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

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 Nov 30, 2017 by meena.p
 
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