若已知方程x²-(tanθ+cotθ)x+1=0有两个实根,且其中一个根是2-根号3,求cos4

x²-(tanθ+cotθ)x+1=0
根据韦达定理,两根之积为1,两根之和为tanθ+cotθ
其中一个根是2-根号3,则另一根是1/(2-根号3)=2+根号3
所以tanθ+cotθ=2--根号3+2+根号3=4
tanθ+1/tanθ=4
(tanθ)²-4tanθ+1=0
(tanθ)²+1=4tanθ
-4tanθ+1=-(tanθ)²
(cosθ)²=1/[(tanθ)²+1]=1/(4tanθ)
cos4θ=2(cos2θ)²-1
=2[2(cosθ)²-1]²-1
=2[1/(2tanθ)-1]²-1
=[4(tanθ)²-4tanθ+1]/[2(tanθ)²]-1
=(-4tanθ+1)/2(tanθ)²+1
=-(tanθ)²/2(tanθ)²+1
=1/2