We prove here that the sine function sin (-x) = - sin x is odd using the unit circle.

We start with the following configuration:

  • unit circle $\mathcal{C}(O,R=1)$
  • definition of the angle $x$
  • definition of the angle $-x$

Now consider the triangles: $(OA_xA)$ and $(OA’_xA’)$.

Proof that sine is an odd function sin(-x) = -sin (x)

Take the definition of the sines of the angles $x$ and $-x$.

In the triangle $(OA_xA)$ :

\[\sin x=\frac{\textrm{opposite}}{\textrm{hypotenuse}}=\frac{\vert OA_y\vert }{R}=\frac{\vert OA_y\vert }{1}=\vert OA_y\vert\]

In the triangle $(OA’_xA’)$:

\[\sin (-x)=\frac{\textrm{opposite}}{\textrm{hypotenuse}}=\frac{\vert OA'_y\vert }{R}=\frac{\vert OA'_y\vert }{1}=\vert OA'_y\vert\]

By construction: $\vert OA_y\vert = -\vert OA’_y\vert $, then we have

\[\forall x\in \mathbb{R},\quad: \sin (-x)=-\sin x\]