N NizDiz Member Joined Nov 5, 2011 Messages 294 Gender Male HSC 2013 Aug 6, 2013 #1 how do u get to sin^2 (x) = 1/2 (1 - cos 2theta)?
O ocatal Active Member Joined Sep 1, 2012 Messages 298 Gender Male HSC 2013 Aug 6, 2013 #2 Shouldn't it be either x or theta. If so, sin^2 (x) = 1 / 2 (1 - (1 - 2sin^2 (x)) sin^2 (x) = 1 / 2 (1 -1 + 2sin^2 (x)) sin^2 (x) = 1 / 2 (2sin^2 (x)) sin^2 (x) = sin^2 (x)
Shouldn't it be either x or theta. If so, sin^2 (x) = 1 / 2 (1 - (1 - 2sin^2 (x)) sin^2 (x) = 1 / 2 (1 -1 + 2sin^2 (x)) sin^2 (x) = 1 / 2 (2sin^2 (x)) sin^2 (x) = sin^2 (x)
talcum New Member Joined Aug 6, 2013 Messages 8 Gender Male HSC 2013 Aug 6, 2013 #3 Assuming x = theta, sub in cos(2ø) = 1 - 2sin^2 (ø) then it just rolls out
N NizDiz Member Joined Nov 5, 2011 Messages 294 Gender Male HSC 2013 Aug 7, 2013 #4 my bad with the x/theta bit. Thanks for helping guys
N NizDiz Member Joined Nov 5, 2011 Messages 294 Gender Male HSC 2013 Aug 7, 2013 #5 Also, can't you evaluate sin^2 x in terms of sin x, like 1/2 (1- sin2x) or something?? How do u get that??
Also, can't you evaluate sin^2 x in terms of sin x, like 1/2 (1- sin2x) or something?? How do u get that??
