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 July 9th, 2018, 01:10 PM #1 Senior Member     Joined: Jun 2017 From: Lima, Peru Posts: 108 Thanks: 2 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 12:45 AM.
 July 9th, 2018, 05:35 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra I would think that noting that $\phi + \psi = \pi - \omega$ would help. Thanks from topsquark
July 9th, 2018, 07:02 PM   #3
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Quote:
 Originally Posted by v8archie I would think that noting that $\phi + \psi = \pi - \omega$ would help.
I tried to follow this suggestion, but I'm still stuck.

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 01:10 AM.

 July 10th, 2018, 01:23 AM #4 Global Moderator   Joined: Dec 2006 Posts: 21,026 Thanks: 2257 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, 12:40 AM   #5
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Quote:
 Originally Posted by skipjack 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.
I tried to catch up from what you had corrected in my calculations but I'm stuck at the beginning, the part where you said you had corrected my last statement.

$$2 \sin \omega (-\cos \omega + \cos (\phi - \psi)) + 2 \cos \omega ( 2 \sin \psi) = 4 \sin \omega \sin \phi \sin \psi$$

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, 01:16 AM   #6
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Revision of earlier post

Quote:
 Originally Posted by skipjack 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.
Well, after going in circles for 30 minutes or so I figured what it was obvious and maybe you were referring to:

(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, 01:17 PM #7 Global Moderator   Joined: Dec 2006 Posts: 21,026 Thanks: 2257 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. Thanks from Chemist116
July 13th, 2018, 05:03 PM   #8
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A brief note

Quote:
 Originally Posted by skipjack 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.
Anyway thanks man. Before opening this thread I was about to throw in the towel with this problem, but it was your post which left me pondering what steps did I overlooked, in the end it was all obvious.

 July 15th, 2018, 01:12 PM #9 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 642 Thanks: 99 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(\pi-2\psi)+\sin(\pi-2\phi)-\sin(2\omega-\pi) \\ &=\sin(\pi-2\psi)+\sin(\pi-2\phi)+\sin(\pi-2\omega) \\ &=\sin(2\psi)+\sin(2\phi)+\sin(2\omega) \end{align}} Thanks from greg1313
 July 15th, 2018, 02:27 PM #10 Global Moderator   Joined: Dec 2006 Posts: 21,026 Thanks: 2257 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 product-to-sum 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*} Thanks from greg1313

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