已找方程x^2+mx+n=0有两个实根x1,x2,a=arctanx1,β=arctanx2

问题描述:

已找方程x^2+mx+n=0有两个实根x1,x2,a=arctanx1,β=arctanx2
(1)当m=3根号3,n=4时,求a+β的值.
(2)m=-sinΘ,n=cosΘ,(0

数学人气:928 ℃时间:2020-06-15 15:23:36
优质解答
(1)由a=arctanx1,β=arctanx2
tanα=x1,tanβ=x2,
由韦达定理:x1+x2=-m
x1*x2=n
则tan(a+β)=(tanα+tanβ)/(1-tanα·tanβ)
=(x1+x2)/(1-x1*x2)
=-m/(1-n)
=-3根号3/(1-4)
=根号3
则a+β=arctan(180*k+60 ) (k为任意整数)
(2)tan(a+β)=(tanα+tanβ)/(1-tanα·tanβ)
=(x1+x2)/(1-x1*x2)
=-m/(1-n)
=sinΘ/(1-cosΘ)
=tan(Θ/2)
所以a+β=180*k+Θ/2 (k为任意整数)
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(1)由a=arctanx1,β=arctanx2
tanα=x1,tanβ=x2,
由韦达定理:x1+x2=-m
x1*x2=n
则tan(a+β)=(tanα+tanβ)/(1-tanα·tanβ)
=(x1+x2)/(1-x1*x2)
=-m/(1-n)
=-3根号3/(1-4)
=根号3
则a+β=arctan(180*k+60 ) (k为任意整数)
(2)tan(a+β)=(tanα+tanβ)/(1-tanα·tanβ)
=(x1+x2)/(1-x1*x2)
=-m/(1-n)
=sinΘ/(1-cosΘ)
=tan(Θ/2)
所以a+β=180*k+Θ/2 (k为任意整数)