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Mathematics Question on Inverse Trigonometric Functions

If
0<x<12 and sin⁡−1xα=cos⁡−1xβ0 < x< \frac{1}{\sqrt2}\ and\ \frac{\sin^{-1}x}{α} = \frac{\cos^{-1}x}{β}
then a value of
sin(2παα+β)sin(\frac{2πα}{α+β})
is

A

4(1−x2)(1−2x2)4\sqrt{(1-x^2)}(1-2x^2)

B

4x(1−x2)(1−2x2)4x\sqrt{(1-x^2)}(1-2x^2)

C

2x(1−x2)(1−4x2)2x\sqrt{(1-x^2)}(1-4x^2)

D

4(1−x2)(1−4x2)4\sqrt{(1-x^2)}(1-4x^2)

Answer

4x(1−x2)(1−2x2)4x\sqrt{(1-x^2)}(1-2x^2)

Explanation

Solution

The correct answer is (B) : 4x(1−x2)(1−2x2)4x\sqrt{(1-x^2)}(1-2x^2)
sin⁡−1xα=cos⁡−1xβ=k\frac{\sin^{-1}x}{α} =\frac{\cos^{-1}x}{β} = k
⇒sin⁡−1x+cos⁡−1x=k(α+β)⇒ \sin^{-1}x+\cos^{-1}x = k(α+β)
⇒α+β=π2k⇒ α+β = \frac{π}{2k}.
Now
2παα+β=2παπ2k\frac{2\pi \alpha}{\alpha + \beta} = \frac{2 \pi \alpha}{\frac{\pi}{2k} }
=4kα=4sin⁡−1x= 4k \alpha = 4\sin^{-1} x
sin⁡(2παα+β)=sin⁡(4sin⁡−1x)\sin(\frac{2\pi \alpha}{\alpha + \beta}) = \sin (4\sin^{-1}x)
Let
sin⁡−1x=θ\sin^{-1}x = \theta
∵x∈(0,12)\because x \in (0,\frac{1}{\sqrt2})
⇒θ∈(0,π4)⇒ \theta \in (0, \frac{\pi}{4})
⇒x=sin⁡θ⇒ x = \sin \theta
⇒cos⁡θ=1−x2⇒ \cos \theta = \sqrt{1-x^2}
⇒sin⁡2θ=2x.1−x2⇒ \sin2 \theta = 2x. \sqrt{1-x^2}
⇒cos⁡2θ=1−4x2(1−x2)⇒ \cos2 \theta = \sqrt{1-4x^2(1-x^2)}
=(2x2−1)2=1−2x2= \sqrt{(2x^2-1)^2}= 1-2x^2
∵cos2θ>0 as 2θ∈(0,π2)\because cos 2\theta > 0\ as\ 2\theta \in (0, \frac{\pi}{2})
⇒sin⁡4θ=2.2x1−x2(1−2x2)⇒ \sin 4 \theta = 2 . 2x\sqrt{1-x^2}(1-2x^2)
=4x1−x2(1−2x2)= 4x\sqrt{1-x^2}(1-2x^2)