Evaluate cos$3\pi$.

Hint: We start to solve the problem by writing $\cos 3\pi $ as $cos\left( 2\pi +\pi \right)$ .We simplify $cos\left( 2\pi +\pi \right)$ by using $cos\left( A+B \right)=cosAcosB-sinAsinB$ formula in trigonometry. We need to substitute the values of required trigonometric functions at angles $\pi $ and $2\pi $ to get the value of $\cos 3\pi $

Complete step by step answer:
In the given question,
We need to evaluate $\cos 3\pi$
$\cos 3\pi$ can be also written as $\cos \left( \pi +2\pi \right)$
The expression $\cos \left( \pi +2\pi \right)$ resembles the expression $cos\left( A+B \right)$
The formula of $cos\left( A+B \right)$ is given by
$cos\left( A+B \right)=cosAcosB-sinAsinB$
Upon substituting we get,
$\Rightarrow cos\left( 2\pi +\pi \right)=cos2\pi .cos\pi - sin2\pi .sin\pi$
Let us now find the value of $\cos \pi$
$\Rightarrow cos\pi =cos\left( \dfrac{\pi }{2}+\dfrac{\pi }{2} \right)$
The value of $cos\left( \dfrac{\pi }{2}+\dfrac{\pi }{2} \right)$ falls into the second quadrant. The value of cosine function in the second quadrant is negative.
$\Rightarrow cos\left( 90+x \right)=-sinx$
$\Rightarrow cos\pi =-sin\left( \dfrac{\pi }{2} \right)$
$\Rightarrow cos\pi =-1$
Let us now find the value of $\cos 2\pi$
$\Rightarrow cos2\pi =cos\left( \pi +\pi \right)$
The value of $cos\left( \pi +\pi \right)$ falls into the third quadrant. The value of cosine function in the third quadrant is negative.
$\Rightarrow cos2\pi =-cos\pi$
$\Rightarrow cos2\pi =-\left( -1 \right)=1$
Let us now find the value of $\sin \pi$
$\Rightarrow \sin \pi =\sin \left( \dfrac{\pi }{2}+\dfrac{\pi }{2} \right)$
The value of $\sin \left( \dfrac{\pi }{2}+\dfrac{\pi }{2} \right)$ falls into the second quadrant. The value of cosine function in the second quadrant is positive
$\Rightarrow \sin \left( 90+x \right)=\cos x$
$\Rightarrow \sin \pi =\cos \left( \dfrac{\pi }{2} \right)$
$\Rightarrow \sin \pi =0$
Let us now find the value of $\sin 2\pi$
$\Rightarrow \sin 2\pi =\sin \left( \pi +\pi \right)$
The value of $\sin \left( \pi +\pi \right)$ falls into the third quadrant. The value of cosine function in the third quadrant is negative.
$\Rightarrow \sin 2\pi =-\sin \pi$
$\Rightarrow \sin 2\pi =-\left( -0 \right)=0$
Substituting the above values in $cos\left( 2\pi +\pi \right)$ , we get,
$\Rightarrow cos\left( 2\pi +\pi \right)=cos2\pi .cos\pi - sin2\pi .sin\pi$
$\Rightarrow cos\left( 2\pi +\pi \right)=\left( 1 \right)\times \left( -1 \right) - 0\times 0$
$\Rightarrow cos\left( 2\pi +\pi \right)=-1$

Hence, the value of $\cos 3\pi$ is equal to -1.

Note: The given question can also be solved using the periodicity of the cosine function. The period of the cosine function is $\pi$ and the values oscillate between +1 and -1 for each period. The graph of cos function at $3\pi$ is below x-axis so the value of $cos\left( 3\pi \right)=-1.$
Must check where the Trigonometric functions become negative in which Quadrant to easily find the values in the given range. Convert the entire Trigonometric equation in either $\cos x$ or $\sin x$ to easily solve the entire expression.