• Apr 5th 2009, 02:47 AM
craigmain
Hi,

I have a proof for $cos( \theta + \delta) = cos \theta cos\delta - sin \theta sin \delta$

Apparently by writing $(\frac{\pi}{2}-\delta)$ for $\delta$ and using $cos(\frac{\pi}{2}-\delta)=sin \delta$ and vice versa

I get
$sin(\theta + \delta)=sin \theta cos \delta + cos \theta sin \delta$. I cannot get this result, not with the signs correct.

$cos(\theta + [\frac{\pi}{2}-\delta])=cos \theta sin \delta - sin \theta cos \delta$
$cos(\frac{\pi}{2}-[\delta - \theta])=...$
we get
$sin(\delta - \theta)=sin \delta cos \theta - cos \delta sin \theta$

I am supposed to get
$sin(\theta + \delta)=sin \theta cos \delta + cos \theta sin \delta$

I am doing something wrong with the signs or something.
• Apr 5th 2009, 05:03 AM
running-gag
Quote:

Originally Posted by craigmain
Hi,

I have a proof for $cos( \theta + \delta) = cos \theta cos\delta - sin \theta sin \delta$

Apparently by writing $(\frac{\pi}{2}-\delta)$ for $\delta$ and using $cos(\frac{\pi}{2}-\delta)=sin \delta$ and vice versa

I get
$sin(\theta + \delta)=sin \theta cos \delta + cos \theta sin \delta$. I cannot get this result, not with the signs correct.

$cos(\theta + [\frac{\pi}{2}-\delta])=cos \theta sin \delta - sin \theta cos \delta$

$cos(\frac{\pi}{2}-[\delta - \theta])=...$
we get
$sin(\delta - \theta)=sin \delta cos \theta - cos \delta sin \theta$

I am supposed to get
$sin(\theta + \delta)=sin \theta cos \delta + cos \theta sin \delta$

I am doing something wrong with the signs or something.

Hi

If your starting point is $\cos( \theta + \delta) = \cos \theta \cos\delta - \sin \theta \sin \delta$

and you want to find $\sin(\theta + \delta)=\sin \theta \cos \delta + \cos \theta \sin \delta$

you can do like this

$\sin(\theta + \delta) = \cos\left(\frac{\pi}{2}-(\theta + \delta)\right)$

$\sin(\theta + \delta) = \cos\left(\left(\frac{\pi}{2}-\theta\right) + (-\delta)\right)$

$\sin(\theta + \delta) = \cos \left(\frac{\pi}{2}-\theta\right) cos(-\delta) - \sin \left(\frac{\pi}{2}-\theta\right) \sin (-\delta)$

$\sin(\theta + \delta) = \sin \theta cos \delta + \cos \theta \sin \delta$
• Apr 5th 2009, 09:40 AM
craigmain
Thanks, I was trying to do it on the wrong side. Silly.
There's an art to stating the obvious that will take a lifetime to learn.