Question

Solve: ${\cos ^{ - 1}}\left( {\cos \left( {\dfrac{{5\pi }}{4}} \right)} \right)$?

Answer 1

Hint: In order to find the solution of the given question, we must know the principal branches of all the inverse trigonometric functions. For ${\sin ^{ - 1}}$ function, the principal value branch is $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$.For ${\cos ^{ - 1}}$ function, the principal value branch is $\left[ {0,\pi } \right]$.For ${\tan ^{ - 1}}$ function, the principal value branch is $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$. We will first find the value of $\cos \left( {\dfrac{{5\pi }}{4}} \right)$ and then the required value of cosine inverse.

Complete step by step answer:
So, in the question, we have,
${\cos ^{ - 1}}\left( {\cos \left( {\dfrac{{5\pi }}{4}} \right)} \right)$.
We have to first find the value of $\cos \left( {\dfrac{{5\pi }}{4}} \right)$.
Now, we know the trigonometric formula $\cos \left( {\pi + x} \right) = - \cos x$. We first split the angle into two parts.
$ \Rightarrow \cos \left( {\dfrac{{5\pi }}{4}} \right) = \cos \left( {\pi + \dfrac{\pi }{4}} \right)$
$ \Rightarrow \cos \left( {\dfrac{{5\pi }}{4}} \right) = - \cos \left( {\dfrac{\pi }{4}} \right)$

Now, we know the value of cosine function for angle $\left( {\dfrac{\pi }{4}} \right)$.
$ \Rightarrow \cos \left( {\dfrac{{5\pi }}{4}} \right) = - \dfrac{1}{{\sqrt 2 }}$
So, now we have, ${\cos ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right)$
According to definition of inverse ratio,
If $\cos x = - \dfrac{1}{{\sqrt 2 }}$,
Then, ${\cos ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right) = x$ where the value of x lies in the range $\left[ {0,\pi } \right]$.

Now, we know that the cosine function is positive in the first and fourth quadrants and negative in the second and in the third quadrant.
So, the angle $x = {\cos ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right)$ must lie either in second quadrant or in third quadrant.
We know that the value of $\cos \left( {\dfrac{{3\pi }}{4}} \right)$ is $ - \dfrac{1}{{\sqrt 2 }}$.
Also, the principal value branch of cosine inverse function is $\left[ {0,\pi } \right]$.
So, $x = {\cos ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right) = \dfrac{{3\pi }}{4}$ .
Therefore, ${\cos ^{ - 1}}\left( {\cos \left( {\dfrac{{5\pi }}{4}} \right)} \right) = \dfrac{{3\pi }}{4}$.

Hence, the value of ${\cos ^{ - 1}}\left( {\cos \left( {\dfrac{{5\pi }}{4}} \right)} \right)$ is $\left( {\dfrac{{3\pi }}{4}} \right)$.

Note:The basic inverse trigonometric functions are used to find the missing angles in right triangles. While the regular trigonometric functions are used to determine the missing sides of the right-angled triangles, using the following formulae:
\[\sin \theta = \left( {\dfrac{{{\text{Opposite Side}}}}{{{\text{Hypotenuse}}}}} \right)\]
\[\Rightarrow \cos \theta = \left( {\dfrac{{{\text{Adjacent Side}}}}{{{\text{Hypotenuse}}}}} \right)\]
\[\Rightarrow \tan \theta = \left( {\dfrac{{{\text{Opposite Side}}}}{{{\text{Adjacent Side}}}}} \right)\]
Besides the trigonometric functions and inverse trigonometric functions, we also have some rules related to trigonometry such as the sine rule and cosine rule. According to the sine rule, the ratio of the sine of two angles is equal to the ratio of the lengths of the sides of the triangle opposite to both the angles. So, $\left( {\dfrac{{\sin A}}{{\sin B}}} \right) = \left( {\dfrac{a}{b}} \right)$.