Trigonometry Trigonometry Math Forum

 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 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. 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*} Thanks from theHeavyLobster October 22nd, 2018, 02:42 AM   #4
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Quote:
 Originally Posted by skipjack 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? October 22nd, 2018, 03:56 AM #5 Global Moderator   Joined: Dec 2006 Posts: 21,110 Thanks: 2326 They were obtained using these standard identities. Thanks from theHeavyLobster October 22nd, 2018, 02:10 PM   #6
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Quote:
 Originally Posted by skipjack 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. Tags angle-sum, cosine, identities, sine Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode

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