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sin4π/3cos5π/6tan(-4π/3)怎么做,求详细过程

来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/07/05 11:38:32
sin4π/3cos5π/6tan(-4π/3)怎么做,求详细过程
sin4π/3乘以cos5π/6乘以tan(-4π/3)=?
sin cos的周期为2π tan周期为π
sin4π=sin(4π-4π)=sin0=0
3cos5π=3cos(5π-4π)=3cosπ=3*(-1)=-3
6tan(-4π/3)=6tan(-4π/3+π)=6tan(-π/3)=-6tan(π/3)=-6√3
再问: sin4π/3乘以cos5π/6乘以tan(-4π/3)=?
再答: sin(π/2+α)=cosα cos(π/2+α)=-sinα sin4π/3=sin(4π/3-2π)=sin(-2π/3)=-sin(2π/3)=-cos(2π/3-π/2)=-cos(π/6)=-√3/2 cos5π/6=-cos(5π/6-π)=-cos(-π/6)=-cosπ/6=-√3/2 tan(-4π/3)=tan(-4π/3+π)=tan(-π/3)=-tan(π/3)=-√3 sin4π/3乘以cos5π/6乘以tan(-4π/3)=-√3/2*(-√3/2)*(-√3)=-3√3/4 tanα ·cotα=1 sinα ·cscα=1 cosα ·secα=1 sinα/cosα=tanα=secα/cscα cosα/sinα=cotα=cscα/secα sin2α+cos2α=1 1+tan2α=sec2α 1+cot2α=csc2α 诱导公式 sin(-α)=-sinα cos(-α)=cosα tan(-α)=-tanα cot(-α)=-cotα sin(π/2-α)=cosα cos(π/2-α)=sinα tan(π/2-α)=cotα cot(π/2-α)=tanα sin(π/2+α)=cosα cos(π/2+α)=-sinα tan(π/2+α)=-cotα cot(π/2+α)=-tanα sin(π-α)=sinα cos(π-α)=-cosα tan(π-α)=-tanα cot(π-α)=-cotα sin(π+α)=-sinα cos(π+α)=-cosα tan(π+α)=tanα cot(π+α)=cotα 背公式,多做题, 不懂请追问,有帮助请采纳,谢谢.