Question

sin2α=sin^2(α+π/4)-cos^2(α+π/4)=2sin^2(a+π/4)-1=1-2cos^2(α+π

Answer 1

1、sin2α=(1/2)[(1+sin2α)-(1-sin2α)]

=(1/2)[(sinα+cosα)^2-(sinα-cosα)^2]

=(1/2){[√2sin(α+π/4)]^2-[√2cos(α+π/4)]^2}

=[sin(α+π/4)]^2-[cos(α+π/4)]^2

=-cos2(α+π/4)

=2[sin(α+π/4)]^2-1

=1-2[cos(α+π/4)]^2

2、cos2α=(cosα)^2-(sinα)^2

=(cosα+sinα)(cosα-sinα)

=2sin(α+π/4)cos(α+π/4)