# trig identity

• Apr 8th 2008, 07:50 PM
Thelema
trig identity
sin(t+s)sin(t-s)=sin^2t-sin^2s
• Apr 8th 2008, 07:54 PM
Mathstud28
here is a long-winded reason
Quote:

Originally Posted by Thelema
sin(t+s)sin(t-s)=sin^2t-sin^2s

if you use your identitys you cansee that we have $[sin(t)cos(s)+sin(s)cost(t)][sin(t)cos(s)-sin(s)cos(t)]$ I think you can see where to go fomr there