已知sinα+cosα=3根号5/5,a∈(0,π/4),sin(β-π/4)=5分之3,β∈(π/4,π/2)
求sin2a和tan2a的值
求cos(a+2β)的值
人气:314 ℃ 时间:2019-11-12 06:27:02
解答
(1)
sinα+cosα=3√5/5,①a∈(0,π/4)
两边平方
sin²α+cos²α+2sinαcosα=9/5
∴1+sin2α=9/5
sin2α=4/5
∵a∈(0,π/4) ∴2α(0,π/2)
∴cos2α=√(1-sin²2α)=3/5
tan2α=sin2α/cos2α=4/3
(2)
(cosα-sinα)²=1-sin2α=1/5
a∈(0,π/4),cosα-sinα>0
∴cosα-sinα=√5/5②
①②==>cosα=2√5/5,sinα=√5/5
sin(β-π/4)=3/5,β∈(π/4,π/2)
sin2β=cos(2β-π/2)=1-2sin²(β-π/4)=7/25
又2β∈(π/2,π)
∴cos2β=-24/25
∴cos(a+2β)
=cosacos2β-sinasin2β
=2√5/5*(-24/25)-√5/5*7/25
=-11√5/25
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