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# Find the value of $\cos\bigg\{\cos^{-1}\bigg(\frac{-\sqrt {3}}{2}\bigg)+\frac{\pi}{6}\bigg\}$

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Toolbox:
• $cos(\pi-\theta)=-cos\theta$
• Principal interval of cos is $[0,\pi]$
• $cos\large\frac{\pi}{6}=\large\frac{\sqrt3}{2}$
• $cos\pi=-1$
By taking $\theta=\large\frac{\pi}{6}$ in the above formula we get
$-\large\frac{\sqrt3}{2}=-cos\large\frac{\pi}{6}=cos(\pi-\large\frac{\pi}{6})$
$cos^{-1}(-\large\frac{\sqrt3}{2})=cos^{-1}cos(\pi-\large\frac{\pi}{6})=\pi-\large\frac{\pi}{6}$

Substituting the values in the given expression, we get
$\Rightarrow\:cos\bigg(cos^{-1}\large\frac{-\sqrt3}{2}+\large\frac{\pi}{6}\bigg)= cos \bigg[ \bigg ( \pi-\large\frac{\pi}{6} \bigg) + \large\frac{\pi}{6} \bigg]$
$= cos\pi = -1$

answered Feb 20, 2013
edited Mar 19, 2013