三角求值三角求值
[tan(π/5)]^2+[tan(2π/5)]^2=?
解 设t=nπ/5 (n=0,1,2,3,4) 是tan(5t)=0的根。
令x=tant，由tan(5t)=0得:tan(3t)=-tan(2t),
即 (3x-x^3)/(1-3x^2)=-2x/(1-x^2),
化简整理为: x^4-10x^2+5=0。
据韦达定理得:
tan(π/5)+tan(2π/5)+tan(3π/5)+tan(4π/5)=0, (1)
tan(π/5)*tan(2π/5)*tan(3π/5)*tan(4π/5)=5, (2)
tan(π/5)*tan(2π/5)+tan(2π/5)*tan(3π/5)+tan(3π/5)*tan(4π/5)+...全部
解 设t=nπ/5 (n=0,1,2,3,4) 是tan(5t)=0的根。
令x=tant，由tan(5t)=0得:tan(3t)=-tan(2t),
即 (3x-x^3)/(1-3x^2)=-2x/(1-x^2),
化简整理为: x^4-10x^2+5=0。
据韦达定理得:
tan(π/5)+tan(2π/5)+tan(3π/5)+tan(4π/5)=0, (1)
tan(π/5)*tan(2π/5)*tan(3π/5)*tan(4π/5)=5, (2)
tan(π/5)*tan(2π/5)+tan(2π/5)*tan(3π/5)+tan(3π/5)*tan(4π/5)+tan(π/5)*tan(3π/5)+tan(π/5)*tan(4π/5)+tan(2π/5)*tan(3π/5)=-10 (3)
(2) [tan(π/5)*tan(2π/5)]^2=5，
故tan(π/5)*tan(2π/5)=√5。
(3) tan(π/5)*[tan(2π/5)+tan(3π/5)+tan(4π/5)]-[tan(2π/5)]^2-tan(π/5)*tan(2π/5)+tan(π/5)*tan(2π/5)=-10
再由tan(π/5)=-[tan(2π/5)+tan(3π/5)+tan(4π/5)]得:
[tan(π/5)]^2+[tan(2π/5)]^2=10。
得证。
一般地，我们有:
tan[π/(2n+1)]*tan[2π/(2n+1)]*…*tan[nπ/(2n+1)]=√(2n+1)
{tan[π/(2n+1)]}^2+{tan[2π/(2n+1)]}^2+…+{tan[nπ/(2n+1)]}^2=n(2n+1)
当n=3时得:
tan(π/7)*tan(2π/7)*…*tan(3π/7)=√7。
[tan(π/7)]^2+[tan(2π/7)]^2+[tan(π/7)]^2=21。
。收起
