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(αsin2α=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……】
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/28 22:00:34
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(αsin2α=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……】
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……】
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(αsin2α=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……】
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)