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October 20th, 2018, 06:19 AM   #1
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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 angle-sum 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.
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October 20th, 2018, 06:26 AM   #2
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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
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October 20th, 2018, 09:40 AM   #3
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You can use the identities given below to achieve some simplification.

$\displaystyle \begin{align*}\sin(e)\sin(y) - \sin(c)\sin(x) &= \sin(e)\sin(e + a) - \sin(e + d)\sin(e + a + d) \\
&= (\cos(a) - \cos(2e + a))/2 - (\cos(a) - \cos(2e + a + 2d))/2 \\
&= (\cos(2e + a + 2d) - \cos(2e + a))/2 \\
&= -\sin(2e + a + d)\sin(d)\end{align*}$

$\displaystyle \begin{align*}\cos(c)\sin(x) - \cos(e)\sin(y) &= \sin(e + a + d)\cos(e + d) - \sin(e + a)\cos(e) \\
&= (\sin(2e + a + 2d) + \sin(a))/2 - (\sin(2e + a) + \sin(a))/2 \\
&= (\sin(2e + a + 2d) - \sin(2e + a))/2 \\
&= \cos(2e + a + d)\sin(d)\end{align*}$
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October 22nd, 2018, 01:42 AM   #4
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Quote:
Originally Posted by skipjack View Post
You can use the identities given below to achieve some simplification.
Thank you! Did you figure these relations out or did you find them somewhere? If it's the latter, could you share your source?
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October 22nd, 2018, 02:56 AM   #5
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They were obtained using these standard identities.
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October 22nd, 2018, 01:10 PM   #6
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Quote:
Originally Posted by skipjack View Post
They were obtained using these standard identities.
These helped me out so much, exactly the identities that I was missing... very unfortunately. Thanks again, with this I can solve everything by myself now.
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