Evaluate: cos^{-1} left( cos frac{7pi}{6} rig | vedclass

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(B) The range of the principal value branch of $\cos^{-1} x$ is $[0, \pi]$.Since $\frac{7\pi}{6} > \pi$,we cannot directly write $\cos^{-1}(\cos 	het

Vedclass · Accepted Answer

$\frac{5\pi}{6}$

Vedclass · Answer

(B) The range of the principal value branch of $\cos^{-1} x$ is $[0, \pi]$.Since $\frac{7\pi}{6} > \pi$,we cannot directly write $\cos^{-1}(\cos 	heta) = 	heta$.We use the property $\cos(\pi + 	heta) = -\cos 	heta$ and $\cos^{-1}(-x) = \pi - \cos^{-1} x$.$\cos^{-1} \left( \cos \frac{7\pi}{6} ight) = \cos^{-1} \left( \cos \left( \pi + \frac{\pi}{6} ight) ight)$$= \cos^{-1} \left( -\cos \frac{\pi}{6} ight)$$= \pi - \cos^{-1} \left( \cos \frac{\pi}{6} ight)$$= \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.

Evaluate: $\cos^{-1} \left( \cos \frac{7\pi}{6} \right)$

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