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.

  • 1
    $\begingroup$ What's your definition of sine and cosine? $\endgroup$ Jun 26 '18 at 5:51
  • $\begingroup$ Sine theta = y, cosine theta = x. $\endgroup$
    – Ethan Chan
    Jun 26 '18 at 5:52
  • 1
    $\begingroup$ "Simply drawing it out" is very elegant. $\endgroup$
    – Doug M
    Jun 26 '18 at 5:55
  • $\begingroup$ Take the derivative of $\sin(-x)+\sin(x)$. $\endgroup$ Jun 26 '18 at 6:07
  • $\begingroup$ @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/… $\endgroup$
    – 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.

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.

  • $\begingroup$ I see, but I still not sure how to prove that if the end point of a radius rotated by theta= (x,y), the end point of a radius rotated by -theta will be equals to (x,-y). $\endgroup$
    – Ethan Chan
    Jun 26 '18 at 5:51
  • $\begingroup$ Revise the definition for trigonometric circle and trigonometric functions, we don’t need any other knowledge to derive these identities. $\endgroup$
    – user
    Jun 26 '18 at 5:54
  • $\begingroup$ Okay, I think I now get y this is the case when theta is <=180 degrees. But after that, I find it kind of hard to visualize why the above is still true by definition. Can you help me clarify? $\endgroup$
    – Ethan Chan
    Jun 26 '18 at 5:58
  • $\begingroup$ make a sketch of the trigonometric circle and try with some different cases $\endgroup$
    – user
    Jun 26 '18 at 6:01

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