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Find the value of the following:\[cos^{-1} \frac {1} {2} +2 sin^{-1}\frac {1} {2} \]

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Ans : \( \frac{\pi}{3}+2.\frac{\pi}{6}=\frac{2\pi}{3} \)
answered Feb 22, 2013 by thanvigandhi_1
 
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Toolbox:
  • The range of the principal value of $\sin^{-1}x$ is $\left [ -\large\frac{\pi}{2}, \large\frac{\pi}{2} \right ]$
  • The range of the principal value of $\cos^{-1}x$ is $\left [ 0,\pi \right ]$
  • In such problems, we find the principal values of each of the inverse trignometric identities and arrive at the final answer.
Let $\sin^{-1}(\large\frac{1}{2}) = x$ $ \Rightarrow \sin x = \large\frac{1}{2}$
$\Rightarrow \sin x = \large\frac{1}{x} = \sin \large\frac{\pi}{6}$
 
\( \therefore\) $x =\large\frac{\pi}{6} \; \epsilon \; \left [ -\large\frac{\pi}{2}, \large\frac{\pi}{2} \right ]$
Let $\cos^{-1}(\large\frac{1}{2}) = y\: $ $ \Rightarrow\:  \cos y = \large\frac{1}{2}$
$\Rightarrow \cos y = \large\frac{1}{2} = \cos \large\frac{\pi}{3}$
 
\( \therefore\) $y = \large\frac{\pi}{3} \; \epsilon \; \left [ 0,\pi \right ]$
$\cos^{-1}(\large\frac{1}{2}) + 2 \sin^{-1}(\large\frac{1}{2}) = y + 2x$
$\Rightarrow \large\frac{\pi}{3} + 2\large\frac{\pi}{6} = \large\frac{\pi}{3} + \large\frac{\pi}{3} = 2 \large\frac{\pi}{3} $
\( \therefore\) $\cos^{-1}(\large\frac{1}{2}) + 2 \sin^{-1}(\large\frac{1}{2}) = 2 \large\frac{\pi}{3} $

 

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