October 20th, 2018, 07:19 AM  #1 
Member Joined: Jul 2018 From: Germany Posts: 37 Thanks: 1  What would you do next?
I want to break this down/ cancel stuff out: $\displaystyle x = e + a + d \\ y = e + a \\ z = e + a/2 \\ c = e + d \\ $ $\displaystyle \frac{v*\cos(c)*\sin(x)}{\sin(a)*\cos(z)} \frac{u*\sin(c)*\sin(x)}{\sin(a)*\cos(z)}+ \frac{u*\sin(e)*\sin(y)}{\sin(a)*\cos(z)} \frac{v*\cos(e)*\sin(y)}{\sin(a)*\cos(z)} $ Bracketing everything doesn't help me, as there are too many variables. I was thinking of breaking apart x, y, z and c with anglesum identities. Do you think that would make sense as the next step? I did so with: $\displaystyle v*\cos(e)*\sin(y) = v*\cos^2(e)*(\sin(a)+\tan(e)*\cos(a)) $ I just don't know if that's going in the right direction, as I'm bloating everything up. Also, is there a way of excluding e from sin(e+a)? So that I'd end up having sin(a) * [...] * e. 
October 20th, 2018, 07:26 AM  #2 
Member Joined: Jul 2018 From: Germany Posts: 37 Thanks: 1 
Oh, and do you think going to rewrite it in terms of complex analysis would be a good approach? https://en.wikipedia.org/wiki/List_o...ntial_function 
October 20th, 2018, 10:40 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 21,110 Thanks: 2326 
You can use the identities given below to achieve some simplification. $\displaystyle \begin{align*}\sin(e)\sin(y)  \sin(c)\sin(x) &= \sin(e)\sin(e + a)  \sin(e + d)\sin(e + a + d) \\ &= (\cos(a)  \cos(2e + a))/2  (\cos(a)  \cos(2e + a + 2d))/2 \\ &= (\cos(2e + a + 2d)  \cos(2e + a))/2 \\ &= \sin(2e + a + d)\sin(d)\end{align*}$ $\displaystyle \begin{align*}\cos(c)\sin(x)  \cos(e)\sin(y) &= \sin(e + a + d)\cos(e + d)  \sin(e + a)\cos(e) \\ &= (\sin(2e + a + 2d) + \sin(a))/2  (\sin(2e + a) + \sin(a))/2 \\ &= (\sin(2e + a + 2d)  \sin(2e + a))/2 \\ &= \cos(2e + a + d)\sin(d)\end{align*}$ 
October 22nd, 2018, 02:42 AM  #4 
Member Joined: Jul 2018 From: Germany Posts: 37 Thanks: 1  
October 22nd, 2018, 03:56 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 21,110 Thanks: 2326 
They were obtained using these standard identities.

October 22nd, 2018, 02:10 PM  #6  
Member Joined: Jul 2018 From: Germany Posts: 37 Thanks: 1  Quote:
 