# problem in general solution

• Mar 19th 2008, 09:21 AM
afeasfaerw23231233
problem in general solution
problem in general solutionAttachment 5431
• Mar 19th 2008, 09:41 AM
topsquark
Quote:

Originally Posted by afeasfaerw23231233
problem in general solutionAttachment 5431

If I am reading them correctly any of these are good.

Note that $\displaystyle 2n \pm 1$ is just a way to denote an arbitrary odd number. So you could simplify this by using
$\displaystyle (2n + 1)\pi \pm y$
and still cover all of your solutions.

-Dan
• Mar 19th 2008, 09:51 AM
TheEmptySet
This may help.
I messed this up. I will try again.
• Mar 19th 2008, 10:14 AM
TheEmptySet
Maybe i'm wrong but

$\displaystyle \cos(x \pm y)=\cos(x) \cos(y) \mp \sin(x) \sin(y)$

How do you make the connection?
• Mar 19th 2008, 10:22 AM
topsquark
Quote:

Originally Posted by TheEmptySet
Maybe i'm wrong but

$\displaystyle \cos(x \pm y)=\cos(x) \cos(y) \mp \sin(x) \sin(y)$

How do you make the connection?

I believe it is meant to read:
$\displaystyle -cos(y) = cos(\pi \pm y)$
which is correct.

-Dan