Question

Find the principal value of the following:
${{\cos }^{-1}}\left( \cos \dfrac{7\pi }{6} \right)$

Answer 1

In this type of question we have to use the concept of the principal value of a function. The principal value of a function is the value from the appropriate range that an inverse function returns. In case of inverse cosine functions the appropriate range in which the principal value lies is $0\text{ to }\pi$. In other words the range for inverse cosine function is $0\text{ to }\pi$. Here, as $\dfrac{7\pi }{6}$ does not lies between $0\text{ and }\pi$ we have to use $\cos \left( \pi +x \right)=\cos \left( \pi -x \right)$. First we express $\dfrac{7\pi }{6}$ in the form of $\left( \pi +x \right)$ and then we use the above rule to obtain the result.

Complete step by step answer:
Now we have to find the principal value of ${{\cos }^{-1}}\left( \cos \dfrac{7\pi }{6} \right)$.
We know that, $\dfrac{7\pi }{6}$ can be expressed as $\left( \pi +\dfrac{\pi }{6} \right)$. Hence, we can write
$\Rightarrow \cos \left( \dfrac{7\pi }{6} \right)=\cos \left( \pi +\dfrac{\pi }{6} \right)$
As we know that, $\cos \left( \pi +x \right)=\cos \left( \pi -x \right)$ we get,
$\Rightarrow \cos \left( \dfrac{7\pi }{6} \right)=\cos \left( \pi -\dfrac{\pi }{6} \right)$
$\Rightarrow \cos \left( \dfrac{7\pi }{6} \right)=\cos \left( \dfrac{5\pi }{6} \right)$
Hence, we can write the given function as
$\Rightarrow {{\cos }^{-1}}\left( \cos \left( \dfrac{7\pi }{6} \right) \right)={{\cos }^{-1}}\left( \cos \left( \dfrac{5\pi }{6} \right) \right)$
$\Rightarrow {{\cos }^{-1}}\left( \cos \left( \dfrac{7\pi }{6} \right) \right)=\dfrac{5\pi }{6}$ which the required principal value as it is between $0\text{ and }\pi$.
Thus the principal value of ${{\cos }^{-1}}\left( \cos \dfrac{7\pi }{6} \right)$ is $\dfrac{5\pi }{6}$ which lies between $0\text{ and }\pi$.

Note: In this type of question students may make mistakes and directly write the principal value as $\dfrac{7\pi }{6}$. Students have to note that ${{\cos }^{-1}}\left( \cos \dfrac{7\pi }{6} \right)\ne \dfrac{7\pi }{6}$ as $\dfrac{7\pi }{6}$ does not lies between $0\text{ and }\pi$. Also one of the students may find the principal value by using the formula $\cos x=\cos \left( 2\pi -x \right)$ as follows:
We know that $\dfrac{7\pi }{6}$ can be expressed as $\left( 2\pi -\dfrac{5\pi }{6} \right)$ hence we can write,
$\Rightarrow \cos \dfrac{7\pi }{6}=\cos \left( 2\pi -\dfrac{5\pi }{6} \right)$
$\Rightarrow {{\cos }^{-1}}\left( \cos \dfrac{7\pi }{6} \right)={{\cos }^{-1}}\left( \cos \left( 2\pi -\dfrac{5\pi }{6} \right) \right)$
$\Rightarrow {{\cos }^{-1}}\left( \cos \dfrac{7\pi }{6} \right)={{\cos }^{-1}}\left( \cos \left( \dfrac{5\pi }{6} \right) \right)$ as $\cos x=\cos \left( 2\pi -x \right)$
$\Rightarrow {{\cos }^{-1}}\left( \cos \dfrac{7\pi }{6} \right)=\dfrac{5\pi }{6}$ which the required principal value as it is between $0\text{ and }\pi$.
Thus the principal value of ${{\cos }^{-1}}\left( \cos \dfrac{7\pi }{6} \right)$ is $\dfrac{5\pi }{6}$.