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)
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 $\displaystyle \sin\left(2t\right)\cos\left(2t\right)=\tfrac{1}{2 }\sin\left(4t\right)$