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# What is the principal value of $\cos^{-1} \bigg( -\large\frac{\sqrt 3}{2} \bigg).$

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• $cos(\pi-\theta)=-cos\theta$
We know that $cos\frac{\pi}{6}=\frac{\sqrt3}{2}$
$\Rightarrow\:-\frac{\sqrt3}{2}=cos(\pi-\frac{\pi}{6})=cos\frac{5\pi}{6}$
$\Rightarrow\: \cos^{-1}-\frac{\sqrt3}{2}=cos^{-1}cos \bigg( \pi-\large\frac{\pi}{6} \bigg)=\large\frac{5\pi}{6}$
edited Mar 21, 2013