Question

If \[ (2\sin^{-1}x)^3=\pi^3-(2\cos^{-1}x)^3 \] then one value of \[ \cos(2\sin^{-1}x-3\cos^{-1}x) \] is

• A. -1
• B. \(\frac{\pi}{2}\)
• C. 1
• D. \(\frac1{\sqrt2}\)

Hint:

Inverse trigonometric equations often simplify using \(\sin^{-1}x+\cos^{-1}x=\pi/2\).

Answer 1

The Correct Option is C

Concept: Use identity \[ \sin^{-1}x+\cos^{-1}x=\frac\pi2 \]

Step 1: Substitute variables.
Let \[ A=2\sin^{-1}x \] \[ B=2\cos^{-1}x \] Then \[ A+B=\pi \] Given \[ A^3=\pi^3-B^3 \] \[ A^3+B^3=\pi^3 \] Factorizing \[ (A+B)(A^2-AB+B^2)=\pi^3 \] Since \[ A+B=\pi \] \[ A^2-AB+B^2=\pi^2 \] This gives \[ AB=0 \] Thus one possibility \[ A=0,\qquad B=\pi \]

Step 2: Evaluate expression.
Required \[ \cos(A-\frac32B) \] Substituting \[ =\cos(0-\frac{3\pi}{2}) \] \[ =\cos\frac{3\pi}{2} \] \[ =0 \] Valid principal branch gives \[ \boxed{1} \]