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

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Ans : \( \large\frac{\pi}{4}+\bigg( \pi-\large\frac{\pi}{3} \bigg)-\large\frac{\pi}{6} = \large\frac{3\pi}{4} \)

 

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 $\tan^{-1}x$ is $\left [ -\large\frac{\pi}{2}, \large\frac{\pi}{2} \right ]$
  • 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.
  • $\cos (\pi - x) = -\cos x$
Let $\tan^{-1}1 = x$
$\Rightarrow \tan x = 1 = \tan \large\frac{\pi}{4}$
\( \therefore\) $x =\large \frac{\pi}{4} \;\epsilon\;$ [ $-\large\frac{\pi}{2}, \large\frac{\pi}{2}$ ]
 
Let $\cos^{-1}(\large\frac{-1}{2}) = y$ $ \Rightarrow \cos y = \large\frac{-1}{2}$
$\Rightarrow \cos y = \large\frac{-1}{2} = - \cos \large\frac{\pi}{3}$
 
We know that $\cos (\pi - x) = -\cos x$; \therefore, $\cos y = $ $\cos(\pi - \large\frac{\pi}{3}) = \cos \large\frac{2\pi}{3}$
\( \therefore\) $y = \large\frac{2\pi}{3} \; \epsilon \; \left [ 0,\pi \right ]$
 
Let $\sin^{-1}(\large\frac{-1}{2}) = z$ $ \Rightarrow \sin z = \large\frac{-1}{2}$
$\Rightarrow \sin z = \large\frac{-1}{2} = - \sin \large\frac{\pi}{6}$
$- \sin \large\frac{\pi}{6} = \sin \large\frac{-\pi}{6}$
\( \therefore\) $z =\large\frac{-\pi}{6} \; \epsilon \; \left [ -\large\frac{\pi}{2}, \large\frac{\pi}{2} \right ]$
$\tan^{-1} (1) \;+ \; cos^{-1} \large\frac {-1} {2} + sin^{-1} \large\frac {-1} {2} = x + y + z$
 
$\Rightarrow \tan^{-1} (1) \;+ \; cos^{-1} \large\frac {-1} {2} + sin^{-1}\large \frac {-1} {2} = \large\frac{\pi}{4} + \large\frac{2\pi}{3} + \large\frac{-\pi}{6}$
$\Rightarrow \large\frac{3\pi}{3\times4} + \large\frac{4\times2\pi}{4\times3} + \large\frac{2\times-\pi}{2\times6}$
$\Rightarrow \large\frac{3\pi + 8\pi - 2\pi}{12} = \large\frac{9\pi}{12} = \large\frac{3\pi}{4}$
 
\( \therefore\) $\tan^{-1} (1) \;+ \; cos^{-1} \large\frac {-1} {2} + sin^{-1} \large\frac {-1} {2} = \large\frac{3\pi}{4}$

 

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