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Mathematics Question on Trigonometric Identities

Let SS be the set of all solutions of the equation cos⁡−1(2x)−2cos⁡−1(1−x2)=π\cos ^{-1}(2 x)-2 \cos ^{-1}\left(\sqrt{1-x^2}\right)=\pi, x∈[−12,12]x \in\left[-\frac{1}{2}, \frac{1}{2}\right] Then ∑x∈S2sin⁡−1(x2−1)\displaystyle\sum_{x \in S} 2 \sin ^{-1}\left(x^2-1\right) is equal to

A

0

B

−2π3\frac{-2 \pi}{3}

C

π−2sin⁡−1(34)\pi-2 \sin ^{-1}\left(\frac{\sqrt{3}}{4}\right)

D

π−sin⁡−1(34)\pi-\sin ^{-1}\left(\frac{\sqrt{3}}{4}\right)

Answer

−2π3\frac{-2 \pi}{3}

Explanation

Solution

The correct answer is (B) : −2π3\frac{-2\pi}{3}
cos−1(2x)−2cos−11−x2=πcos^{−1}(2x)−2cos^{−1} \sqrt{1−x^{2}} =π
cos−1(2x)−cos−1(2(1−x2)−1)=πcos^{−1}(2x)−cos^{−1} (2(1−x^{2})-1) =π
cos−1(2x)−cos−1(1−2x2)=πcos^{−1}(2x)−cos^{−1} (1−2x^{2}) =π
−cos−1(1−2x2)=π−cos−1(2x)−cos^{−1} (1−2x^{2}) =π-cos^{−1}(2x)
Taking cos both sides we get
cos(−cos−1(1−2x2))=cos(π−cos−1(2x))cos(−cos^{−1} (1−2x^{2})) = cos(π-cos^{−1}(2x))
1−2x2=−2x1−2x^2=−2x
2x2−2x−1=02x^{2}−2x−1=0
On solving, x=1−32,1+32\text{On solving, } x= \frac{1− \sqrt3}{2}, \frac{1+ \sqrt3}{2}
As x=[−12,12],x=1+32=rejected\text{As }x=[\frac{−1}{2}, \frac{1}{2}],x= \frac{1+ \sqrt3}{2} = \text{rejected}
So x=1−32⇒x2−1=−32\text{So }x= \frac{1−\sqrt3}{2} ⇒x^2 −1=− \frac{\sqrt3}{2}
=2sin−1(x2−1)=2sin−1(−32)=−2π3=2sin^{−1}(x^{2}−1)=2sin^{−1}(\frac{−\sqrt{3}}{2})= \frac{−2π}{3}