三角证明题（你从来没见过的）(tan12°)^2+(tan24°)^2+(tan48°)^2+(tan84°)^2=92

Question

三角证明题（你从来没见过的）
(tan12°)^2+(tan24°)^2+(tan48°)^2+(tan84°)^2=92

Answer 1

令α=kπ/15,k=1,...,7.
那么,显然tan 7α + tan 8α=0.
考虑到tan 7α = tan (3α + 4α),tan 8α = tan (2*4α),利用两角和以及倍角公式展开并化简得
tan 3α + 3 tan 4α - 3 tan 3α tan^2 4α - tan^3 4α = 0.
令tan α = x,利用三倍角公式和四倍角公式,
tan 3α = (3x - x^3) / (1 - 3x^2),
tan 4α = (4x 4x^3) / (1 - 6x^2 + x^4),
得
(3x - x^3) / (1 - 3x^2) + 3 (4x 4x^3) / (1 - 6x^2 + x^4) - 3 (3x - x^3) / (1 - 3x^2) * (4x 4x^3)^2 / (1 - 6x^2 + x^4)^2 - (4x 4x^3)^3 / (1 - 6x^2 + x^4)^3 = 0.
化简,得
x(x^14 - 105 x^12 + 1365 x^10 - 5005 x^8 + 6435 x^6 - 3003 x^4 + 455 x^2 - 15) = 0.
由于x≠0,
故x^14 - 105 x^12 + 1365 x^10 - 5005 x^8 + 6435 x^6 - 3003 x^4 + 455 x^2 - 15 = 0.
这说明,tan^2 π/15,...,tan^2 7π/15是方程
x^7 - 105 x^6 + 1365 x^5 - 5005 x^4 + 6435 x^3 - 3003 x^2 + 455 x - 15 = 0
的七个根.根据韦达定理,知道
tan^2 π/15 + ...,+ tan^2 7π/15 = 105.
又因为
tan^2 3π/15 = tan^2 π/5 = 5-2√5,
tan^2 5π/15 = tan^2 π/3 = 3,
tan^2 6π/15 = tan^2 2π/15 = 5+2√5,
在上式中减去这三项,立得
tan^2 π/15 + tan^2 2π/15 + tan^2 4π/15 + tan^2 7π/15 = 92.
证毕.
顺便,从韦达定理还可以得出这四个平方之积等于1.