作业帮cos(α-55)cos(α+5)+sin(α-55)sin（α+5）化简
来源：学生作业帮助网 编辑：作业帮 时间：2024/07/17 13:06:37
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cos(α-55)cos(α+5)+sin(α-55)sin（α+5）化简
cos(α-55)cos(α+5)+sin(α-55)sin（α+5）化简
cos(α-55)cos(α+5)+sin(α-55)sin（α+5）化简
cos(α-55)cos(α+5)+sin(α-55)sin（α+5）
=(cosacos55+sinasin55)(cosacos5-sinasin5)+(sinacos55-cosasin55)(sinacos5+cosasin5)
=(cosacos55+sinasin55)(cosasin55-sinacos55)+(sinacos55-cosasin55)(sinasin55+cosacos55)
=(cosacos55+sinasin55)(cosasin55-sinacos55+sinacos55-cosasin55)
=(cosacos55+sinasin55)*0
=0
cosα
求证：|cosα+cosβ|
5cos^2a+4cos^2β=4cosα则cos^2a+cos^2β
求和：cos(0)+cos(α)+cos(2α)+cos(3α)+……cos(nα)
5cos^2α=1 cosα=?
cos(α+5π/12)cos(α+π/6)+cos(π/12-α)cos(π/3-α) 化简
若sinα+cosβ=2/3 cosα+cosβ=5/6求cos(α-β)
[sin(5π-α)cos(-π-α)]/[cos(α-π)cos(π/2+α)]化简
cos(α-55)cos(α+5)+sin(α-55)sin（α+5）化简
化简:cos(60-α)-cos(60+α)
已知5cos²α+4cos²β=4cosα则cos²α+cos²β范围
5cos^2 α +4cos^2 β =4cosα,求cos^2 α +cos^2 β的取值范围?
已知5(cosα)^2+4(cosβ)^2=4cosα,则(cosα)^2+(cosβ)^2的取值范围是?
cosα*tanα
sin(α+π/3)+sinα=负5分之4根号3 α∈（-π/2,0）求cosα怎样从9/4*sin²α=9/4(1-cos²α)=3/4*cos²a+12/5*cosx+48/25化为cos²α+4/5*cosα-11/100=0呀sin(α+π/3)+sinα=-4√3/5sinαcosπ/3+cosαsinπ/3+sinα=-4√3/53/2*si
若cosα
证明cos(α+β)
化简:cos(-π-α)