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

1 Answer

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 by thanvigandhi_1
edited Nov 30, 2017 by meena.p
 

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