If 0<x<12024 and sin−1xα=cos−1xβ, then a value of sin[2παα+β] is

sin−1xα=cos−1xβ=k⇒α=sin−1xK,β=cos−1xK sin(2π αα+β)=sin(2πsin−1xsin−1x+cos−1x)=sin(2πsin−1x(π2)) =sin(4sin−1x)=sin(2(2sin−1x)) =sin(2(sin−1(2x1−x2)))=2(2x1−x2)1−(2x1−x2)2 =4x1−x21−4x2+4x4=4x1−x2(1−2x2)

If 0<x<12024 and sin−1xα=cos−1xβ, then a value of sin[2παα+β] is

sin−1xα=cos−1xβ=k⇒α=sin−1xK,β=cos−1xK sin(2π αα+β)=sin(2πsin−1xsin−1x+cos−1x)=sin(2πsin−1x(π2)) =sin(4sin−1x)=sin(2(2sin−1x)) =sin(2(sin−1(2x1−x2)))=2(2x1−x2)1−(2x1−x2)2 =4x1−x21−4x2+4x4=4x1−x2(1−2x2)

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If $0 < x < \frac{1}{\sqrt{2024}}$ and $\frac{\sin^{- 1} x}{α} = \frac{\cos^{- 1} x}{β},$ then a value of $\sin [\frac{2 π α}{α + β}]$ is

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

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

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

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

answer is B.

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Detailed Solution

$\frac{\sin^{- 1} x}{α} = \frac{\cos^{- 1} x}{β} = k \Rightarrow α = \frac{\sin^{- 1} x}{K}, β = \frac{\cos^{- 1} x}{K} \sin (\frac{2 π α}{α + β}) = \sin (\frac{2 π \sin^{- 1} x}{\sin^{- 1} x + \cos^{- 1} x}) = \sin (\frac{2 π \sin^{- 1} x}{(\frac{π}{2})}) = \sin (4 \sin^{- 1} x) = \sin (2 (2 \sin^{- 1} x)) = s i n (2 (\sin^{- 1} (2 x \sqrt{1 - x^{2}}))) = 2 (2 x \sqrt{1 - x^{2}}) \sqrt{1 - {(2 x \sqrt{1 - x^{2}})}^{2}} = 4 x \sqrt{1 - x^{2}} \sqrt{1 - 4 x^{2} + 4 x^{4}} = 4 x \sqrt{1 - x^{2}} (1 - 2 x^{2})$