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求值

不查表求:[tan(π/5)]^2+[tan(2π/5)]^2的值。

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2013-08-25

0 0

    cos(π/5)cos(2π/5) =4sin(π/5)cos(π/5)cos(2π/5)/[4sin(π/5)] =sin(4π/5)/[4sin(π/5)] =1/4, ∴[tan(π/5)]^2+[tan(2π/5)]^2 =[tan(π/5)+tan(2π/5)]^2-2tan(π/5)tan(2π/5) ={sin(3π/5)/[cos(π/5)cos(2π/5)]}^2-2sin(π/5)sin(2π/5)/[cos(π/5)cos(2π/5)] =[4sin(2π/5)]^2-8sin(π/5)sin(2π/5) =16[sin(2π/5)]^2-8sin(π/5)sin(2π/5) =8[1-cos(4π/5)]+4[cos(3π/5)-cos(π/5)] =8+4[cos(π/5)+cos(3π/5)] =8+8cos(π/5)cos(2π/5) =8+8/4 =10。
    。

2013-08-25

45 0

    由sin36°=cos54°,2sin18°cos18°=4(cos18°)^3-3cos18°,得 4(sin18°)^2+2sin18°-1=0, ∴ sin18°=(√5-1)/4, cos36°=1-2(sin18°)^2=(√5+1)/4, sin18°-cos36°=-1/2。
    由半角公式,可得 原式=[(1-cos72°)/(1+cos72°)]+[(1+cos36°)/(1-cos36°)] =[2+(sin18°-cos36°)]/[1+1。
  5(sin18°-cos36°)]=10。

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