求证：(tan(π/5))^2+(tan(2π/5))^2=2tan(π/5)tan(2π/5)
求证：(tan(π/5))^2+(tan(2π/5))^2=2tan(π/5)tan(2π/5)
上面的问题打错，这个才对
求证：(tan(π/5))^2+(tan(2π/5))^2=2(tan(π/5)tan(2π/5))^2
对不起，已修改
tan(A+B)=(tanA+tanB)(1-tanAtanB)
tanA+tanB=tan(A+B)(1-tanAtanB)
同理 tanA-tanB=tan(A-B)(1+tanAtanB)
(tan(π/5))^2+(tan(2π/5))^2-2tan(π/5)tan(2π/5)
=[tan(π/5)-tan(2π/5)]²
=tan²(π/5)[1+tan(π/5)tan(2π/5)]²
=tan²(π/5){1+sin(π/5)sin(2π/5)/[cos(π/5)cos(2π/5)]}²
=tan²(π/5){【cos(π/5)cos(2π/5)+sin(π/5)sin(2π/5)】/[cos(π/5)cos(2π/5)]}²
=tan²(π/5){cos(π/5)/[cos(π/5)cos(2π/5)]}²
=tan²(π/5)[1/cos(2π/5)]²
=tan(π/5)tan(2π/5) tan(π/5)/[sin(2π/5)cos(2π/5)]
=tan(π/5)tan(2π/5) tan(π/5)/[sin(4π/5)/2]
=2tan(π/5)tan(2π/5) tan(π/5)/[sin(π/5)]
=2tan(π/5)tan(2π/5)/[cos(π/5)]
2(tan(π/5)tan(2π/5))^2-2tan(π/5)tan(2π/5)
=2tan(π/5)tan(2π/5)[tan(π/5)tan(2π/5)-1]
=2tan(π/5)tan(2π/5) {sin(π/5)sin(2π/5)/[cos(π/5)cos(2π/5)]-1}
=2tan(π/5)tan(2π/5) {[sin(π/5)sin(2π/5)-cos(π/5)cos(2π/5)]/[cos(π/5)cos(2π/5)]}
=2tan(π/5)tan(2π/5) {-cos(3π/5)]/[cos(π/5)cos(2π/5)]}
=2tan(π/5)tan(2π/5) {cos(2π/5)]/[cos(π/5)cos(2π/5)]}
=2tan(π/5)tan(2π/5) /cos(π/5)
所以 (tan(π/5))^2+(tan(2π/5))^2-2tan(π/5)tan(2π/5)=2(tan(π/5)tan(2π/5))^2-2tan(π/5)tan(2π/5)
所以 (tan(π/5))^2+(tan(2π/5))^2=2(tan(π/5)tan(2π/5))^2