My Math Forum Cancelling sums in fractions

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 October 1st, 2018, 08:29 AM #1 Member   Joined: Jul 2018 From: Germany Posts: 37 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 interference-patterns: Last edited by skipjack; October 1st, 2018 at 02:57 PM.
 October 1st, 2018, 01:15 PM #2 Global Moderator   Joined: May 2007 Posts: 6,786 Thanks: 708 "Both sides of the subtraction"?? I suggest you check for typos.
October 1st, 2018, 02:16 PM   #3
Member

Joined: Jul 2018
From: Germany

Posts: 37
Thanks: 1

Quote:
 Originally Posted by mathman "Both sides of the subtraction"?? I suggest you check for typos.
I meant those two sides:

I don't know if defining the sides was the issue or if my description of a subtraction having two sides is incorrect.

 October 1st, 2018, 02:34 PM #4 Member   Joined: Jul 2018 From: Germany Posts: 37 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 03:00 PM.

 Tags cancelling, cosine, fractions, sine, sums

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