Question

If n is an integer, prove that $\cos \left( n\pi +\theta \right)={{\left( -1 \right)}^{n}}\cos \theta$.

Answer 1

Hint: We will break the question into two cases, one for n = odd number and the other is for n = even number. And then we will look at in which quadrant cos is negative and positive. And with this much information we will prove that $\cos \left( n\pi +\theta \right)={{\left( -1 \right)}^{n}}\cos \theta $

Complete step-by-step solution -
Let’s solve this question.
We know that cos is positive in the $1^{st}$ and $4^{th}$ quadrant.
And cos is negative in the $2^{nd}$ and $3^{rd}$ quadrant.
Now if n is odd then $\cos \left( n\pi +\theta \right)$ will either lie in $2^{nd}$ or $3^{rd}$ quadrant.
And hence the value will be negative, and we will get the value of $\cos \left( n\pi +\theta \right)=-\cos \theta ..........(1)$
Now if the value of n is even then $\cos \left( n\pi +\theta \right)$ will either lie in $1^{st}$ or $4^{th}$ quadrant.
And hence the value will be positive, and we will get the value of $\cos \left( n\pi +\theta \right)=\cos \theta ...........(2)$
We know that the value of ${{\left( -1 \right)}^{n}}$ is 1 for n = even, and it is -1 for n = odd.
Therefore, from the equation (1) and (2) we can say that,
$\cos \left( n\pi +\theta \right)={{\left( -1 \right)}^{n}}\cos \theta $
Hence, we have proved the given statement.

Note: The student should know that in which quadrant the value of cos is positive and which quadrant it is negative. The idea of breaking the question into two parts and then proving them separately and after that again combining them to get the answer is very important.