C crex New Member Joined Apr 13, 2013 Messages 28 Gender Male HSC N/A Aug 17, 2013 #1 Hey guys, how do you prove this? Thanks! Last edited: Aug 17, 2013
Solution BoS Revolutionary Joined Aug 3, 2012 Messages 120 Gender Undisclosed HSC N/A Aug 17, 2013 #2 A little rusty eheh but... Sin2x = 2SinxCosx So if you consider: Sin2[(a-b)/2] = 2Sin[(a-b)/2]Cos[(a-b)/2] Therfore: Sin(a-b) = (as above)
A little rusty eheh but... Sin2x = 2SinxCosx So if you consider: Sin2[(a-b)/2] = 2Sin[(a-b)/2]Cos[(a-b)/2] Therfore: Sin(a-b) = (as above)
C crex New Member Joined Apr 13, 2013 Messages 28 Gender Male HSC N/A Aug 17, 2013 #3 Solution said: A little rusty eheh but... Sin2x = 2SinxCosx So if you consider: Sin2[(a-b)/2] = 2Sin[(a-b)/2]Cos[(a-b)/2] Therfore: Sin(a-b) = (as above) Click to expand... oh tyty
Solution said: A little rusty eheh but... Sin2x = 2SinxCosx So if you consider: Sin2[(a-b)/2] = 2Sin[(a-b)/2]Cos[(a-b)/2] Therfore: Sin(a-b) = (as above) Click to expand... oh tyty
B Bobbo1 Member Joined Apr 19, 2011 Messages 971 Gender Male HSC 2010 Aug 17, 2013 #4 yep simple double angle
