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Evaluate \(cos\bigg(cos^{-1}\frac{-\sqrt{3}}{2}+\frac{\pi}{6}\bigg)\)

1 Answer

Toolbox:
  • \(-cos\theta=cos(\pi-\theta)\)
  • Principal interval of cos is \([0,\pi]\)
  • \(cos\large\frac{\pi}{6}=\large\frac{\sqrt{3}}{2}\)
  • \(cos\pi=-1\)
\(-\large\frac{\sqrt{3}}{2}=-cos\large\frac{\pi}{6}\)
By taking \(\theta=\large\frac{\pi}{6},\:-cos\large\frac{\pi}{6}=cos(\pi-\large\frac{\pi}{6})\)
\(cos^{-1}-\large\frac{\sqrt{3}}{2}=cos^{-1}(-cos\large\frac{\pi}{6})=cos^{-1}(cos(\pi-\large\frac{\pi}{6}))\)
\(=(\pi-\large\frac{\pi}{6})\)
 
The given expression becomes
\( cos\big(cos^{-1}(-\large\frac{\sqrt{3}}{2})+\large\frac{\pi}{6}\big)=cos( \pi-\large\frac{\pi}{6}+\large\frac{\pi}{6})\)
\( = cos\pi=-1\)

 

answered Mar 2, 2013 by thanvigandhi_1
edited Mar 19, 2013 by thanvigandhi_1
 
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