Question

三角函数 (4 19:19:25)

Answer 1

sin2x

=2sinx*cosx

=2sinx/cosx*cos^2 x

=2tanx/(1/cos^2x)

=2tanx/[(sin^2x+cos^2x)/cos^2x]

=2tanx/(1+tan^2x)

2tanx/（1+tan^2x） =3/5

sin2x=3/5

Cos2A=CosA^2-SinA^2=1-2SinA^2

sin^2(π/4 +x)

=[1-cos2(π/4 +x)]/2

=[1-cos(π/2 +2x)]/2

=[1+3/5]/2

=4/5