The value of cos^{-1}(cos frac{5pi}{3}) + sin | vedclass

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(A) We know that $\cos^{-1}(\cos x) = x$ for $x \in [0, \pi]$ and $\sin^{-1}(\sin x) = x$ for $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.First,simplify $

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(A) We know that $\cos^{-1}(\cos x) = x$ for $x \in [0, \pi]$ and $\sin^{-1}(\sin x) = x$ for $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.First,simplify $\cos^{-1}(\cos \frac{5\pi}{3})$:$\cos \frac{5\pi}{3} = \cos(2\pi - \frac{\pi}{3}) = \cos(\frac{\pi}{3})$.Since $\frac{\pi}{3} \in [0, \pi]$,$\cos^{-1}(\cos \frac{5\pi}{3}) = \frac{\pi}{3}$.Next,simplify $\sin^{-1}(\sin \frac{5\pi}{3})$:$\sin \frac{5\pi}{3} = \sin(2\pi - \frac{\pi}{3}) = \sin(-\frac{\pi}{3})$.Since $-\frac{\pi}{3} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,$\sin^{-1}(\sin \frac{5\pi}{3}) = -\frac{\pi}{3}$.Adding these values:$\frac{\pi}{3} + (-\frac{\pi}{3}) = 0$.

The value of $\cos^{-1}(\cos \frac{5\pi}{3}) + \sin^{-1}(\sin \frac{5\pi}{3})$ is

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