Why my approach to prove this trigonometric identity does not work?

Jun 2017
276
6
Lima, Peru
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?. :rolleyes:
 
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v8archie

Math Team
Dec 2013
7,709
2,677
Colombia
I would think that noting that $\phi + \psi = \pi - \omega$ would help.
 
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Jun 2017
276
6
Lima, Peru
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!!
 
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skipjack

Forum Staff
Dec 2006
21,296
2,375
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.
 
Jun 2017
276
6
Lima, Peru
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.
 
Jun 2017
276
6
Lima, Peru
Revision of earlier post

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. :)
 

skipjack

Forum Staff
Dec 2006
21,296
2,375
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.
 
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Jun 2017
276
6
Lima, Peru
A brief note

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. :)
 

jks

Jul 2012
642
99
DFW Area
For the fun of it, using complex exponentials:

\(\displaystyle \large \sin(x)=\frac{e^{ix}-e^{-ix}}{2i}\)

With $\large \omega+\phi+\psi=\pi$:

\(\displaystyle \large {\begin{align} 4\sin(\omega)\sin(\phi)\sin(\psi)
&=\frac{4}{-8i}\ (e^{i\omega}-e^{-i\omega})(e^{i\phi}-e^{-i\phi})(e^{i\psi}-e^{-i\psi}) \\
&=\frac{4}{-8i}\ (e^{i(\omega+\phi)}-e^{i(\omega-\phi)}-e^{-i(\omega-\phi)}+e^{-i(\omega+\phi)})(e^{i\psi}-e^{-i\psi}) \\
&=\frac{4}{-8i}\ (\cancel{e^{i(\omega+\phi+\psi)}}-e^{i(\omega+\phi-\psi)}-e^{i(\omega-\phi+\psi)}+e^{i(\omega-\phi-\psi)}-e^{-i(\omega-\phi-\psi)}+e^{-i(\omega-\phi+\psi)}+e^{-i(\omega+\phi-\psi)}-\cancel{e^{-i(\omega+\phi+\psi)}}) \\
&=\frac{4}{-8i} \ 2i(-\sin(\underbrace{\omega+\phi}_{\pi-\psi}-\psi)-\sin(\underbrace{\omega-\phi+\psi}_{\omega+\psi=\pi-\phi})+\sin(\omega \underbrace{-\phi-\psi}_{\omega-\pi})) \\
&=\sin(\pi-2\psi)+\sin(\pi-2\phi)-\sin(2\omega-\pi) \\
&=\sin(\pi-2\psi)+\sin(\pi-2\phi)+\sin(\pi-2\omega) \\
&=\sin(2\psi)+\sin(2\phi)+\sin(2\omega)
\end{align}}\)
 
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skipjack

Forum Staff
Dec 2006
21,296
2,375
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,
$\begin{align*}4\sin(\phi)\sin(\omega)\sin(\psi) &= 2(\cos(\phi - \omega) - \cos(\phi + \omega))\sin(\psi) \\
&= 2\cos(\phi - \omega)\sin(\psi) - 2\cos(\phi + \omega)\sin(\psi) \\
&= \sin(\phi - \omega + \psi) - \sin(\phi - \omega - \psi) - \sin(\phi + \omega + \psi) + \sin(\phi + \omega - \psi) \\
&= \sin2\omega + \sin2\phi + \sin2\psi \end{align*}$
 
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