trig functions (1 Viewer)

[NizDiz]

how do u get to sin^2 (x) = 1/2 (1 - cos 2theta)?

 

[ocatal]

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]

Assuming x = theta,
sub in cos(2ø) = 1 - 2sin^2 (ø)
then it just rolls out

 

[NizDiz]

my bad with the x/theta bit. Thanks for helping guys

 

[NizDiz]

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??