# Simplification of answer involving Cosine

I have manged to get my answer down to the first line in the picture but I have tried all ways and can't seem to simplify it to the second line.

Thank you

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$\cos(x\pi)=(-1)^x$ if $x\in\mathbb{Z}$
Does that help?

What I don't get is how to I simplify from
$\frac{-(cos(n*pi+pi)}{(n+1)}$+$\frac{-(cos(n*pi-pi)}{(n-1)}$

to
$\frac{(cos(n*pi)}{(n+1)}$-$\frac{(cos(n*pi)}{(n-1)}$

to
$\frac{-(2 cos(pi n)}{(n^2-1)}$

haruspex
Homework Helper
Gold Member
You may be looking for something more complicated than necessary.
What I don't get is how to I simplify from
$\frac{-(cos(n*π+π)}{(n+1)}$+$\frac{-(cos(n*π-π)}{(n-1)}$
to
$\frac{(cos(n*π)}{(n+1)}$-$\frac{(cos(n*π)}{(n-1)}$
How would you simplify cos(θ+π) and cos(θ-π)?
to
$\frac{-(2 cos(π n)}{(n^2-1)}$
How would combine a/(n+1) - a/(n-1) into a single fraction?

You may be looking for something more complicated than necessary.

How would you simplify cos(θ+π) and cos(θ-π)?

How would combine a/(n+1) - a/(n-1) into a single fraction?
Yup got it, totally forgot the trig identity

cos(A+ B)= cos(A)cos(B)- sin(A)sin(B)

Thank you

You do not actually need that identity for this, the rather simple identities $\cos(\pi+\theta)=-\cos(\theta)$ and $\cos(\theta-\pi)=\cos(\pi-\theta)=-\cos(\theta)$ are enough.