Why is $\cos(x)-i\sin(x) = \cos(-x)+i\sin(-x)=\operatorname*{cis}(-x)$?

I do not understand why: $$\cos(x)-i\sin(x) = \cos(-x)+i\sin(-x)$$

The last step i understand. (It is one of the solutions of a quadratic equation if it matters)

• Because $\cos x$ is an even function, i.e., $f(x)=f(-x)$ is the definition for such a function. (And also, $\sin -x=-\sin x$, since $\sin x$ is an odd function.) – abiessu May 13 '17 at 14:53
• what is this $sen(x)$? – samjoe May 13 '17 at 14:57
• – Shaun May 13 '17 at 14:59
• Yes, sorry i forgot sen is not used in english, and thanks for the mathjax tutorial – Alex May 13 '17 at 15:00
• @samjoe $sen(x)$ is the way the Spanish write $\sin(x)$ because Sine in Spanish is Seno. So it is abbreviated to $sen(x)$ instead of $\sin(x)$ – Malcolm May 13 '17 at 15:32

$$\cos (-x)=\cos (x)$$
And, $$\sin (-x)=-\sin (x)$$
Because $\cos$ is even and $\sin$ is odd. In detail, it's because $$\cos(z)=\frac{e^{iz}+e^{-iz}}{2}$$ and $$\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}.$$