8sin(π/7)sin(2π/7)sin(3π/7) = √7 64sin²(π/7)sin²(2π/7)sin²(3π/7) = 7 8[1-cos(2π/7)][1-cos(4π/7)][1-cos(6π/7)] = 7 8[1-(cos(2π/7)+cos(4π/7)+cos(6π/7))+cos(2π/7)cos(4π/7)+cos(4π/7)• •cos(6π/7)+cos(6π/7)cos(2π/7)-cos(2π/7)cos(4π/7)cos(6π/7)] = 7 而cos(2π/7)cos(4π/7)cos(6π/7)=cos(2π/7)cos(3π/7)cos(π/7)=1/8 cos(2π/7)cos(4π/7)+cos(4π/7)cos(6π/7)+cos(6π/7)cos(2π/7) =cos(2π/7)+cos(4π/7)+cos(6π/7) 左边=cos(4π/7)[cos(2π/7)+cos(6π/7)]+cos(6π/7)cos(2π/7) =cos(4π/7)×2cos(4π/7)cos(2π/7)+cos(6π/7)cos(2π/7) =cos(2π/7)×2cos²(4π/7)+cos(6π/7)cos(2π/7) =cos(2π/7)(1+cos(8π/7))+cos(6π/7)cos(2π/7) =cos(2π/7)[1+cos(8π/7)+cos(6π/7)] =cos(2π/7)+cos(2π/7)×2cosπcos(π/7) =cos(2π/7)-2cos(2π/7)cos(π/7) =cos(2π/7)-(cos(3π/7)+cos(π/7)) =cos(2π/7)+cos(4π/7)+cos(6π/7)=右边．证毕。
问：高一数学题设0<α<β<П/2,0<θ<П/2, 求证:[sin(θ+α)]/[sin(β+θ)]>sinα/sinβ
答：原式=[sin(θ+α)]sinβ>sinα[sin(β+θ)],把他们展开，就可以看到很明显的思路了。详情>>
答：详情>>
