Question

How do you find the exact value of $\cos 5\pi$?

Answer 1

The above question is based on trigonometric functions which shows the relationship between the angles and sides of the triangle. The expression $\cos 5\pi$ is a cosine trigonometric function which can be solved by applying trigonometric formulas with the help of right-angle triangles.

Complete step by step solution:

Whenever we solve the trigonometric functions, we solve it with the help of a right-angle triangle.

In the right angle triangle, there are three sides which are named as hypotenuse, adjacent side and opposite side.

Below shown is the right-angle triangle,

The side which is opposite ${90^o}$ is called hypotenuse and the side opposite to the angle is opposite side and adjacent to the angle is adjacent side. So, the formula for cosine trigonometric function is the adjacent side divided by hypotenuse.

To solve the term $\cos 5\pi$,we will understand by considering a unit circle in a cartesian plane which is divided into four quadrants. In the below cartesian plane.

$\cos \pi$ here stands for cos${180^o}$.The first quadrant varies from 0 to 90 degree and second the quadrant varies from 90 to 180 degrees. Since we are considering a unit circle the value will be negative one.

Negative sign is used because $\pi$ lies on the negative x-axis.

$\cos \pi = - 1$

To find $\cos 5\pi$,the cosine function is repeated after every period of $\pi$i.e.

So, we can write it has

$\cos 5\pi = \cos (\pi + 4\pi ) = -\cos 4\pi$(trigonometric identities i.e. $cos(\pi+\theta) = -cos\theta)$

We know that,

$cosn\pi = 1$, when n is even.

$cosn\pi = -1$, when n is odd.

So, $cos 4\pi = 1$

Now, $\cos 5\pi = -\cos 4\pi = -1$

Therefore, $\cos 5\pi = - 1$ .

Note: In the cartesian plane, every trigonometric function has positive or negative values based on the quadrant which it is present in. For example, in the second quadrant the cosine function and secant function are both negative so this is why we get -1 for cos180 since it is in the second quadrant.