Write the value of sin $[\frac{π}{3} - \sin^{- 1} (- \frac{1}{2})]$

Solution

$\sin [\frac{π}{3} - \sin^{- 1} (\frac{- 1}{2})] Let \sin^{- 1} (\frac{- 1}{2}) = x \Rightarrow (\frac{- 1}{2}) = sinx \Rightarrow sinx = - \sin \frac{π}{6} = \sin (- \frac{π}{6}) = \sin (2 π - \frac{π}{6}) \Rightarrow x = 2 π - \frac{π}{6}$

$∴ \sin [\frac{π}{3} - \sin^{- 1} (\frac{- 1}{2})] = \sin [\frac{π}{3} - (2 π - \frac{π}{6})] = \sin [- \frac{9 π}{6}] = - \sin [\frac{3 π}{2}] = - \sin [π + \frac{π}{2}] = - [- \sin \frac{π}{2}] = + \sin \frac{π}{2} = 1 Thus, \sin [\frac{π}{3} - \sin^{- 1} (\frac{- 1}{2})] = 1$

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Inverse Trigonometric Functions

Write the value of sin $[\frac{π}{3} - \sin^{- 1} (- \frac{1}{2})]$

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Write the value of sin $[\frac{π}{3} - \sin^{- 1} (- \frac{1}{2})]$