Question

sin(2cos ^-1 (-3/5)) is equal to (a)6/25 (b)24/25 (c)4/5 (d) –24/25

Answer 1

We have if

$\cos \theta =x\\ \theta =\cos ^{-1} x\\ \cos 2\theta=2\cos^2 \theta -1\\ \cos 2\theta=2x ^2-1\\ 2\theta =\cos ^{-1} (2x ^2-1)\\ 2\cos ^{-1} x=\cos ^{-1} (2x ^2-1)$

Here $x=-\frac{3}{5}$

$2\cos ^{-1} (-\frac{3}{5})=\cos ^{-1} (2(-\frac{3}{5}) ^2-1)\\ 2\cos ^{-1} (-\frac{3}{5})=\cos ^{-1} (-\frac{7}{25}) \\ \sin [2\cos ^{-1} (-\frac{3}{5})]=\sin[\cos ^{-1} (-\frac{7}{25}) ]\\ \sin [2\cos ^{-1} (-\frac{3}{5})]=\sin[\sin ^{-1}(-\sqrt{1- (\frac{7}{25}) ^2})] \\ \sin [2\cos ^{-1} (-\frac{3}{5})]=\sin[\sin ^{-1}(-\sqrt{ (\frac{24}{25}) ^2})] \\ \sin [2\cos ^{-1} (-\frac{3}{5})]=\sin[\sin ^{-1}((-\frac{24}{25}) )] \\ \sin [2\cos ^{-1} (-\frac{3}{5})]=-\frac{24}{25}$

The correct choice is (d)