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

To solve the given problem, let's start by understanding the equation provided:

\(\frac{\sin^{-1}x}{α} = \frac{\cos^{-1}x}{β} \)

We know that:

\(\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}\)

Using this identity, we can express one in terms of the other. Substituting for \(\sin^{-1}x\) in terms of \(\cos^{-1}x\) , we get:

\(\sin^{-1}x = \frac{\pi}{2} - \cos^{-1}x \)

Substitute the above into the original equation:

\(\frac{\frac{\pi}{2} - \cos^{-1}x}{α} = \frac{\cos^{-1}x}{β} \)

Multiply throughout by \(αβ\) to clear the denominators:

\(β(\frac{\pi}{2} - \cos^{-1}x) = α\cos^{-1}x \)

This simplifies to:

\(β\frac{\pi}{2} = (α + β)\cos^{-1}x \)

Therefore, we find:

\(\cos^{-1}x = \frac{β\frac{\pi}{2}}{α + β} \)

We know that:

\(\cos^{-1}x = θ \Rightarrow x = \cos θ\)

We need to calculate:

\(\sin(\frac{2πα}{α+β}) \)

Using \(\sin(2θ) = 2\sin θ \cos θ \):

Since \(2θ = \frac{2πα}{α+β}\), we use this to express the sine in terms of cosine, known as:

\(\sin(\frac{2πα}{α+β}) = 2 \sin(\frac{πα}{α+β}) \cos(\frac{πα}{α+β})\)

Rationalizing and calculating yields:

\(4x\sqrt{(1-x^2)}(1-2x^2)\),

Given options match this result. Thus, the correct answer is:

\(4x\sqrt{(1-x^2)}(1-2x^2)\)