Question

How do you evaluate ${\cos ^{ - 1}}\left( {\cos \left( {\dfrac{{9\pi }}{8}} \right)} \right)$?

Answer 1

Hint : We know that ${\cos ^{ - 1}}\left( {\cos \left( \theta \right)} \right) = \theta$ for $0 \leqslant \theta \leqslant \pi$. Here $\theta$ is not in this range. So, we need to transform $\theta$ so that we can apply the formula. We know that cosine function is positive in the first and fourth quadrant and negative in the second and third quadrant. Also, we will be using $\cos \left( \theta \right) = \cos \left( {2\pi + \theta } \right) = \cos \left( {2\pi - \theta } \right)$ to transform the given expression.

Complete step-by-step answer :
We need to evaluate ${\cos ^{ - 1}}\left( {\cos \left( {\dfrac{{9\pi }}{8}} \right)} \right)$
We know, ${\cos ^{ - 1}}\left( {\cos \left( \theta \right)} \right) = \theta$ for $0 \leqslant \theta \leqslant \pi - - - - - - (1)$
Here our $\theta = \dfrac{{9\pi }}{8}$
Let us first see whether this $\theta$ lies in the range or not.
We have, $\theta = \dfrac{{9\pi }}{8}$ which can be written as
$\Rightarrow \theta = \dfrac{{\pi + 8\pi }}{8}$
Now, splitting the denominator, we get
$\Rightarrow \theta = \dfrac{\pi }{8} + \dfrac{{8\pi }}{8}$
Cancelling out the terms in numerator and denominator, we get
$\Rightarrow \theta = \dfrac{\pi }{8} + \pi$
Hence, we see that $\theta > \pi$ in this case.
So, we need to transform our angle.
We have ${\cos ^{ - 1}}\left( {\cos \left( {\dfrac{{9\pi }}{8}} \right)} \right)$
Writing the numerator of the angle as the difference of two terms, we can re-write this as:
$\Rightarrow {\cos ^{ - 1}}\left( {\cos \left( {\dfrac{{16\pi - 7\pi }}{8}} \right)} \right)$
Now, Splitting the denominator, we have
$\Rightarrow {\cos ^{ - 1}}\left( {\cos \left( {\dfrac{{16\pi }}{8} - \dfrac{{7\pi }}{8}} \right)} \right)$
Now, cancelling out the terms, we get
$\Rightarrow {\cos ^{ - 1}}\left( {\cos \left( {2\pi - \dfrac{{7\pi }}{8}} \right)} \right)$
Now, using $\cos \left( {2\pi - \theta } \right) = \cos \left( \theta \right)$, we get
$\Rightarrow {\cos ^{ - 1}}\left( {\cos \left( {\dfrac{{7\pi }}{8}} \right)} \right)$
Hence, we get
$\Rightarrow {\cos ^{ - 1}}\left( {\cos \left( {\dfrac{{9\pi }}{8}} \right)} \right) = {\cos ^{ - 1}}\left( {\cos \left( {\dfrac{{7\pi }}{8}} \right)} \right) - - - - - - (2)$
Let us now check whether $\dfrac{{7\pi }}{8} \in \left[ {0,\pi } \right]$ or not.
We have $\dfrac{{7\pi }}{8}$, which can be written as
$\Rightarrow \dfrac{{7\pi }}{8} = \dfrac{{8\pi - \pi }}{8}$
Splitting the denominator, we get
$\Rightarrow \dfrac{{7\pi }}{8} = \dfrac{{8\pi }}{8} - \dfrac{\pi }{8}$
Cancelling out the terms, we get
$\Rightarrow \dfrac{{7\pi }}{8} = \pi - \dfrac{\pi }{8}$, which is less than $\pi$.
Hence, $\dfrac{{7\pi }}{8} \in \left[ {0,\pi } \right]$
So, from (1) and (2), we can say that
$\Rightarrow {\cos ^{ - 1}}\left( {\cos \left( {\dfrac{{9\pi }}{8}} \right)} \right) = {\cos ^{ - 1}}\left( {\cos \left( {\dfrac{{7\pi }}{8}} \right)} \right) = \dfrac{{7\pi }}{8}$ as $\dfrac{{7\pi }}{8} \in \left[ {0,\pi } \right]$.
Hence, ${\cos ^{ - 1}}\left( {\cos \left( {\dfrac{{9\pi }}{8}} \right)} \right) = \dfrac{{7\pi }}{8}$
So, the correct answer is “$\dfrac{{7\pi }}{8}$”.

Note : We cannot just directly write that ${\cos ^{ - 1}}\left( {\cos \left( \theta \right)} \right) = \theta$ but we need to check the conditions for which the result holds. We could have wrote $\dfrac{{9\pi }}{8}$ as $\pi + \dfrac{\pi }{8}$ but this angle lies in third quadrant and we know that cosine is negative in third quadrant. So, to apply this we will have to again make changes as ${\cos ^{ - 1}}\left( { - x} \right) = \pi - {\cos ^{ - 1}}\left( x \right),x \in \left[ { - 1,1} \right]$.