Tricky trig question electrical

[anonwater]

Hello, I'm stuck with this derivation:

[MATH]\frac{1}{2}\left ( \cos( \theta -\gamma) +\cos (2\omega t+\theta +\gamma ) \right )=....[/MATH][MATH]\text{should become....}=\frac{1}{2}\left [ \cos( \theta -\gamma)(1+ \cos (2\omega t+2\theta)) + \sin( \theta -\gamma)\sin (2\omega t+2\theta) \right ][/MATH]
My attempt of solution:
If I start by assuming [MATH]\alpha=\omega t+\theta \quad and \quad \beta=\omega t +\gamma[/MATH]and then [MATH]\alpha -\beta=\theta -\gamma \text{ and } \alpha +\beta=2\omega t+\theta +\gamma [/MATH]
*I do it so I can use the trig forumla: [MATH]\cos (\alpha +\beta)=\cos (\alpha)\cos (\beta)-\sin(\alpha)\sin(\beta)[/MATH]
then by using the first row I have*: [MATH] \frac{1}{2}\left ( \cos( \alpha -\beta) +\cos (\alpha +\beta) \right )=\frac{1}{2}\left ( \cos( \alpha -\beta) + \cos (\alpha)\cos (\beta)-\sin(\alpha)\sin(\beta) \right )[/MATH]
But I get stuck here, I assume I am supposed to get the terms sin and for example [MATH]\sin(2\alpha)[/MATH] somehow?

Thank you

 

[Steven G]

I am confused! Why would you expand cos(α+β), but not expand cos(α-β)?

 

[anonwater]

I am confused! Why would you expand cos(α+β), but not expand cos(α-β)?

My idea was to not expand cos(α-β) because the final expression is supposed to have 3 terms in it and cos(α-β) is one of them.
since [MATH]\cos( \alpha-\beta)=\cos( \theta -\gamma)[/MATH]

 

[JeffM]

Hello, I'm stuck with this derivation:

[MATH]\frac{1}{2}\left ( \cos( \theta -\gamma) +\cos (2\omega t+\theta +\gamma ) \right )=....[/MATH][MATH]\text{should become....}=\frac{1}{2}\left [ \cos( \theta -\gamma)(1+ \cos (2\omega t+2\theta)) + \sin( \theta -\gamma)\sin (2\omega t+2\theta) \right ][/MATH]
My attempt of solution:
If I start by assuming [MATH]\alpha=\omega t+\theta \quad and \quad \beta=\omega t +\gamma[/MATH]and then [MATH]\alpha -\beta=\theta -\gamma \text{ and } \alpha +\beta=2\omega t+\theta +\gamma [/MATH]
*I do it so I can use the trig forumla: [MATH]\cos (\alpha +\beta)=\cos (\alpha)\cos (\beta)-\sin(\alpha)\sin(\beta)[/MATH]
then by using the first row I have*: [MATH] \frac{1}{2}\left ( \cos( \alpha -\beta) +\cos (\alpha +\beta) \right )=\frac{1}{2}\left ( \cos( \alpha -\beta) + \cos (\alpha)\cos (\beta)-\sin(\alpha)\sin(\beta) \right )[/MATH]
But I get stuck here, I assume I am supposed to get the terms sin and for example [MATH]\sin(2\alpha)[/MATH] somehow?

Thank you

You had a sensible idea, but you could choose a more promising set of changes of variable.

[MATH]\text {Let } \kappa = \theta - \gamma,\ \lambda = 2 \omega t + \theta + \gamma, \text { and } \psi = 2 \omega t + 2 \theta.[/MATH]
[MATH]\psi - \kappa = 2 \omega t + 2 \theta - (\theta - \gamma) = 2 \omega t + \theta + \gamma = \lambda.[/MATH]
[MATH]\therefore \cos( \theta -\gamma) +\cos (2\omega t+\theta +\gamma ) =[/MATH]
[MATH]cos( \kappa ) + cos( \lambda) = cos ( \kappa ) + cos(\psi - \kappa) =[/MATH]
[MATH]cos ( \kappa ) + cos( \psi ) cos( \kappa ) + sin ( \psi) sin( \kappa ) = [/MATH]
[MATH]cos( \kappa) \{1 + cos ( \psi) \} + sin ( \kappa ) \sin ( \psi) =[/MATH]
[MATH]cos( \theta - \gamma ) \{1 + cos ( 2 \omega t + 2 \theta\} + sin( \theta - \gamma) sin( 2 \omega t + 2 \theta).[/MATH]
You want to end up with [MATH]2 \omega t + 2 \theta[/MATH]. Use that knowledge when figuring out what change of variable is handy.

 

[Steven G]

My idea was to not expand cos(α-β) because the final expression is supposed to have 3 terms in it and cos(α-β) is one of them.
since [MATH]\cos( \alpha-\beta)=\cos( \theta -\gamma)[/MATH]

Good point. For the record, [MATH]\cos( \alpha-\beta)=\cos( \theta -\gamma)[/MATH] is not one of the terms in the final expression. Do you know what a term is? How about a factor?

Actually that final expression has exactly one term and if you consider what 1/2 is being multiplied by, that part has exactly two terms.

 

[anonwater]

You had a sensible idea, but you could choose a more promising set of changes of variable.

[MATH]\text {Let } \kappa = \theta - \gamma,\ \lambda = 2 \omega t + \theta + \gamma, \text { and } \psi = 2 \omega t + 2 \theta.[/MATH]
[MATH]\psi - \kappa = 2 \omega t + 2 \theta - (\theta - \gamma) = 2 \omega t + \theta + \gamma = \lambda.[/MATH]
[MATH]\therefore \cos( \theta -\gamma) +\cos (2\omega t+\theta +\gamma ) =[/MATH]
[MATH]cos( \kappa ) + cos( \lambda) = cos ( \kappa ) + cos(\psi - \kappa) =[/MATH]
[MATH]cos ( \kappa ) + cos( \psi ) cos( \kappa ) + sin ( \psi) sin( \kappa ) = [/MATH]
[MATH]cos( \kappa) \{1 + cos ( \psi) \} + sin ( \kappa ) \sin ( \psi) =[/MATH]
[MATH]cos( \theta - \gamma ) \{1 + cos ( 2 \omega t + 2 \theta\} + sin( \theta - \gamma) sin( 2 \omega t + 2 \theta).[/MATH]
You want to end up with [MATH]2 \omega t + 2 \theta[/MATH]. Use that knowledge when figuring out what change of variable is handy.

Good point. For the record, [MATH]\cos( \alpha-\beta)=\cos( \theta -\gamma)[/MATH] is not one of the terms in the final expression. Do you know what a term is? How about a factor?

Actually that final expression has exactly one term and if you consider what 1/2 is being multiplied by, that part has exactly two terms.

Thank you so much for the help!
And Jomo yes you are right, I was probably thinking of the final expression all written out 1/2cos+1/2coscos+1/2sinsin which is three terms, if that make sense.