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

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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.

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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.

Indeed changing

$$\theta \to -\theta \implies P(x,y)\to P(x,-y)$$

since the operation $\theta \to -\theta$ corresponds to a reflection with respect to the $x$ axis.