Inverse Trigonometric Functions Question 7
Question 7 - 01 February - Shift 1
Let $S$ be the set of all solutions of the equation $\cos ^{-1}(2 x)-2 \cos ^{-1}(\sqrt{1-x^{2}})=\pi, \quad x \in[-\frac{1}{2}, \frac{1}{2}]$. Then $\sum _{x \in S} 2 \sin ^{-1}(x^{2}-1)$ is equal to
(1) 0
(2) $\frac{-2 \pi}{3}$
(3) $\pi-\sin ^{-1}(\frac{\sqrt{3}}{4})$
(4) $\pi-2 \sin ^{-1}(\frac{\sqrt{3}}{4})$
Show Answer
Answer: (2)
Solution:
Formula: Identity of double angle of sine and cosine inverse function.
$\cos ^{-1}(2 x)-2 \cos ^{-1} \sqrt{1-x^{2}}=\pi$
$\cos ^{-1}(2 x)-\cos ^{-1}(2(1-x^{2})-1)=\pi$
$\cos ^{-1}(2 x)-\cos ^{-1}(1-2 x^{2})=\pi$
$-\cos ^{-1}(1-2 x^{2})=\pi-\cos ^{-1}(2 x)$
Taking cos both sides we get
$\cos (-\cos ^{-1}(1-2 x^{2}))=\cos (\pi-\cos ^{-1}(2 x))$
$1-2 x^{2}=-2 x$
$2 x^{2}-2 x-1=0$
On solving, $x=\frac{1-\sqrt{3}}{2}, \frac{1+\sqrt{3}}{2}$
As $x=[-1 / 2,1 / 2], x=\frac{1+\sqrt{3}}{2}=$ rejected
So $x=\frac{1-\sqrt{3}}{2} \Rightarrow x^{2}-1=-\sqrt{3} / 2$
$=2 \sin ^{-1}(x^{2}-1)=2 \sin ^{-1}(\frac{-\sqrt{3}}{2})=\frac{-2 \pi}{3}$
