Show that $\cos \pi = -1$ How can I proof that $\cos \pi = -1$. I know that this is the answer if I type it in my calculator. If I draw the unit cirlce, then the answer is also clear, but is there an more mathematical way to show that $\cos \pi = -1$ ?
To add some more context: This is question from a highschool student that I teach. Normally, I try to learn here how she can prove to herself the tricks she has to memorize . And she likes that, but with trigonometry I'm not that skilled, and I have no idea how to proof that $\cos \pi = -1$.
 A: One possible way to define $\cos \alpha$ is to say that 

$\cos \alpha=$ the projection of a unit vector making an angle $\alpha$ with $x$-axis, on this same axis. 

Now if $\alpha=\pi$, the directions of our vector and $x$-axis are opposite, so that the projection is $-1$.
A: Take a virtual rope of length $\alpha$ and bind one of its ends on point $(1,0)$. Roll it counterclockwise around the unit circle and have a look at the other end. It will be located at point $\left(\cos\alpha,\sin\alpha\right)$. Doing this for $\alpha=\pi$ you will end up at point $(-1,0)$, so apparantly $\cos\pi=-1$ and $\sin\pi=0$. This works always. If $\alpha$ is negative you must roll it up clockwise. Realize here that the perimeter of the unit circle is $2\pi$. It is a real nice trick to memorize as you express it.
A: The definition of $\cos()$ (and $\sin()$) with the greatest scope and simplicity at the high school level is to start with the unit circle, parameterized by $\theta$ in the standard way. For $\theta\ge 0$ this is simply the arc length counter-clockwise from $(x,y)=(1,0)$. Then simply define:


*

*$\cos(\theta)=$ the x-coordinate of the point $\theta$ on the unit circle.

*$\sin(\theta)=$ the y-coordinate of the point $\theta$ on the unit circle.


Real-valued trigonometry is then more or less transparent from this perspective.
The circumference of the unit circle is $2\pi$. So when you go half way around starting at $(1,0)$, you end up at $(-1,0)$ having walked a circular distance of $\pi$. So $\cos(\pi)=-1$.
A: Can you put $B=A$ in 
$$\cos(A+B)=\cos A\cos B-\sin A\sin B$$
to find $$\cos2A=2\cos^2A-1$$ 
and  $\cos\frac\pi2=0$?
Or can you put $A=B=\frac\pi2$ in $$\cos(A+B)=\cos A\cos B-\sin A\sin B$$
A: You can use $$e^{i\pi}=-1$$ which gives you $\cos \pi =-1.$ 
A: \begin{equation*}
\cos(\pi-\theta) = -\cos \theta\
\end{equation*}
If $\theta=0$ then $\cos\pi=-\cos 0=-1$.
