3

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 $\cos \pi = -1$.

90intuition
  • 2,622
  • 5
    How do you define $\cos x$? – Michael Albanese Oct 04 '13 at 18:43
  • 2
    And, a bit more trickily, how do you define $\pi$? – dfeuer Oct 04 '13 at 18:46
  • It depends on what definition of the cosine function you use. Sometimes it's defined by a power series and sometimes it's defined as the solution of the differential equation $y''=-y$ satisfying the initial conditions $y(0)=1$ and $y'(0)=0$, and sometimes it's defined in other ways. And there's also the question of which definition of $\pi$ you're using. – Michael Hardy Oct 04 '13 at 18:47
  • See also http://math.stackexchange.com/questions/63102/how-to-prove-periodicity-of-sinx-or-cosx-starting-from-the-taylor-seri – dfeuer Oct 04 '13 at 18:56

6 Answers6

9

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 $-1$.

Start wearing purple
  • 53,234
  • 13
  • 164
  • 223
3

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.

drhab
  • 151,093
2

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 $\cos(\pi)=-1$.

Matt
  • 730
  • 5
  • 14
1

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 $$\cos(A+B)=\cos A\cos B-\sin A\sin B$$

lab bhattacharjee
  • 274,582
1

You can use $$e^{i\pi}=-1$$ which gives you $\cos \pi =-1.$

Shobhit
  • 6,902
1

\begin{equation*} \cos(\pi-\theta) = -\cos \theta\ \end{equation*}

If $\theta=0$ then $\cos\pi=-\cos 0=-1$.

kiss my armpit
  • 3,896