# Math Help - trig identity

1. ## trig identity

sin(t+s)sin(t-s)=sin^2t-sin^2s

2. ## here is a long-winded reason

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