Can someone tell me if this is correct

sin(2t)cos(2t) = 1/4sin(4t)

This is my reasoning

sin(2t) = 2sin(t)cos(t) yes?

sin(4t) = 4sin(t)cos(t) yes?

therefore 1/4sin(4t) = sin(t)cos(t)

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- Apr 4th 2009, 09:23 PMpberarditrig identity
Can someone tell me if this is correct

sin(2t)cos(2t) = 1/4sin(4t)

This is my reasoning

sin(2t) = 2sin(t)cos(t) yes?

sin(4t) = 4sin(t)cos(t) yes?

therefore 1/4sin(4t) = sin(t)cos(t) - Apr 4th 2009, 09:29 PMChris L T521
You're almost there...but you're answer is incorrect.

Note that $\displaystyle \sin\left(2t\right)\cos\left(2t\right)\sim\sin\alp ha\cos\alpha$.

From the double angle identity, $\displaystyle 2\sin\alpha\cos\alpha=\sin\left(2\alpha\right)$. Thus, $\displaystyle \sin\alpha\cos\alpha=\tfrac{1}{2}\sin\left(2\alpha \right)$.

Since we let $\displaystyle \alpha=2t$, we now see that $\displaystyle \sin\left(2t\right)\cos\left(2t\right)=\tfrac{1}{2 }\sin\left(4t\right)$