Question

Write the value of cos\[\left( 2 \sin^{- 1} \frac{1}{3} \right)\]

Answer 1

Let \[y = \sin^{- 1} \frac{1}{3}\]

Then, \[\sin{y} = \frac{1}{3}\]

Now,

 \[\cos{y} = \sqrt{1 - \sin^2 y}\]

\[\Rightarrow \cos{y} = \sqrt{1 - \frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}\]

\[\cos\left( 2 \sin^{- 1} \frac{1}{3} \right) = \cos(2y)\]
\[ = \cos^2 y - \sin^2 y \left[ \because \cos 2x = \cos^2 x - \sin^2 x \right]\]
\[ = \left( \frac{2\sqrt{2}}{3} \right)^2 - \left( \frac{1}{3} \right)^2 \]
\[ = \frac{8}{9} - \frac{1}{9}\]
\[ = \frac{7}{9}\]

∴ \[\cos\left( 2 \sin^{- 1} \frac{1}{3} \right) = \frac{7}{9}\]