Prove $\sin (-\alpha ) = -\sin(\alpha)$ and $\cos(-\alpha) = \cos(\alpha)$

Question

I want to demonstrate these basic trigonometric identities but I have no idea how to do it

$\sin (-\alpha ) = -\sin(\alpha)$

$\cos(-\alpha) = \cos(\alpha)$

help me please

thank you

Answer 1

Here is a little hint: look at the unit circle (draw it) and look at the definition of sin and cos for a given point

Answer 2

You could use the Taylor expansions for sin($\alpha$) and cos($\alpha$).

$sin(\alpha)=\alpha-\alpha^3/3!+\alpha^5/5!+...$

$cos(\alpha)=1-\alpha^2/2!+\alpha^4/4!+...$

So,

$sin(-\alpha)=-\alpha+\alpha^3/3!-\alpha^5/5!+... =(-1)(\alpha-\alpha^3/3!+\alpha^5/5!+...)=-sin(\alpha)$

$cos(-\alpha)=1-(-\alpha)^2/2!+(-\alpha)^4/4!+...=1-\alpha^2/2!+\alpha^4/4!+...=cos(\alpha)$

Answer 3

Consider the right angled triangles, $\triangle OPA$ and $\triangle OPA'$,

$$\sin \alpha = \frac{PA}{OA} = \frac{y}{\sqrt{x^2+y^2}}$$ $$\cos \alpha = \frac{OP}{OA} = \frac{x}{\sqrt{x^2+y^2}}$$ $$\sin(-\alpha) = \frac{PA'}{OA'} = \frac{-y}{\sqrt{x^2+y^2}} = -\sin \alpha$$ $$\cos(-\alpha) = \frac{OP}{OA'} = \frac{x}{\sqrt{x^2+y^2}} = \cos \alpha$$

You can also prove this by expanding taylor series of $\sin \alpha$ and $\cos \alpha$.

Hope this helps.

Answer 4

$$\cos(-\alpha)+i\sin(-\alpha)=e^{-i\alpha}=\frac1{e^{i\alpha}}=\frac1{\cos\alpha+i\sin\alpha}=\frac{\cos\alpha-i\sin\alpha}{\cos^2\alpha+\sin^2\alpha}=\cos\alpha-i\sin\alpha.$$

By equating the real and imaginary parts,

$$\cos(-\alpha)=\cos\alpha,\\\sin(-\alpha)=-\sin\alpha.$$