Question

What are the exact values of $\cos \left( {\dfrac{{3\pi }}{4}} \right)$ radians and $\sin \left( {\dfrac{{3\pi }}{4}} \right)$ radians?

Answer 1

To find the exact values of $\cos \left( {\dfrac{{3\pi }}{4}} \right)$ radians and $\sin \left( {\dfrac{{3\pi }}{4}} \right)$ radians, we will use the concept of reference angle. Using the concept of reference angle, we will write $\cos \left( {\dfrac{{3\pi }}{4}} \right) = \cos \left( {\pi - \dfrac{\pi }{4}} \right)$ and $\sin \left( {\dfrac{{3\pi }}{4}} \right) = \sin \left( {\pi - \dfrac{\pi }{4}} \right)$. Then using the value of standard angles, we will find the value of $\cos \left( {\dfrac{{3\pi }}{4}} \right)$ radians and $\sin \left( {\dfrac{{3\pi }}{4}} \right)$ radians.

Complete step by step answer:
There are six functions of an angle commonly used in trigonometry namely sine, cosine, tangent, cotangent, secant and cosecant. According to the question find the exact values of $\cos \left( {\dfrac{{3\pi }}{4}} \right)$ radians and $\sin \left( {\dfrac{{3\pi }}{4}} \right)$ radians. As we know that the reference angle is the acute angle with the x-axis. Using the concept of reference angle, we will find the exact values of $\cos \left( {\dfrac{{3\pi }}{4}} \right)$ radians and $\sin \left( {\dfrac{{3\pi }}{4}} \right)$ radians.

Let us consider the original angle is given by $\theta$ and the auxiliary value is given by $\alpha$.
For the first quadrant, we have $\theta = \alpha$.
For the second quadrant, we have $\theta = \pi - \alpha$.
For the third quadrant, we have $\theta = \pi + \alpha$.
For the fourth quadrant, we have $\theta = 2\pi - \alpha$.
Consider $\cos \left( {\dfrac{{3\pi }}{4}} \right)$. $\cos \left( {\dfrac{{3\pi }}{4}} \right)$ is in the second quadrant.
Therefore, $\cos \left( {\dfrac{{3\pi }}{4}} \right) = \cos \left( {\pi - \dfrac{\pi }{4}} \right)$

In the second quadrant, $\cos$ is negative. So,
$\Rightarrow \cos \left( {\pi - \dfrac{\pi }{4}} \right) = - \cos \left( {\dfrac{\pi }{4}} \right)$
$\therefore \cos \left( {\dfrac{{3\pi }}{4}} \right) = - \dfrac{1}{{\sqrt 2 }}$
Now, consider $\sin \left( {\dfrac{{3\pi }}{4}} \right)$. $\sin \left( {\dfrac{{3\pi }}{4}} \right)$ lies in the second quadrant.
Therefore, $\sin \left( {\dfrac{{3\pi }}{4}} \right) = \sin \left( {\pi - \dfrac{\pi }{4}} \right)$.
In the second quadrant, $\sin$ is positive.
So,
$\Rightarrow \sin \left( {\pi - \dfrac{\pi }{4}} \right) = \sin \left( {\dfrac{\pi }{4}} \right)$
$\therefore \sin \left( {\dfrac{{3\pi }}{4}} \right) = \dfrac{1}{{\sqrt 2 }}$

Therefore, the exact values of $\cos \left( {\dfrac{{3\pi }}{4}} \right)$ radians and $\sin \left( {\dfrac{{3\pi }}{4}} \right)$ radians are $- \dfrac{1}{{\sqrt 2 }}$ and $\dfrac{1}{{\sqrt 2 }}$ respectively.

Note: In the first quadrant, all trigonometric functions are positive. In the second quadrant, $\sin$ and $\cos ec$ are positive. In the third quadrant, $\tan$ and $\cot$ are positive. In the fourth quadrant, $\cos$ and $\sec$ are positive. Also, note that here we have used values of some standard angles i.e., $\sin \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}$ and $\cos \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}$.