problem in general solution
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Originally Posted by afeasfaerw23231233 problem in general solution 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
I messed this up. I will try again.
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?
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
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