For -1 ≤ x ≤ 1, what is the least value of (sin^-1 x)^2 + (cos^-1 x)^2?

Question 118

HardOriginal question

For $-1 \le x \le 1$ , what is the least value of $\left(\sin^{-1} x\right)^{2} + \left(\cos^{-1} x\right)^{2}$ ?

\frac{\pi^{2}}{16}

Correct answerYour answer

\frac{\pi^{2}}{8}

Correct answerYour answer

\frac{\pi^{2}}{4}

Correct answerYour answer

\frac{5 \pi^{2}}{4}

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer B

Correct answer: B

Let $s = \sin^{-1} x$ , in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ ; then $\cos^{-1} x = \frac{\pi}{2} - s$ .
$s^{2} + \left(\frac{\pi}{2} - s\right)^{2} = 2\left(s - \frac{\pi}{4}\right)^{2} + \frac{\pi^{2}}{8}$ .
It is least when $s = \frac{\pi}{4}$ , that is at $x = \frac{1}{\sqrt{2}}$ : the least value is $\frac{\pi^{2}}{8}$ .

Worth noting

$\frac{\pi^{2}}{4}$ makes one term 0, at $x = 0$ or $x = 1$ , but the other is then $\left(\frac{\pi}{2}\right)^{2}$ ; two equal parts give less.

Ten questions from Inverse Trigonometric Functions, at your own pace.