The value of cos^{-1}(frac{1}{2}) + 2sin^{-1} | vedclass

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Question

(D) We know that $\cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$ because $\cos(\frac{\pi}{3}) = \frac{1}{2}$.We also know that $\sin^{-1}(\frac{1}{2}) = \fra

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$\frac{2\pi}{3}$

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(D) We know that $\cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$ because $\cos(\frac{\pi}{3}) = \frac{1}{2}$.We also know that $\sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}$ because $\sin(\frac{\pi}{6}) = \frac{1}{2}$.Substituting these values into the expression:$\cos^{-1}(\frac{1}{2}) + 2\sin^{-1}(\frac{1}{2}) = \frac{\pi}{3} + 2(\frac{\pi}{6})$$= \frac{\pi}{3} + \frac{\pi}{3}$$= \frac{2\pi}{3}$.

The value of $\cos^{-1}(\frac{1}{2}) + 2\sin^{-1}(\frac{1}{2})$ is equal to

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