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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

∵sin(π/4+a)sin(π/4-a)={cos[(π/4+a)-(π/4-a)]-cos[(π/4+a)+(π/4-a)]}/2=(cos2a-cosπ/2)/2=(cos2a)/2=1/6∴cos2a=1/3又∵a∈(π/2,π),则2a∈(π,2π)∴sin2a=-√(1-cos²2a)=-√(1-1/9)=-(2√2)/3∴sin4a=2sin2acos2a=2×(-2√2)/3×1/3=-(4√2)/9

sin(π/6-α)=1/4, 则cos(π/3-2α)=cos[2(π/6-α)] =1-2 sin²(π/6-α) =7/8. 所以cos(2α-π/3) =7/8. sin(π/6+2α)= sin((2α-π/3)+ π/2) = cos(2α-π/3)==7/8. 如果你认可我的回答,敬请及时采纳, ~如果你认可我的回答,请及时点击【采纳为满意回...

sin(π/6-α)=1/4, 则cos(π/3-2α)=cos[2(π/6-α)] =1-2 sin²(π/6-α) =7/8. 所以cos(2α-π/3) =7/8. sin(π/6+2α)= sin((2α-π/3)+ π/2) = cos(2α-π/3)==7/8.

sin(π/4+x)sin(π/4-x)=sin(π/4+x)cos[π/2-(π/4-x)]=sin(π/4+x)cos(π/4+x)=1/2*2sin(π/4+x)cos(π/4+x)=1/2*sin(π/2+2x)=1/2*cos2x=1/6从而cos2x=1/3;又x∈(π/2,π),则2x∈(π,2π);则sin2x

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

sin(π/6+2a)=sin(π/2-π/3+2a)=cos(π/3-2a)=1-2sin^2(π/6-a)=1-2(1/16)=7/8

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:∵α∈(π/2,π) ∴cosα

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