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Inverse Trigonometric Functions

Question

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Evaluate: $\sin [\frac{π}{3} - \sin^{- 1} (\frac{1}{2})]$

Solution

We know that domain and range of the principal value branch of function sin^-1 is defined as:

$\sin^{- 1} : [- 1, 1] \to [- \frac{π}{2}, \frac{π}{2}] ∴ \sin [\frac{π}{3} - \sin^{- 1} (\frac{- 1}{2})] = \sin [\frac{π}{3} - (- \frac{π}{6})] = \sin [\frac{π}{3} + \frac{π}{6}] = \sin [\frac{π}{2}] = 1$

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Inverse Trigonometric Functions

Evaluate: $\sin [\frac{π}{3} - \sin^{- 1} (\frac{1}{2})]$

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Inverse Trigonometric Functions

Evaluate: sinπ3 - sin-1 12

Some More Questions From Inverse Trigonometric Functions Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Evaluate: $\sin [\frac{π}{3} - \sin^{- 1} (\frac{1}{2})]$