Hi I know the formula
$\displaystyle sin(\alpha \pm n)\pi=(-1)^nsin\alpha \pi$
but what about
$\displaystyle cos(\alpha \pm n)\pi=?$
$\displaystyle cos(\alpha \mp n)\pi=?$
$\displaystyle sin(\alpha \mp n)\pi=?$
cos(a+ b)= cos(a)cos(b)- sin(a)sin(b).
In particular, if $\displaystyle b= n\pi$ and $\displaystyle a= \alpha$,
$\displaystyle cos(\alpha+ n\pi)= cos(\alpha)cos(n\pi)- sin(\alpha)sin(n\pi)$.
Of course, $\displaystyle sin(n\pi)= 0$, for all n, and $\displaystyle cos(n\pi)= (-1)^n$.
Therefore, $\displaystyle cos(\alpha+ n\pi)= (-1)^ncos(\alpha)$.
For $\displaystyle cos(\alpha- n\pi)$, use the fact that cosine is an even function: $\displaystyle cos(-n\pi)= cos(n\pi)= (-1)^n$
there is no difference at all between $\displaystyle \alpha\pm n\pi$ and $\displaystyle \alpha\mp n\pi$. They both mean exactly the same thing.
$\displaystyle cos(\alpha\pm{n}){\pi}=cos(\alpha{\pi})cos(\pm{n}{ \pi})-sin(\alpha{\pi})sin(\pm{n}{\pi})
$
$\displaystyle =cos(\alpha{\pi})cos(\pm{n}{\pi})$
If n=1, this is
$\displaystyle cos(\alpha{\pi})cos(\pm{\pi})=-cos(\alpha{\pi})$
If n=2, it's
$\displaystyle cos(\alpha{\pi})$
so the power of $\displaystyle -1$ is n