Evaluate: cos^{-1}(cos frac{4pi}{3}) + sin^{- | vedclass

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(C) Step $1$: Simplify $\cos^{-1}(\cos \frac{4\pi}{3})$. Since $\frac{4\pi}{3}$ is not in the range $[0, \pi]$, we write $\cos \frac{4\pi}{3} = \cos(2

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$\frac{\pi}{3}$

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(C) Step $1$: Simplify $\cos^{-1}(\cos \frac{4\pi}{3})$. Since $\frac{4\pi}{3}$ is not in the range $[0, \pi]$, we write $\cos \frac{4\pi}{3} = \cos(2\pi - \frac{2\pi}{3}) = \cos \frac{2\pi}{3}$. Thus, $\cos^{-1}(\cos \frac{2\pi}{3}) = \frac{2\pi}{3}$.Step $2$: Simplify $\sin^{-1}(\sin \frac{4\pi}{3})$. Since $\frac{4\pi}{3}$ is not in the range $[-\frac{\pi}{2}, \frac{\pi}{2}]$, we write $\sin \frac{4\pi}{3} = \sin(\pi + \frac{\pi}{3}) = -\sin \frac{\pi}{3} = \sin(-\frac{\pi}{3})$. Thus, $\sin^{-1}(\sin(-\frac{\pi}{3})) = -\frac{\pi}{3}$.Step $3$: Add the results: $\frac{2\pi}{3} + (-\frac{\pi}{3}) = \frac{\pi}{3}$.

Evaluate: $\cos^{-1}(\cos \frac{4\pi}{3}) + \sin^{-1}(\sin \frac{4\pi}{3}) = \dots$

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