Question

How do you solve $\cos \left( {\dfrac{{9\pi }}{4}} \right)$?

Answer 1

First of all, we will write the angle in the form of 2 pi + something and then, using this we get back into first quadrant with a smaller angle and thus get the required answer.

Complete step-by-step answer:
Here, we are given that we need to solve $\cos \left( {\dfrac{{9\pi }}{4}} \right)$.
The angle inside the cosine function is $\dfrac{{9\pi }}{4}$.
We can clearly see that we can write $\dfrac{{9\pi }}{4}$ as $2\pi + \dfrac{\pi }{4}$.
Therefore, we have $\cos \left( {\dfrac{{9\pi }}{4}} \right)$ equals to $\cos \left( {2\pi + \dfrac{\pi }{4}} \right)$.
Now, we know that the angle 2 pi means, we have covered the whole axis once and came back again in the first quadrant.
And, we also know that $\cos \left( {2\pi + \theta } \right) = \cos \theta$.
Therefore, we can write the following for sure:-
$\Rightarrow \cos \left( {\dfrac{{9\pi }}{4}} \right) = \cos \left( {2\pi + \dfrac{\pi }{4}} \right) = \cos \dfrac{\pi }{4}$
Now, we know that $\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$.

Hence, the answer is $\cos \left( {\dfrac{{9\pi }}{4}} \right) = \dfrac{1}{{\sqrt 2 }}$.

Note:
The students must commit to memory the following formula:-
$\cos \left( {2\pi + \theta } \right) = \cos \theta$
The students must note that if there angle is of type $n\pi + \theta$, where n is a positive integer, then cosine or sine or any angle remains cosine or sine of that angle respective but if the angle is of the form $\dfrac{{n\pi }}{2} + \theta$, where n is a positive integer, then sine of any angle changes into cosine and cosine of any angle changes in sine. Here, we have n = 2 in the first type of angle and thus, we get the cosine of the angle as cosine form only and since every trigonometric ratio is positive in the first quadrant and thus the answer is positive as well. Now, if it would have been in the fourth quadrant, it would still have been positive because cosine is either positive in the first or fourth quadrant.
In this you have basically first revolved around the all coordinate axis once and came back in the first quadrant again to get the required answer.