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已知sin(π/4+α)sin(π/4%α)=1/6 α∈(π/2,π),求tAn4α

sin(π/4+α)sin(π/4-α)=[sin(π/4)cosa+cos(π/4)sina][sin(π/4)cosa-cos(π/4)sina] =[sin(π/4)cosa]^2-[cos(π/4)sina]^2 =1/2[(cosa)^2-(sina)^2]=1/6 (cosa)^2-(sina)^2=1/3 而(cosa)^2+(sina)^2=1 所以(cosa)^2=2/3 而α∈(π,3/2π) cosα

解:∵α∈(π/2,π) ∴cosα

又是你。。。

sin(π/4-a)=cos[π/2-(π/4-a)]=cos(π+a) sin(π/4+a)sin(π/4-a)=sin(π/4+a)cos(π/4+a)=sin(π/2+2a)/2 所以sin(π/2+2a)=-cos2a=-1/3 sin2a=2倍根号2/3 sin4a=2sin2acos2a=-4倍根号2/9

-√6/3 根据公式:(cosx-sinx)²=1-2sinxcosx=1-1/3=2/3 所以 cosx-sinx=+-√6/3 因为 π/4

∵ α∈(0, π 2 ) ,sinα= 1 2 ∴α= π 6 ∴ cos(α+ π 6 ) =cos π 3 = 1 2 故答案为: 1 2

因为cos(α+π/6)=sin(α+π/6+π/2)=sin(α+2π/3) 所以sin(α+2π/3)*cos(α+π/6)=sin(α+2π/3)*sin(α+2π/3)=1/6 cos( 2α+4π/3)=1-2sin(α+2π/3)*sin(α+2π/3)=2/3 cos( 2α+4π/3)=cos2αcos4π/3-sin2αsin4π/3=-1/2cos2α+根号3/2sin2α=2/3 ...

由已知得cosa√3/2+sina*1/2+sina=4/5*√3,cosa*1/2+sina*√3/2=4/5,cosasinπ/6+sinacosπ/6=4/5,sin(a+π/6)=4/5

解:1.原式=(1+tanα/2)(1-sinα)/(1-tanα/2)=(cosα/2+sinα/2)(1-2sinα/2cosα/2)/(sinα/2-cosα/2) = (cosα/2+sinα/2)(sinα/2-cosα/2)??/(sinα/2-cosα/2)=(cosα/2+sinα/2)(sinα/2-cosα/2) =sin??α/2-cos??α/2=-cosα 2.原式=sin6°sin42°sin66°cos12...

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