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

AI Thread Summary
The discussion centers on finding equivalent identities for the expression $\sin^2(\omega t)$. Two key identities are highlighted: $\sin^2(\theta) = 1 - \cos^2(\theta)$ and $\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}$. Participants clarify that by substituting $\theta = \omega t$, these identities can be directly applied to $\sin^2(\omega t)$. This substitution allows for the expressions $\sin^2(\omega t) = 1 - \cos^2(\omega t)$ and $\sin^2(\omega t) = \frac{1 - \cos(2\omega t)}{2}$. The conversation emphasizes the importance of recognizing the substitution to utilize the identities effectively.
Drain Brain
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is there an equivalent identity for $\sin^2\omega t?$

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

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

$$\sin^2(\theta)=\frac{1-\cos(2\theta)}{2}$$
 
MarkFL said:
It depends on what you want the identity to be in terms of...two identities that spring to mind are:

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

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

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?
 
Drain Brain said:
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:

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

$$\sin^2(\omega t)=\frac{1-\cos(2\omega t)}{2}$$
 
Drain Brain said:
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
 
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