一叶飞棕
幼苗
共回答了20个问题采纳率:95% 举报
根据倍角公式,cosx=2(cosx/2)^2-1,(cosx/2)^2=(1+cosx)/2,sinx=2sin(x/2)cos(x/2)
∴f(x)=((1+cosx)/2-(1/2)sinx-1/2
=(1/2)(cosx-sinx).(1)
=(√2/2)[(√2/2)cosx-(√2/2)sinx]
=(√2/2)cos(π/4+x),
f(α)=(√2/2)cos(π/4+α)=3√2/10,
cos(π/4+α)=3/5,
根据余弦倍角公式,
cos[2(π/4+α)]=2*(3/5)^2-1=18/25-1=-7/25,
cos[2(π/4+α)]=cos(π/2+2α)=-cos[π-(π/2+2α)]=-cos(π/2-2α)=-sin2α,
∴sin2α=7/25,
为什么这样复杂?
从(1)式(1/2)(cosα-sinα)=3√2/10,
两边平方,1-sin2α=18/25,即可得sin2α=7/25,
两边同加1,1+sin2α=32/25,
(sinα)^2+2sinαcosα+(cosα)^2=32/25,
(sinα+cosα)^2=32/25,
sinα+cosα=±4√2/5,.(2)
cosα-sinα=3√2/5.(3)
(2)-(3)式,
2sinα=√2/5,sinα=√2/10,
或,2sinα=-7√2/5,sinα=-7√2/10.
1年前
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