Trig IDs

[Vulcan]

I came across this in a text book and wanted to understand how the solution comes about:



But I cannot get anywhere near the solution given. Any help would be much appreciated.

 

[Cubist]

I think you made a typo on your 2^nd equation. I assume it should look like this...

$\displaystyle \mathit{VI} \cos\left(\theta\right)\cdot \left(1+\cos\left( \color{red} 2 \color{black} \omega t\right)\right)+\mathit{VI} \sin\left(\theta\right)\cdot \sin\left(2\omega t\right)$

--

You found the correct two trig identities to use! You need to use them one after the other:-
- First use product of cosines: cos * cos -> (cos + cos)/2
- then use the other identity on the result cos(A-B) -> cos*cos - sin*sin

 

[Vulcan]

I think you made a typo on your 2^nd equation. I assume it should look like this...

$\displaystyle \mathit{VI} \cos\left(\theta\right)\cdot \left(1+\cos\left( \color{red} 2 \color{black} \omega t\right)\right)+\mathit{VI} \sin\left(\theta\right)\cdot \sin\left(2\omega t\right)$

--

You found the correct two trig identities to use! You need to use them one after the other:-
- First use product of cosines: cos * cos -> (cos + cos)/2
- then use the other identity on the result cos(A-B) -> cos*cos - sin*sin

Thank you so much for your reply, yes you are correct on the type. I have tried to do as you say but just come back to the original equation. See below:

Any further light you can shed on my error would be very welcome!

 

[Cubist]

Good, but on your 3rd line simplify both the angles before moving on, since

$\displaystyle \omega t-\left(\omega t-\theta\right) \to \theta$
$\displaystyle \omega t+\left(\omega t-\theta\right) \to 2\omega t-\theta$

Then use the identity cos(A-B) -> cos*cos - sin*sin

 

[Vulcan]

Good, but on your 3rd line simplify both the angles before moving on, since

$\displaystyle \omega t-\left(\omega t-\theta\right) \to \theta$
$\displaystyle \omega t+\left(\omega t-\theta\right) \to 2\omega t-\theta$

Then use the identity cos(A-B) -> cos*cos - sin*sin

Thank you so much for your help, I haven't done any trig for ages but I should have seen that!