Question

If
\(0 < x< \frac{1}{\sqrt2}\ and\ \frac{\sin^{-1}x}{α} = \frac{\cos^{-1}x}{β} \)
then a value of 
\(sin(\frac{2πα}{α+β}) \)
is

• A. \(4\sqrt{(1-x^2)}(1-2x^2)\)
• B. \(4x\sqrt{(1-x^2)}(1-2x^2)\)
• C. \(2x\sqrt{(1-x^2)}(1-4x^2)\)
• D. \(4\sqrt{(1-x^2)}(1-4x^2)\)

Answer 1

The Correct Option is B

The correct answer is (B) : \(4x\sqrt{(1-x^2)}(1-2x^2)\)
\(\frac{\sin^{-1}x}{α} =\frac{\cos^{-1}x}{β} = k\) 
\(⇒ \sin^{-1}x+\cos^{-1}x = k(α+β)\)
\(⇒ α+β = \frac{π}{2k}\). 
Now
\(\frac{2\pi \alpha}{\alpha + \beta} = \frac{2 \pi \alpha}{\frac{\pi}{2k} }\)
\(= 4k \alpha = 4\sin^{-1} x\)
\(\sin(\frac{2\pi \alpha}{\alpha + \beta}) = \sin (4\sin^{-1}x)\)
Let
\(\sin^{-1}x = \theta \)
\(\because x \in (0,\frac{1}{\sqrt2})\)
\(⇒ \theta \in (0, \frac{\pi}{4})\)
\(⇒ x = \sin \theta\)
\(⇒ \cos \theta = \sqrt{1-x^2}\)
\(⇒ \sin2 \theta = 2x. \sqrt{1-x^2}\)
\(⇒ \cos2 \theta = \sqrt{1-4x^2(1-x^2)}\)
\(= \sqrt{(2x^2-1)^2}= 1-2x^2\)
\(\because cos 2\theta > 0\ as\ 2\theta \in (0, \frac{\pi}{2})\)
\(⇒ \sin 4 \theta = 2 . 2x\sqrt{1-x^2}(1-2x^2)\)
\(= 4x\sqrt{1-x^2}(1-2x^2)\)