高一数学三角部分，求解题过程……谢谢!
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cos(π/9)cos(2π/9)cos(4π/9)=sin(π/9)*[cos(π/9)cos(2π/9)cos(4π/9)]/sin(π/9)
=1/2*[sin(2π/9)cos(2π/9)cos(4π/9)]/sin(π/9)
=1/8sin(8π/9)/sin(π/9)
=1/8
log2(cos(π/9)cos(2π/9)cos(4π/9))=log2 1/8=-3
=1/2*[sin(2π/9)cos(2π/9)cos(4π/9)]/sin(π/9)
=1/8sin(8π/9)/sin(π/9)
=1/8
log2(cos(π/9)cos(2π/9)cos(4π/9))=log2 1/8=-3
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