tan(13π/3)+sin(-5π/2)+cos(-23π/6)
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/08/07 19:23:13
tan(13π/3)+sin(-5π/2)+cos(-23π/6)
原式=tan(13π/3) - sin(5π/2) + cos(23π/6)
=tan(4π + π/3) - sin(2π + π/2) + cos(4π - π/6)
=tan(π/3) - sin(π/2) + cos(π/6)
=√3 - 1 + √3/2
=(3√3)/2 - 1
=tan(4π + π/3) - sin(2π + π/2) + cos(4π - π/6)
=tan(π/3) - sin(π/2) + cos(π/6)
=√3 - 1 + √3/2
=(3√3)/2 - 1
tan(13π/3)+sin(-5π/2)+cos(-23π/6)
cosπ/3+tanπ/4+3tan²π/6+sinπ/2+cosπ+sin3π/2等于?.
化简:[sin(π+α)*cos(π-α)*tan(π-α)]/[cos(π/2+α)*tan(3π/2-α)*tan(
已知tanθ=3,求(3cosθ-5sin^2θcosθ)/sin(π-θ)的值
已知tan(α+π/4)=3,求cosα+2sinα/cosα-sinα的值
已知tan(3π+α)=2,求:1、(sinα+cosα)²;2、sinα-cosα/2sinα+cosα
已知tan(π-α)=2,求sinα-2sinαcosα-cosα/4cosα-3sinα的值
tan(a-4π)cos(π+a)sin^2(a+3π)/tan(3π+a)cos^2(2/5π+2)=0.5,求tan
化简sin³(﹣α)cos(2π+α)tan(2π-α)/sin(α-2π)cos(α-3π/2)-tan(π
求证tan(2π-α)sin(-2π-α)cos(6π-α)/cos(α-π)sin(5π-α)=-tanα
求证(tan(2π-θ)sin(2π-θ)cos(6π-θ))/((-cosθ)sin(5π+θ))=tanθ
已知tanα/2=2,1.求tan(α+π/4)的值 2.(6sinα+cosα)/(3sinα-2cosα)的值