What is the equivalent identity for $\sin^2\omega t$?

[Drain Brain]

is there an equivalent identity for $\sin^2\omega t?$

please tell me.REGARDS!

 

[MarkFL]

It depends on what you want the identity to be in terms of...two identities that spring to mind are:

\(\displaystyle \sin^2(\theta)=1-\cos^2(\theta)\)

\(\displaystyle \sin^2(\theta)=\frac{1-\cos(2\theta)}{2}\)

 

[Drain Brain]

It depends on what you want the identity to be in terms of...two identities that spring to mind are:

\(\displaystyle \sin^2(\theta)=1-\cos^2(\theta)\)

\(\displaystyle \sin^2(\theta)=\frac{1-\cos(2\theta)}{2}\)

I'm familiar with what you've posted. But $\sin^2(\omega t)$ doesn't fit the form of just $sin^2(\theta)$. I'm confused, how am I suppose to use those identities above?

 

[I like Serena]

I'm familiar with what you've posted. But $\sin^2(\omega t)$ doesn't fit the form of just $sin^2(\theta)$. I'm confused, how am I suppose to use those identities above?

Substitute $\theta=\omega t$ to get:

\(\displaystyle \sin^2(\omega t)=1-\cos^2(\omega t)\)

\(\displaystyle \sin^2(\omega t)=\frac{1-\cos(2\omega t)}{2}\)

 

[MarkFL]

I'm familiar with what you've posted. But $\sin^2(\omega t)$ doesn't fit the form of just $sin^2(\theta)$. I'm confused, how am I suppose to use those identities above?

With $\theta=\omega t$ it fits the form exactly. :D