Question

Verify identity:
sin4t=4sintcost(1-2sinˆ2t)
Got as far as:
LHS
sin(2x2t)=
2sin2tcos2t=
and i'm stuck....

Answer 1

Very hard to tell what the original problem is.

Answer 2

\[\sin 4t=4\sin t \cos t(1-2\sin ^2t)\]

Answer 3

Is that it?

Answer 4

yes

Answer 5

that is the original problem

Answer 6

\[\sin 4t=\sin (2t+2t)=\sin 2t \cos 2t+\cos 2t \sin 2t=2\sin 2t \cos 2t=\]
\[2(2\sin t \cos t)(1-2\sin ^2t)=4\sin t \cos t(1-2\sin ^2t)\]