# Simplifying with trig identities

1. May 15, 2015

### earthloop

1. The problem statement, all variables and given/known data

Hi, I am currently working through a textbook, and the following simplification is given for an example question:

I can't seem to work out how they have moved from cos(pi+n*pi) to cos(pi)cos(n*pi) so easily? Is there a simple trick I have missed? I understand the identity that separates out the sine and cosine terms (-cos(a+b) = sin(a)sin(b)-cos(a)cos(b)) but I'm having very little luck in getting the textbooks answer from that.

2. Relevant equations

cos(a+b) = cos(a)cos(b)-sin(a)sin(b)
cos(a-b) = cos(a)cos(b)+sin(a)sin(b)
-cos(a+b) = sin(a)sin(b)-cos(a)cos(b)
-cos(a-b) = -sin(a)sin(b)-cos(a)cos(b)

3. The attempt at a solution
applying the identity to the cosine/sine part of the equation ONLY (ignoring 2/(1-n^2) ) where a = pi :

$-\frac{\left(n + 1\right)\, \left(\cos\!\left(a\, n\right)\, \cos\!\left(a\right) + \sin\!\left(a\, n\right)\, \sin\!\left(a\right)\right) + \left(n - 1\right)\, \left(\cos\!\left(a\, n\right)\, \cos\!\left(a\right) - \sin\!\left(a\, n\right)\, \sin\!\left(n\right)\right)}{n^2 - 1}$

Am I on the right track?

Thanks

2. May 15, 2015

### Staff: Mentor

Look at the cos(a+b) identity, what happens when you have sin(n*pi)?

3. May 15, 2015

### HallsofIvy

n is assumed to be an integer here, correct?

4. May 15, 2015

### earthloop

jedishrfu... sorry not sure what you mean? Can you give a bit more detail?
Hallsofivy...yes n is an integer

5. May 15, 2015

### earthloop

oh! Thanks guys.... Completely couldn't see the wood for the trees :)

sin(n*pi) = 0

Doh!

Cheers

6. May 15, 2015

### Staff: Mentor

Sometimes the wood is fossilized. :-)