sin2α=sin^2(α+π/4)-cos^2(α+π/4)=2sin^2(a+π/4)-1=1-2cos^2(α+π/4); _数学解答

sin2α=sin^2(α+π/4)-cos^2(α+π/4)=2sin^2(a+π/4)-1=1-2cos^2(α+π/4); cos2α=2sin(α+π/4)cos(α
sin2α=sin^2(α+π/4)-cos^2(α+π/4)=2sin^2(a+π/4)-1=1-2cos^2(α+π/4)
cos2α=2sin(α+π/4)cos(α+π/4)
没有看懂……】

人气：412 ℃　时间：2020-01-27 12:00:14

解答

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)