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October 1st, 2018, 09:29 AM  #1 
Member Joined: Jul 2018 From: Germany Posts: 34 Thanks: 1  Cancelling sums in fractions
Help me do the impossible, I'm looking at this: $\displaystyle \frac{\sin(a+e)}{b}*\frac{u*\sin(e)  v*\sin(e)}{\sin(a)*\cos(e+a)+(1\cos(a))*\cos(e+a)}\frac{\cos(e+a)}{b}*\frac{u*\sin(e)  v*\cos(e)}{\sin(a)*\sin(a+e) + (1\cos(a))*\cos(a+e)} $ As you can see, both sides of the subtraction are very similar, I need to reduce 3 of these subtractions, if you have any hints on how to shorten the formulas, I'd be thankful. I'm almost certain that the outcome should be very simple actually, these formulas are part of a bigger one that describes interferencepatterns: Last edited by skipjack; October 1st, 2018 at 03:57 PM. 
October 1st, 2018, 02:15 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,641 Thanks: 625 
"Both sides of the subtraction"?? I suggest you check for typos.

October 1st, 2018, 03:16 PM  #3 
Member Joined: Jul 2018 From: Germany Posts: 34 Thanks: 1  
October 1st, 2018, 03:34 PM  #4 
Member Joined: Jul 2018 From: Germany Posts: 34 Thanks: 1 
And I did make some progress: $\displaystyle \frac{(1\cos(a))*\cos(a+e)}{\sin(a)}=\cos(a+e)*(\csc(a)\cot(a)) $ and I'm trying to simplify this atm: $\displaystyle \frac{(u*\sin(e)v*\cos(e))*\cos(a+e)}{\sin(a)} $ I think it has to do something with the golden ratio, as I tried visualizing it I saw that the graph peaks when $\displaystyle e=phi=0.618...$, that's at least pretty interesting to me. the rectangle represents $\displaystyle \sin(e)*\cos(e+a)$ and the line that goes Y represents $\displaystyle \frac{\sin(e)*\cos(e+a)}{\sin(a)}$ Last edited by skipjack; October 1st, 2018 at 04:00 PM. 

Tags 
cancelling, cosine, fractions, sine, sums 
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