sin2x = (4tanxcos^2x - 2tanx)/1-tan^2x

Question

Prove the trigonometry problem above, it is proving the identity of sin 2x. I know sin2x=2sinxcosx and 1-tan^2x=cos2x/cos^2x.

Answer 1

sin(2x) = {2tan(x)[2cos²(x) - 1)]}/(1 - tan²(x))

sin(2x) = (2tan(x))/(1 - tan²(x))•(2cos²(x) - 1)

Using the double angle identity tan(2x) = 2tan(x)/(1 - tan²(x)) we can replace the first half of the right side with tan(2x):

sin(2x) = tan(2x)•(2cos²(x) - 1)

Using the double angle identity cos(2x) = 2cos²(x) - 1 we can replace the last half of the right side with cos(2x):

sin(2x) = tan(2x)•cos(2x)

Using tan(2x) = sin(2x)/cos(2x) we can replace tan(2x):

sin(2x) = sin(2x)/cos(2x)•cos(2x) and then cancelling cos(2x) on top and bottom we get:

sin(2x) = sin(2x)