求值cot15cot25cot35cot85

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