a∈（π／2,π）,
所以
sina>0
sina=4/5
cos(a+π/3)
=cosacos(π/3)-sinasin(π/3)
=(-3/5)*(1/2)-(4/5)*(√3/2)
=-3/10-4√3/10
=-3/10-2√3/5
已知cosa=-3\/5,a∈(π\/2,π),求cos(a+三分之π)
cosa=-3／5,a∈（π／2,π）,所以 sina>0 sina=4\/5 cos(a+π\/3)=cosacos(π\/3)-sinasin(π\/3)=(-3\/5)*(1\/2)-(4\/5)*(√3\/2)=-3\/10-4√3\/10 =-3\/10-2√3\/5
已知cosa=-3\/5,a∈(π\/2,π),求sin(a+π\/3)的值
解：∵a∈（π\/2，π）且cosa=-3\/5 ∴sina=√(1-cos²a)=4\/5 于是 sin(a+π\/3)=sinacosπ\/3+cosasinπ\/3 =4\/5×1\/2+(-3\/5)×（√3\/2）=4\/10-3√3\/10 =(4-3√3)\/10
若cosa=-3\/5,a属于(π\/2,π),则cos(a+π\/3)=
cosa=-3\/5,a属于(π\/2,π),, sina=4\/5 cos(a+π\/3)=cosacosπ\/3-sinasinπ\/3=1\/2x(-3\/5)-4\/5x(√3\/2)=-(3+4√3)\/10
己知cosa=-3\/5,a属于(丌\/2,丌),求sin(a+丌\/3)的值?
cosa=-3\/5 a属于(丌\/2,丌)sina=√(1-cos^2a)=4\/5 sin(a+丌\/3)=sinacos(π\/3)+cosasin(π\/3)=4\/5*1\/2-3\/5*√3\/2 = (4-3√3)\/10
已知cosα= -3\/5 ,α∈(π\/2,π),求sin(α+π\/3)的值
由a的取值范围可知 sinа=4\/5, 则sin(а+п\/3)=sinаcos(п\/3)+cosаsin(п\/3)则可得为 (2\/5-3√3\/10)希望回答能够帮助到你~
已知cos a=-3\/5,a属於(兀\/2,兀),求sin (a+兀\/3)的值
你参考看看～
已知cosα=负的五分之三,α属于(π\/2,π),求sin(α+π\/3)的值
cosα=-3\/5 α属于(π\/2,π),sinα=根号（1-cos^2）=4\/5 sin(α+π\/3)=sinacosπ\/3+cosasinπ\/3 =4\/5x1\/2-3\/5x根号3\/2 =(4-3根号3)\/10
cos⊙=-3\/5,⊙∈(π\/2,π),求cos(⊙+π\/3)的值
sinθ=4\/5 所以cos(θ+π\/3)= =-3\/5 ×1\/2- 4\/5×√3 \/2 =- （3+4√3）\/10
已知cosa=-3\/5,a属于(π\/2,π),求sin2a,cos2a,tan2a.求详细过程和解题思...
cosa=-3\/5,a属于（π\/2,π)sina=4\/5 sin2a =2sinacosa =2*(4\/5)*(-3\/5)=-24\/25 cos2a =2cos^2a-1 =2*(-3\/5)^2-1 =18\/25-1 =-7\/25 tan2a =sin2a\/cos2a =(24\/25)\/(-7\/25)=-24\/7
已知cosα=-3\/5,α∈(π\/2),求cos(π\/4-α)的ŀ
解 由cosα=-3\/5, α∈(π\/2, π)可知：sinα=4\/5, cosα=-3\/5 又sin(π\/4)=cos(π\/4)=(√2)\/2 ∴结合三角函数诱导公式可得：cos(π\/4-α)=cos(π\/4)cosα+sin(π\/4)sinα =[(√2)\/2][cosα+sinα]=(√2)\/10 ...
