfrac{\pi}{4})^2+\tfrac{\pi^2}{8}$. For $x\in[-\tfrac{\sqrt{3}}{2},\tfrac{1}{\sqrt{2}}]$, $\alpha\in[-\tfrac{\pi}{3},\tfrac{\pi}{4}]$. Maximum at $\alpha=-\tfrac{\pi}{3}$ (farthest from vertex $\tfrac{\pi}{4}$).
Maximum $=\tfrac{29}{36}\pi^2$. $m+n=65$.
Correct Answer: 65