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# The principal value of $\cos^{-1}\bigg(-\Large {\frac{1}{2}}\normalsize \bigg)$ is_____________.

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

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## 1 Answer

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Toolbox:
• The range of the principal value of $\cos^{-1}x$ is $\left [ 0,\pi \right ]$
• $\cos (\pi - x) = -cos\; x$
Ans : $cos^{-1} \bigg( cos \bigg( \pi - \large\frac{\pi}{3} \bigg) \bigg)=2\large\frac{\pi}{3}$

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 Feb 18, 2013
edited Mar 16, 2013

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