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  • 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
    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)
    沒有看懂……】
    數(shù)學(xué)人氣:538 ℃時(shí)間:2020-01-27 12:00:14
    優(yōu)質(zhì)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)
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