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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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  • \(cos(\pi-\theta)=-cos\theta\)
  • 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 by thanvigandhi_1
edited Mar 19, 2013 by thanvigandhi_1
 

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