Question

1.已知tanα和tan(π/4-α)是方程x^2+px+q=0的两个根,若3tanα=tan(π/4-α)求q,p

Answer 1

1

tanα和tan(π/4-α)是方程x^2+px+q=0的两个根,

∴tanα+tan(π/4-α)=-p/2;

tanα·tan(π/4-α)=q/2;

3tanα=tan(π/4-α),则

4tanα=-p/2;p=-8tanα

3(tanα)^2=q/2;q=6(tanα)^2.

3tanα=tan(π/4-α)

=(tanπ/4-tanα)/(1+tanπ/4·tanα)

=(1-tanα)/(1+tanα)

∴tanα=(-2±√7)/3

代入p=-8tanα,q==6(tanα)^2即可救出

2

已知tanα=√3(tanαtanβ+m),又α,β都是钝角,求α+β的值

tanα=√3(tanαtanβ+m),

tanα(1-√3tanβ)=√3m;

tanα[(√3+tanβ)/tan(π/3+β)]=√3m;

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