# Simplification of answer involving Cosine

1. Nov 23, 2012

### 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

#### Attached Files:

• ###### 1213.png
File size:
1.6 KB
Views:
73
2. Nov 23, 2012

### Millennial

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

3. Nov 23, 2012

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

4. Nov 23, 2012

### haruspex

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?

5. Nov 23, 2012

### zack7

Yup got it, totally forgot the trig identity

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

Thank you

6. Nov 24, 2012

### 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.