The principal value of cos^{-1}[cos(-680^{cir | vedclass

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Question

(C) We know that $\cos(-	heta) = \cos(	heta)$. Therefore,$\cos(-680^{\circ}) = \cos(680^{\circ})$.We can express $680^{\circ}$ as $720^{\circ} - 40^

Vedclass · Accepted Answer

$\frac{2\pi}{9}$

Vedclass · Answer

(C) We know that $\cos(-	heta) = \cos(	heta)$. Therefore,$\cos(-680^{\circ}) = \cos(680^{\circ})$.We can express $680^{\circ}$ as $720^{\circ} - 40^{\circ}$,which is $4\pi - 40^{\circ}$.Since $\cos(2n\pi - 	heta) = \cos(	heta)$,we have $\cos(680^{\circ}) = \cos(40^{\circ})$.The principal value branch of $\cos^{-1}(x)$ is $[0, \pi]$.Thus,$\cos^{-1}[\cos(40^{\circ})] = 40^{\circ}$.Converting $40^{\circ}$ to radians: $40 	imes \frac{\pi}{180} = \frac{4\pi}{18} = \frac{2\pi}{9}$.Therefore,the correct option is $C$.

The principal value of $\cos^{-1}[\cos(-680^{\circ})]$ is equal to: . . . . . . .

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