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

解:∵α∈(π/2,π) ∴cosα

∵ sin( π 6 +2α) =cos[ π 2 -( π 6 +2α) ]=cos( π 3 -2α )=cos2( π 6 -α )=1-2 sin 2 ( π 6 -α) = 7 8 故答案为: 7 8

cos[π/2-(π/6+α)]=sin(π/6+α)=1/3 所以cos(π/3-α)=1/3 利用倍角公式 则cos2(π/3-α)=cos(2π/3-2α)=2[cos(π/3-α)]^2-1=2*(1/9)-1=-7/9

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

解: ∵π/3<a<π ∴π/2<a+π/6<7π/6 ∵sin(a+π/6)=1/3, ∴π/2<a+π/6<π, ∴cos(a+π)=-√[1-sin²(a+π/6)]=-2√2/3, sin(π/12-a) =sin[π/4-(a+π/6)] =ain(π/4)cos(a+π/6)-cos(π/4)sin(a+π/6) =√2/2×(-2√2/9)-√2/2×1/3 =-2/9-√2/6 =-(4...

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-√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

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 ...

f(x)=cosxsin(x+π/6)-cos2x-1/4, =cosx(√3/2sinx+1/2cosx)-cos2x-1/4, =√3/2sinxcosx+1/2(cosx)^2-cos2x-1/4, =√3/4sin2x+1/4(1+cos2x)-cos2x-1/4, =√3/4sin2x-3/4cos2x =√3/2(1/2sin2x-√3/2cos2x) =√3/2sin(2x-π/3) 2x-π/3∈[2kπ-π/2,2kπ+π/2]单...

cos(π/6-a)=1/3 展开得到√3/2 *cosa +1/2 *sina=1/3 故sina=2/3 -√3 cosa 平方即(sina)^2=4/9 -4/√3 cosa+3(cosa)^2 所以0=-5/9 -4/√3 cosa+4(cosa)^2 解得cosa=(4√3 +8√2)/24 或(4√3 -8√2)/24 cos(π/6+a)+sin(2π/3-a) 展开得到 √3/2 *cosa -1/2...

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