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Cos(一兀/2)等于多少

三角函数公式。

sin2x=cos(2x-pi/2)=2*(3/5)^2-1=-7/25

原题应是:设tana=3,则(sin(a-π)+cos(π-a))/(sin(π/2-a)+cos(π/2+a)) (sin(a-π)+cos(π-a))/(sin(π/2-a)+cos(π/2+a)) =((-sina)+(-cosa))/(cosa-sina) =-(cosa+sina)/(cosa-sina) =-(1+tana)/(1-tana) =-(1+3)/(1-3) =2 希望能帮到你!

解答: tan(兀一a)×cos³(一a一兀)tan(一a一2兀) =(-tana)*(-cosa)³*(-tana) =-tan²a*cos³a =-sin²a/cos²a *cos³a =-sin²a*cosa

说明:此题应该是: 已知0

sin(π-a)=sina cos(2π-a)=cosa tan(-a-π)= - tan(π+a)= - tana f(a)=sina*cosa*(-sina)/(cosa)= - (sina)^2

1. Cos2x=cos(x+x) = cosxcosx-sinxsinx =(cosx)平方 - (sinx)平方 =(cosx + sinx)(cosx-sinx) 2.Tan(π/4+x) = (1+tanx)(1-tanx) = (1+sinx/cosx)(1-sinx/cosx) =((cosx+sinx)/cosx)((cosx-sinx)cosx) =(cosx+sinx)(cosx-sinx)/cosx 所以:1/2 =...

T=6πymax=1ymin=-1

由cos(a+π/6)=5/13 a∈(0,π/2) 所以a+π/6∈(π/6,2π/3), sin(a+π/6)=12/13 cosa=(a+π/6-π/6) =cos(a+π/6)cosπ/6+sin(a+π/6)sinπ/6 =√3/2×5/13+1/2×12/13 =(5√3+12)/26

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