4sin^2(x/2)cos^2(x/2) = sin^2(x)
thanks a lot!
2007-12-06 04:32:28 · 3 answers · asked by Nate 6 in Science & Mathematics ➔ Mathematics
whoa hold on how did you go from
= [2 sin(x/2)cos(x/2) ]^2
to
= [ sin (x/2 + x/2) ]^2
???
2007-12-06 04:45:14 · update #1
Use half angle formulas
sin(x/2) = +/- sqrt((1-cosx)/2)
cos(x/2) = +/- sqrt((1+cosx)/2)
Then sin^2(x/2) = (1-cosx)/2 and cos^2(x/2) = (1+cosx)/2
Their product is 1-cos^2x /4 = sin^2x/4
4*sin^2x/4 = sin^2x
sin^2x = sin^2x
2007-12-06 04:56:01 · answer #1 · answered by ironduke8159 7 · 0⤊ 0⤋
sin^2(x)
= [2sin(x/2)cos(x/2)]^2
= 4sin^2(x/2)cos^2(x/2)
------
Ideas to save time: You go from the right side to the left side.
2007-12-06 12:39:14 · answer #2 · answered by sahsjing 7 · 0⤊ 0⤋
4sin^2(x/2)cos^2(x/2)
= [2 sin(x/2)cos(x/2) ]^2
= [ sin (x/2 + x/2) ]^2
= sin^2(x)
QED
2007-12-06 12:39:09 · answer #3 · answered by harry m 6 · 0⤊ 0⤋