Proof that $\cos(\pi/3)=1/2$
Is it possible to prove that $\cos(\pi/3)=1/2$ without using any trigonometric identities?
To be specific, my calculus-book simply states that with an angle of $\pi/3$ on the unit-circle the points (0,0) and (1,0) are the vertices of an equilateral triangle with edge length 1. But how can one reach such conclusion?
Thanks!
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$\begingroup$Start out with an equilateral triangle of side length $a$.
Then drop a perpendicular from one of the vertices to the side opposite to that vertex. By symmetry you have split the triangle into two equal right triangles. Now consider one of these right triangles, one of the angles is $\pi/3$. Use the definition of cosine (as base/hypotenuse) to get the value of $\cos \pi/3$.
$\endgroup$ 2 $\begingroup$Let $O=(0,0)$, $A=(1,0)$ and $P$ the point on the unit circle such that $\angle AOP=60°$. By construction you have $OA=OP=1$ so that $OAP$ is an isosceles triangle with base $AP$. The sum of the angles in $A$ and $P$ is $180°-60°=120°$ but as they are equal we find that $\angle PAO=\angle APO=\angle AOP=60°$ Hence $ABC$ is an equilateral triangle.
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