2sinx -sin2x/2sinx+sin2x=tg^2x/2

Question

2sinx -sin2x/2sinx+sin2x=tg^2x/2

Answer 1

Решение:
(2sinХ-Sin2X)/((2sinХ+Sin2X) = (2sinx-2SinXCosX)/(2sinx+2SinXCosX) =
2sinx(1-cosx)/2sinx(1+cosX) = (1-cos(2x/2))/(1+cos(2x/2)) =
(1-cos^2(x/2)+sin^2(x/2))/(1+cos^2(x/2)-sin^2(x/2)) =
2sin^2(x/2)/2cos^2(x/2) = tg^2(x/2)