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Find the principal values of the following: \[ cos^{-1} \bigg(\large\frac { \sqrt 3} {2}\bigg) \]

$\begin{array}{1 1} -\frac{\pi}{6} \\ \frac{\pi}{6} \\ \frac{\pi}{\sqrt 3} \\ \frac{\pi}{2} \end{array} $

2 Answers

Ans : \( cos^{-1} \bigg( cos\large\frac{\pi}{6} \bigg) = \large\frac{\pi}{6} \)

 

answered Feb 22, 2013 by thanvigandhi_1
edited Mar 15, 2013 by thanvigandhi_1
 
Toolbox:
  • The range of the principal value of $\cos^{-1}x$ is $\left [ 0,\pi \right ]$
Let $\cos^{-1}(\large\frac{\sqrt 3}{2}) = x$ $ \Rightarrow \cos x = \large\frac{\sqrt 3}{2}$
We know that the range of the principal value of $\cos^{-1}x$ is $\left [ 0,\pi \right ]$
Therefore, $\cos x = \large\frac{\sqrt 3}{2} = \cos \large\frac{\pi}{6}$
$\Rightarrow x = \large\frac{\pi}{6}$, where $x \;\epsilon \; \left [ 0,\pi \right ]$
Therefore, the principal value of $\cos^{-1}(\large\frac{\sqrt 3}{2})$ is $\large\frac{\pi}{6}$

 

answered Mar 2, 2013 by balaji.thirumalai
edited Mar 15, 2013 by thanvigandhi_1
 
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