July 9th, 2018, 02:10 PM  #1 
Member Joined: Jun 2017 From: Lima, Peru Posts: 52 Thanks: 1 Math Focus: Calculus  Why my approach to prove this trigonometric identity does not work?
The problem is as follows: $\textrm{Prove:}$ $\sin 2\omega+\sin 2\phi+\sin 2\psi= 4 \sin\omega \sin\phi \sin\psi$ $\textrm{Given this condition:}$ $$\omega +\phi+ \psi= \pi$$ In my attempt to solve this problem I did what I felt obvious and that was to take the sine function to the whole condition as shown below: $$\sin \left (\omega +\phi+ \psi \right )= \sin \left ( \pi \right )$$ Grouping: $$\sin \left (\omega + \left(\phi+ \psi\right) \right )= \sin \left ( \pi \right )$$ $$\sin \left (\omega + \left(\phi+ \psi\right) \right )= 0 $$ $$\sin \left(\omega\right) \cos \left(\phi+ \psi\right) + \cos \left(\omega\right) \sin \left(\phi+ \psi\right) = 0 $$ $$\sin \omega \left(\cos \phi \cos \psi  \sin \phi \sin \psi\right) + \cos \omega \left(\sin \phi \cos \psi + \cos \phi \sin \psi \right) = 0 $$ $$\sin \omega \cos \phi \cos \psi  \sin \omega \sin \phi \sin \psi + \cos \omega \sin \phi \cos \psi + \cos \omega \cos \phi \sin \psi = 0$$ $$\sin \omega \cos \phi \cos \psi + \cos \omega \sin \phi \cos \psi + \cos \omega \cos \phi \sin \psi = \sin \omega \sin \phi \sin \psi$$ $$\sin \omega \left(\cos \phi \cos \psi\right) + \cos \omega \left(\sin \phi \cos \psi + \cos \phi \sin \psi\right) = \sin \omega \sin \phi \sin \psi$$ Using product to sum identity: $$\sin \omega \left(\frac{1}{2}(\cos (\phi + \psi) + \cos \left(\phi \psi\right))\right) + \cos \omega \left(\left(\sin \phi \cos \psi + \cos \phi \sin \psi\right)\right) = \sin \omega \sin \phi \sin \psi$$ $$\sin \omega \left(\frac{1}{2}(\cos (\phi + \psi) + \cos \left(\phi \psi\right))\right) + \cos \omega \left(\frac{1}{2} \left(\sin (\phi + \psi) + \sin (\phi\psi) + \sin(\phi+\psi)  \sin(\phi\psi)\right)\right)= \sin \omega \sin \phi \sin \psi$$ Simplifying similar terms: $$\sin \omega \left(\frac{1}{2}(\cos (\phi + \psi) + \cos \left(\phi \psi\right))\right) + \cos \omega \left(\frac{1}{2} \left( 2 \sin (\phi + \psi) \right)\right)= \sin \omega \sin \phi \sin \psi $$ Then multiplying all by $4$ $$2 \sin \omega \left((\cos (\phi + \psi) + \cos \left(\phi \psi\right))\right) + 2 \cos \omega \left( \left( 2 \sin (\phi + \psi) \right)\right)= 4 \sin \omega \sin \phi \sin \psi$$ And that's how far I went, I feel I'm almost there but I don't know if I am in the right path or maybe not. Can somebody instruct me what is it missing or what should I do to prove this identity?. Last edited by skipjack; July 10th, 2018 at 01:45 AM. 
July 9th, 2018, 06:35 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,502 Thanks: 2511 Math Focus: Mainly analysis and algebra 
I would think that noting that $\phi + \psi = \pi  \omega$ would help.

July 9th, 2018, 08:02 PM  #3  
Member Joined: Jun 2017 From: Lima, Peru Posts: 52 Thanks: 1 Math Focus: Calculus  Quote:
By continuing from what I left off in the previous post: $$2 \sin \omega (\cos (\pi\omega)+\cos(\phi\psi))+2 \cos \omega (2 \sin(\pi\omega))= 4 \sin \omega \sin \phi \sin \psi$$ $$2 \sin \omega (\cos \omega + \cos (\phi  \psi)) + 2 \cos \omega ( 2 \sin \omega) = 4 \sin \omega \sin \phi \sin \psi$$ So far that's where I'm at and I don't see how I can relate with the double angles. Any help!! Last edited by skipjack; July 10th, 2018 at 02:10 AM.  
July 10th, 2018, 02:23 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,870 Thanks: 1833 
I corrected "$\pi\psi$" to "$\pi  \omega$" for you. Gathering like terms in your last statement gives $$2 \sin \omega \cos (\phi  \psi)) + 2 \cos \omega \sin \omega = 4 \sin \omega \sin \phi \sin \psi$$ which leads to $$\sin(\omega + \phi  \psi) + \sin(\omega  \phi + \psi) + \sin(2\omega) = 4\sin\omega\sin\phi\sin\psi$$ and so $$\sin(\pi  2\psi) + \sin(\pi  2\phi) + \sin(2\omega) = 4\sin\omega\sin\phi\sin\psi$$ etc. 
July 12th, 2018, 01:40 AM  #5  
Member Joined: Jun 2017 From: Lima, Peru Posts: 52 Thanks: 1 Math Focus: Calculus  Quote:
$$2 \sin \omega (\cos \omega + \cos (\phi  \psi)) + 2 \cos \omega ( 2 \sin \psi) = 4 \sin \omega \sin \phi \sin \psi$$ How does it become into? $$2 \sin \omega \cos (\phi  \psi)) + 2 \cos \omega \sin \omega = 4 \sin \omega \sin \phi \sin \psi$$ If I follow the rest it checks. But this initial step is where I'm stuck. Can you help me with this a bit more explicit? Could it be this the statement you were referring to? or this? $$2 \sin \omega \left((\cos (\phi + \psi) + \cos \left(\phi \psi\right))\right) + 2 \cos \omega \left( \left( 2 \sin (\phi + \psi) \right)\right)= 4 \sin \omega \sin \phi \sin \psi$$ When I multiplied by $4$ the whole expression I end up with $4$ next to the double angle of the sine. Wouldn't make sense that became into $2 \sin 2 \omega$ ?. I'm confused.  
July 12th, 2018, 02:16 AM  #6  
Member Joined: Jun 2017 From: Lima, Peru Posts: 52 Thanks: 1 Math Focus: Calculus  Revision of earlier post Quote:
(sorry about earlier typo: should had been $2 \cos \omega ( 2 \sin \omega)$) $$2 \sin \omega (\cos \omega + \cos (\phi  \psi)) + 2 \cos \omega ( 2 \sin \omega) = 4 \sin \omega \sin \phi \sin \psi$$ $$2 \sin \omega \cos \omega + 2 \sin \omega \cos (\phi  \psi)) + 4 \cos \omega ( \sin \psi) = 4 \sin \omega \sin \phi \sin \psi$$ therefore: $$\sin \omega \cos (\phi  \psi)) + 2 \cos \omega ( \sin \omega) = 4 \sin \omega \sin \phi \sin \psi$$ and the rest: $$\sin (\omega + \phi  \psi) + \sin (\omega  \phi + \psi) + \sin 2 \omega = 4 \sin \omega \sin \phi \sin \psi$$ then by inserting the known equation: $$\omega = \pi  \phi  \psi$$ $$\sin (\pi  \phi  \psi + \phi  \psi) + \sin (\pi  \phi  \psi  \phi + \psi) + \sin 2 \omega = 4 \sin \omega \sin \phi \sin \psi$$ $$\sin (\pi  2 \psi ) + \sin (\pi  2 \phi) + \sin 2 \omega = 4 \sin \omega \sin \phi \sin \psi$$ Then solving the sum of angles in the function: $$\sin (2\psi ) + \sin (2 \phi) + \sin (2 \omega) = 4 \sin \omega \sin \phi \sin \psi$$ and I thought you were referring to this.  
July 12th, 2018, 02:17 PM  #7 
Global Moderator Joined: Dec 2006 Posts: 19,870 Thanks: 1833 
Yes, you had typed $\sin \psi$ instead of $\sin \omega$. As you were puzzled prior to posting that typo, you perhaps relied on some private notes that contained a slip of that nature or to the uncorrected version of your earlier post. The working given in your original post can be shortened slightly. 
July 13th, 2018, 06:03 PM  #8  
Member Joined: Jun 2017 From: Lima, Peru Posts: 52 Thanks: 1 Math Focus: Calculus  A brief note Quote:
 
July 15th, 2018, 02:12 PM  #9 
Senior Member Joined: Jul 2012 From: DFW Area Posts: 628 Thanks: 92 Math Focus: Electrical Engineering Applications 
For the fun of it, using complex exponentials: $\displaystyle \large \sin(x)=\frac{e^{ix}e^{ix}}{2i}$ With $\large \omega+\phi+\psi=\pi$: $\displaystyle \large {\begin{align} 4\sin(\omega)\sin(\phi)\sin(\psi) &=\frac{4}{8i}\ (e^{i\omega}e^{i\omega})(e^{i\phi}e^{i\phi})(e^{i\psi}e^{i\psi}) \\ &=\frac{4}{8i}\ (e^{i(\omega+\phi)}e^{i(\omega\phi)}e^{i(\omega\phi)}+e^{i(\omega+\phi)})(e^{i\psi}e^{i\psi}) \\ &=\frac{4}{8i}\ (\cancel{e^{i(\omega+\phi+\psi)}}e^{i(\omega+\phi\psi)}e^{i(\omega\phi+\psi)}+e^{i(\omega\phi\psi)}e^{i(\omega\phi\psi)}+e^{i(\omega\phi+\psi)}+e^{i(\omega+\phi\psi)}\cancel{e^{i(\omega+\phi+\psi)}}) \\ &=\frac{4}{8i} \ 2i(\sin(\underbrace{\omega+\phi}_{\pi\psi}\psi)\sin(\underbrace{\omega\phi+\psi}_{\omega+\psi=\pi\phi})+\sin(\omega \underbrace{\phi\psi}_{\omega\pi})) \\ &=\sin(\pi2\psi)+\sin(\pi2\phi)\sin(2\omega\pi) \\ &=\sin(\pi2\psi)+\sin(\pi2\phi)+\sin(\pi2\omega) \\ &=\sin(2\psi)+\sin(2\phi)+\sin(2\omega) \end{align}}$ 
July 15th, 2018, 03:27 PM  #10 
Global Moderator Joined: Dec 2006 Posts: 19,870 Thanks: 1833 
Using $\sin2\phi = \sin(\pi  2\phi) = \sin(\omega + \phi + \psi  2\phi) = \sin(\phi  \omega  \psi)$ (and similar results involving $2\omega$ and $2\psi$), $\sin(\phi +\omega + \psi) = \sin(\pi) = 0$, and the standard producttosum identities, $\begin{align*}4\sin(\phi)\sin(\omega)\sin(\psi) &= 2(\cos(\phi  \omega)  \cos(\phi + \omega))\sin(\psi) \\ &= 2\cos(\phi  \omega)\sin(\psi)  2\cos(\phi + \omega)\sin(\psi) \\ &= \sin(\phi  \omega + \psi)  \sin(\phi  \omega  \psi)  \sin(\phi + \omega + \psi) + \sin(\phi + \omega  \psi) \\ &= \sin2\omega + \sin2\phi + \sin2\psi \end{align*}$ 

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