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Question: The minimum value of (sin<sup>–1</sup>x)<sup>3</sup> + (cos<sup>–1</sup>x)<sup>3</sup> is equal to-...

The minimum value of (sin–1x)3 + (cos–1x)3 is equal to-

A

π332\frac{\pi^{3}}{32}

B

5π332\frac{5\pi^{3}}{32}

C

9π332\frac{9\pi^{3}}{32}

D

11π332\frac{11\pi^{3}}{32}

Answer

π332\frac{\pi^{3}}{32}

Explanation

Solution

Let I = (sin–1 x)3 + (cos–1 x)3

= (sin–1 x + cos–1 x) [(sin–1 x)2 + (cos–1 x)2 –(sin–1 x) (cos–1 x)]

= π2\frac { \pi } { 2 } [(sin⁡–1x+cos⁡−1x)2–3sin⁡–1x(π2−sin⁡–1x)]\left\lbrack (\sin^{–1}{}x + \cos^{- 1}x)^{2}–3\sin^{–1}x\left( \frac{\pi}{2} - \sin^{–1}x \right) \right\rbrack

= π2\frac { \pi } { 2 } [(π24−3π2sin⁡–1x +3(sin⁡−1x)2)]\left\lbrack \left( \frac{\pi^{2}}{4} - \frac{3\pi}{2}\sin^{–1}{}x\ + 3(\sin^{- 1}x)^{2} \right) \right\rbrack

= π2\frac { \pi } { 2 } [3(sin⁡−1x)2–π2(sin⁡−1x)+π216−π216)+π24]\left\lbrack 3\left( \sin^{- 1}{}x)^{2}–\frac{\pi}{2}(\sin^{- 1}x) + \frac{\pi^{2}}{16} - \frac{\pi^{2}}{16} \right) + \frac{\pi^{2}}{4} \right\rbrack

= π2\frac { \pi } { 2 } (π216+3(sin⁡−1x−π4)2)\left( \frac{\pi^{2}}{16} + 3\left( \sin^{- 1}{}x - \frac{\pi}{4} \right)^{2} \right)

Now, π4\frac{\pi}{4}³ sin–1 x –π4\frac{\pi}{4}³ –3π4\frac{3\pi}{4}

̃ 9π216\frac{9\pi^{2}}{16} ³ (sin⁡−1x−π4)2\left( \sin^{- 1}x - \frac{\pi}{4} \right)^{2}³ 0

̃ I ³ π2\frac{\pi}{2}· π216\frac{\pi^{2}}{16}= π332\frac{\pi^{3}}{32}

Hence (1) is correct answer.