Question

show that: sin(4θ=(cosθ)(4sinθ-8sin^3θ

Answer 1

\[\cos \theta(4\sin \theta -8\sin^{3} \theta)\]
\[=4\sin \theta \cos \theta (1-2\sin^{2} \theta)\]
cos double angle identity
\[=4\sin \theta \cos \theta \cos (2\theta)\]
sine double angle identity
\[=2\sin (2\theta) \cos(2\theta)\]
\[=\sin (4\theta)\]

Answer 2

Thank you

Answer 3

yw :)