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Home > Mathematics > Trigonometry > 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 $(OA’_xA’)$.
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
$$ \forall x\in \mathbb{R},\quad: \sin (-x)=-\sin x $$