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).
来源:学生作业帮 编辑:搜搜考试网作业帮 分类:数学作业 时间:2024/06/05 05:26:42
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).
0<α<π/2
0+π/3<α+π/3<π/2+π/3
π/3<α+π/3<5π/6
cos(π/3+α)=-3/5 ,
sin(π/3+α)=4/5 ,
π/2+π/3<β+π/3<π+π/3
5π/6<β+π/3<4π/3
cos(π/3+α)=-3/5 ,
sin(2π/3-β)=5/13
sin(π-π/3-β)=5/13
sin(π/3+β)=5/13
cos(π/3+β)=-12/13
cos(β-α)
=cos[π/3+β-(π/3+α)]
=cos(π/3+β)cos(π/3+α)+sin(π/3+β)sin(π/3+α)
=-12/13*(-3/5)+5/13*4/5
=36/65+20/65
=56/65
0+π/3<α+π/3<π/2+π/3
π/3<α+π/3<5π/6
cos(π/3+α)=-3/5 ,
sin(π/3+α)=4/5 ,
π/2+π/3<β+π/3<π+π/3
5π/6<β+π/3<4π/3
cos(π/3+α)=-3/5 ,
sin(2π/3-β)=5/13
sin(π-π/3-β)=5/13
sin(π/3+β)=5/13
cos(π/3+β)=-12/13
cos(β-α)
=cos[π/3+β-(π/3+α)]
=cos(π/3+β)cos(π/3+α)+sin(π/3+β)sin(π/3+α)
=-12/13*(-3/5)+5/13*4/5
=36/65+20/65
=56/65
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).
已知0<α<π/2,且3sinα=4cosα求(sin^2α+2sinαcosα)/(3cos^2α-1)求cos^2α
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cos(α-β/2)=-3/5,sin(α/2-β)=12/13,且α∈(π/2,π),β∈(0,π/2),求cos
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<β<π,sinα=3/5,cos(α+β)=-4/5求cosβ
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
已知cos(π/2-α)-2cosα=sinπ,且cosα<0,则sinα的值为?
已知α∈(0,π/2),且2sinα-sinαcosα-3cosα=0.求[sin(α+π/4)]/[sin2α+cos
已知α∈(0,π/2),且2sin²α-sinαcosα-3cos²α=0,求[sin(α+π/4)
已知tanα=2,sinα+cosα<0,求[sin(2π-α)*sin(π+α)*cos(-π+α)]/[sin(3π