Simplification of answer involving Cosine

[zack7]

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

 

[Millennial]

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

 

[zack7]

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]

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?

 

[zack7]

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

 

[Millennial]

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.