Trig IDs
- Thread starter Vulcan
- Start date
Cubist
Senior Member
- Joined
- Oct 29, 2019
- Messages
- 1,686
I think you made a typo on your 2nd 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
\(\displaystyle \mathit{VI} \cos\left(\theta\right)\cdot \left(1+\cos\left( \color{red} 2 \color{black} \omega t\right)\right)+\mathit{VI} \sin\left(\theta\right)\cdot \sin\left(2\omega t\right) \)
--
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
Senior Member
- Joined
- Oct 29, 2019
- Messages
- 1,686
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
\(\displaystyle \omega t-\left(\omega t-\theta\right) \to \theta \)
\(\displaystyle \omega t+\left(\omega t-\theta\right) \to 2\omega t-\theta \)
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!