不查表求:[tan(π/5)]^2+[tan(2π/5)]^2的值。
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。
。
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。