已知cosα=1/7,cos(α-β)=13/14,且0< β <α<π/2
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/24 00:36:31
已知cosα=1/7,cos(α-β)=13/14,且0< β <α<π/2
cos(π+2α)tan(π-2α)sin(π/2-2α)
(1)求 ---------------------------------------- 的值
cos(π/2+2α)
(2)求角β
cos(π+2α)tan(π-2α)sin(π/2-2α)
(1)求 ---------------------------------------- 的值
cos(π/2+2α)
(2)求角β
(1)cos(π+2α)tan(π-2α)sin(π/2-2α)/cos(π/2+2α)
=[-cos(2α)][-tan(2α)]cos(2α)/[-sin(2α)] (应用诱导公式)
=-cos²(2α)[sin(2α)/cos(2α)]/sin(2α)
=-cos(2α)
=1-2cos²α (应用倍角公式)
=1-2*(1/7) (∵cosα=1/7)
=5/7;
(2)∵cosα=1/7,cos(α-β)=13/14,且0< β <α<π/2
∴sinα=√(1-cos²α)=4√3/7,sin(α-β)=√(1-cos²(α-β))=3√3/14
∵sinβ=sin(α-(α-β))
=sinαcos(α-β)-cosαsin(α-β)
=(4√3/7)(13/14)-(1/7)(3√3/14)
=√3/2
∴β=π/3.
=[-cos(2α)][-tan(2α)]cos(2α)/[-sin(2α)] (应用诱导公式)
=-cos²(2α)[sin(2α)/cos(2α)]/sin(2α)
=-cos(2α)
=1-2cos²α (应用倍角公式)
=1-2*(1/7) (∵cosα=1/7)
=5/7;
(2)∵cosα=1/7,cos(α-β)=13/14,且0< β <α<π/2
∴sinα=√(1-cos²α)=4√3/7,sin(α-β)=√(1-cos²(α-β))=3√3/14
∵sinβ=sin(α-(α-β))
=sinαcos(α-β)-cosαsin(α-β)
=(4√3/7)(13/14)-(1/7)(3√3/14)
=√3/2
∴β=π/3.
已知cosα=1/7,cos(α+β)=-11/14,且α、β∈(0,2/Π),求cosβ的值
已知cos(π/2-α)-2cosα=sinπ,且cosα<0,则sinα的值为?
已知sin(3π-α)=√2cos(3π/2+β)和√3cos(-α)=-√2cos(π+β),且0<α<π,0<β<π
已知0<α<π/2,且3sinα=4cosα求(sin^2α+2sinαcosα)/(3cos^2α-1)求cos^2α
已知π/2<β<α<3π/4,且cos( α-β)=12/13,cos(α+β)=-4/5,求sin2α
已知sin(3π-α)=根号二cos(3π/2+β),根号三cos(-α)=负根号二cos(π+β),且0
已知COS(2α-β)=-11/14 SIN(α-2β)=4倍根号3/7 且π/4<α<π/2,0<β<π/4 求COS
已知sin(5π-α)=根号2×cos[(2÷7)+β]和根号3×cos(-α)=根号2×cos(π+α),且0<α<π
怎么求cosβ的值已知cosα=1/7,cos(α+β)=-11/14,且α∈(0,π/2),α+β∈(π/2,π),求
已知cos(π/3+α)=-3/5 ,sin(2π/3-β)=5/13,且0<α<π/2<β<π,求cos(β-α).
1、已知cos(π/3+α)= -3/5,sin(2π/3-β)=3/15,且0<α<π/2<β<π,求cos(β-α)
已知α.β均为锐角,且cos(α+β)=12/13,cos(2α+β)=3/5,求cosα.