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Sine is an odd function sin(-x)=-sin x

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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 $(OA_xA)$ :

$$ \sin x=\frac{\textrm{opposite}}{\textrm{hypotenuse}}=\frac{|OA_y|}{R}=\frac{|OA_y|}{1}=|OA_y| $$

In the triangle $(OA’_xA’)$:

$$ \sin (-x)=\frac{\textrm{opposite}}{\textrm{hypotenuse}}=\frac{|OA’_y|}{R}=\frac{|OA’_y|}{1}=|OA’_y| $$

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

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

Also in this section

  1. Cosine is even function cos(-x)=cos x
  2. Sine is an odd function sin(-x)=-sin x