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Write the Value of Cos\[\Left( 2 \Sin^{- 1} \Frac{1}{3} \Right)\] - Mathematics

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प्रश्न

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

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उत्तर

Let \[y = \sin^{- 1} \frac{1}{3}\]

Then, \[\sin{y} = \frac{1}{3}\]

Now,

 \[\cos{y} = \sqrt{1 - \sin^2 y}\]

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

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

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

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पाठ 4: Inverse Trigonometric Functions - Exercise 4.15 [पृष्ठ ११७]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 4 Inverse Trigonometric Functions
Exercise 4.15 | Q 15 | पृष्ठ ११७

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