Question

Heyy,

how do you get from:

Asin(3wt + b) * Bsin(wt)
to
(AB/2)[cos(2w - b) - cos(4wt + b)]

Thank you x

Answer 1

Asin(3wt + b) * Bsin(wt)=AB [ sin(3wt + b) *sin(wt)]
=AB/2[2 sin(3wt + b) *sin(wt)]
=(AB/2)[cos(2w - b) - cos(4wt + b)]

Answer 2

how did you get from the 2nd line to the third line?
ta x

Answer 3

Here is a list of trig identities.
http://www.sosmath.com/trig/Trig5/trig5/trig5.html
and attached is a short derivation for sin(a)sin(b)