Question

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

Answer 1

Hint: 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 \[\dfrac{5\pi }{6}\].