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

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

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Ans : \( cos^{-1} \bigg( cos \bigg( \pi - \large\frac{\pi}{3} \bigg) \bigg)=2\large\frac{\pi}{3}\)

 

answered Feb 22, 2013 by thanvigandhi_1
edited Mar 15, 2013 by thanvigandhi_1
 
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Toolbox:
  • The range of the principal value of $\cos^{-1}x$ is $\left [ 0,\pi \right ]$
  • $\cos (\pi - x) = -cos\; x$
Let $\cos^{-1}(\large\frac{-1}{2}) = x$ $ \Rightarrow \cos x = \large\frac{-1}{2}$
We know that the range of the principal value of $\cos^{-1}x$ is $\left [ 0,\pi \right ]$
Therefore, $\cos x = \large\frac{-1}{2} = -\cos \large\frac{\pi}{3}$
Because $\cos (\pi - x) = -cos\; x$, $\cos x = cos (\pi - \large\frac{\pi}{3}) = \cos \large\frac{2\pi}{3}$
$\Rightarrow x = \large\frac{2 \pi}{3}$, where $x \;\epsilon \; \left [ 0,\pi \right ]$
Therefore, the principal value of $\cos^{-1} (\large\frac{-1}{2})$ is $\large\frac{2\pi}{3}$
answered Mar 2, 2013 by balaji.thirumalai
edited Apr 18 by meena.p
 
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