# Proving the Trig Identities $\sin(-\theta)=-\sin(\theta)$ and $\cos(-\theta)=\cos(\theta)$

I want to prove the trig identities $\sin(-\theta)=-\sin(\theta)$ and that $\cos(-\theta)=\cos(\theta)$. I realize I can prove it by drawing out the radius when it is rotated by $\theta$, and the radius when it is rotated by $-\theta$ for each magnitude of rotation.

However, I want I more elegant proof. I want to show that since $\theta$ rotates the radius counterclockwise, and -theta rotates the radius clockwise, by definition, a radius rotated by $-\theta$ will be equals to a radius rotated by $\theta$ reflected vertically. I am not sure how to show this, though. Can someone help me? If this can proved, this give me a far more elegant way to prove that $\sin(-\theta)=-\sin(\theta)$ and that $\cos(-\theta)=\cos(\theta)$ than by simply drawing it out.

• What's your definition of sine and cosine? – Lord Shark the Unknown Jun 26 '18 at 5:51
• Sine theta = y, cosine theta = x. – Ethan Chan Jun 26 '18 at 5:52
• "Simply drawing it out" is very elegant. – Doug M Jun 26 '18 at 5:55
• Take the derivative of $\sin(-x)+\sin(x)$. – Michael Hoppe Jun 26 '18 at 6:07
• @EthanChan Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… – user Aug 5 '18 at 20:59

The best way to prove is directly by the definition of $\cos \theta$ and $\sin \theta$ as coordinates of the point $P(x,y)$ on the trigonometric circle.
$$\theta \to -\theta \implies P(x,y)\to P(x,-y)$$
since the operation $\theta \to -\theta$ corresponds to a reflection with respect to the $x$ axis.