Question

Why are $\cos \left(\mathrm{\pi }\right)$ and $\cos \left(-\mathrm{\pi }\right)$ both equal to $-1$?

Answer 1

Determine the value of $\cos \left(\mathrm{\pi }\right)$ and $\cos \left(-\mathrm{\pi }\right)$.

We know that $\cos \left(\mathrm{\pi }\right)=\cos \left(180°\right)$ and $\cos \left(0\right)=1$.

Also, $\cos \left(180°\right)$ lies in the second quadrant where the cosine function is negative.

$\begin{array}{rcl}\cos \left(180°\right)& =& \cos \left(180+0\right)\\ & \Rightarrow & \cos \left(180°\right)=-\cos \left(0\right)\\ \therefore \cos \left(180°\right)& =& -1\end{array}$

By using the trigonometry identity, $\cos \left(\theta \right)=\cos \left(-\theta \right)$.

Thus, $\cos \left(\mathrm{\pi }\right)=\cos \left(-\mathrm{\pi }\right)=-1$.

Hence, using the identity $\cos \left(\theta \right)=\cos \left(-\theta \right)$, the value of $\cos \left(\mathrm{\pi }\right)$ and $\cos \left(-\mathrm{\pi }\right)$ are equal to $-1$.