If $\cos(n\pi)= (-1)^n$, then why is $-\frac{4}{n}\cos(n\pi) = \frac{4}{n}(-1)^{n+1}$?

Question

If $\cos(n\pi)= (-1)^n$, then why is $-\frac{4}{n}\cos(n\pi) = \frac{4}{n}(-1)^{n+1}$?

I lectures has this written down in one of the solutions to an exercise however I'm not sure how he got $\frac{4}{n}(-1)^{n+1}$? I thought it would just be $-\frac{4}{n}. (-1)^n$? Am I missing something?

Thanks.

Answer 1

Those two things are equal, indeed $$-\frac{4}{n}(-1)^n=\frac{4}{n}(-1)^1(-1)^n=\frac{4}{n}(-1)^{n+1}.$$

Answer 2

$-4/n\cdot(-1)^n=(-1)\cdot4/n\cdot(-1)^n=4/n\cdot(-1)\cdot(-1)^n=4/n\cdot(-1)^{n+1}$

Answer 3

Note that $$(-\frac{4}{n})\cos(n\pi)=\frac{4}{n}(-1)\cdot (-1)^n=\frac{4}{n}(-1)^{n+1}$$