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Evaluate Sin \[\Left( \Frac{1}{2} \Cos^{- 1} \Frac{4}{5} \Right)\] - Mathematics

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प्रश्न

Evaluate sin

\[\left( \frac{1}{2} \cos^{- 1} \frac{4}{5} \right)\]

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उत्तर

We know that

\[\cos^{- 1} x = 2 \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}}\]
\[ \tan^{- 1} x = \sin^{- 1} \frac{x}{\sqrt{1 + x^2}}\]

\[\therefore \sin\left( \frac{1}{2} \cos^{- 1} \frac{4}{5} \right) = \sin\left( \frac{1}{2}2 \tan^{- 1} \sqrt{\frac{1 - \frac{4}{5}}{1 + \frac{4}{5}}} \right)\]
\[ = \sin\left( \tan^{- 1} \sqrt{\frac{\frac{1}{5}}{\frac{9}{5}}} \right)\]
\[ = \sin\left( \tan^{- 1} \frac{1}{3} \right)\]
\[ = \sin\left\{ \sin^{- 1} \left( \frac{\frac{1}{3}}{\sqrt{1 + \frac{1}{9}}} \right) \right\}\]
\[ = \sin\left( \sin^{- 1} \frac{1}{\sqrt{10}} \right)\]
\[ = \frac{1}{\sqrt{10}} \left[ \because \sin\left( \sin^{- 1} x \right) = x \right]\]

∴ \[\sin\left( \frac{1}{2} \cos^{- 1} \frac{4}{5} \right) = \frac{1}{\sqrt{10}}\]

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

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 4 Inverse Trigonometric Functions
Exercise 4.15 | Q 17 | पृष्ठ ११७

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