Question

How do you verify the identity `sin4x= 4sinxcosx(1-sin^2x)`?

Answer 1

Remember
`sin(2theta) = 2*sin(theta)*cos(theta)`
and
`cos(2theta) = 1 - 2*sin^2(theta)`

So
`sin(4x)`

`= 2*sin(2x)*cos(2x)`

`=2 * (2 * sin(x) * cos(x) )* (1-2 * sin^2(x))`

`=4sin(x)cos(x)(1-sin^2(x))`