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为什么%tAn11/4π=tAnπ/4

-tan(11π/4) =-tan(3π-π/4) =-tan(-π/4) =tan(π/4)

sin0-cos(-7π/3)+tan(-5π/4)+cosπ/2+sin11π/6+tanπ =sin0-cos(π/3)-tan(π/4)+cosπ/2-sin1π/6+tan0 =0-1/2-1+0-1/2+0 =-2

你好!!! 依题意有,用韦达定律 tanα+tan(π/4-α)=-P,tanαtan(π/4-α)=q. ∴tan[α+(π/4-α)]=[tanα+tan(π/4-α)]/[1-tanαtan(π/4-α)] →tan(π/4)=-p/(1-q) →p-q=-1 …… (1) 而p+q=11 …… (2) 解(1)、(2)得, p=5,q=6. 希望能够帮助你!!!

cos11π/4+tan(-7π/6)+sin(-π)+cos(-5π/6) =cos3π/4+tan(-π/6)+sin(π)+cos(5π/6) =-√2/2-√3/3+0-√3/2 =-√2/2-5√3/6

这是三角比三角函数sin是三角形的正弦 cos是三角形的余弦 tan是三角形的正切 万能公式 sin(α+β)=sinαcosβ+cosαsinβ sin(α-β)=sinαcosβ-cosαsinβ cos(α+β)=cosαcosβ-sinαsinβ cos(α-β)=cosαcosβ+sinαsinβ tanα+tanβ tan(α...

1、sin(-4/5π)cos(-5/3π)-tan(-11/6π)-(cos7/6 π×3tan4/3 π)/sin2/3 π =-sin(4/5π)cos(5/3π)+tan(11/6π)-(cos7/6 π×3tan4/3 π)/sin2/3 π=-sin(1/5π)cos(1/3π)-tan(1/6π)+(cos1/6 π×3tan1/3 π)/sin1/3 π=-(1/2)sin(1/5π)-3...

看它们位于哪个象限,再根据三角函数的定义判定

利用正切函数的单调性比较下列各组中两个函数值的大小 (1)tan(-π/5)与tan(-3π/7): (2)tan1519°与tan1493° (3)tan(6又9/π11)与tan(-...

tanα+1/tanα=-10/3,tana=-3,或tana=-1/3 [5sin²a/2+8sina/2 cosa/2+11cos²a/2-8]/√2sin(a-π/4) =[5sin²a/2+5cos²a/2+8sina/2 cosa/2+6cos²a/2-8]/√2[sinacos(π/4)-cosasin(π/4)] =[-3+4sina+6cos²a/2]/(sina-cos...

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