Question

设tanα和tanβ是方程x^2+px+q=0的两根

设tanα和tanβ是方程x^2+px+q=0的两根.试以p , q表示sin(α+β)^2+psin(α+β)cos(α+β)+q(cos(α+β))^2

Answer 1

tanα+tanβ=-p
tanαtanβ=q
tan(α+β)=-p/(1-q)
sin(α+β)=[-p/(1-q)]cos(α+β)
容易得到
cos²(α+β)=(1-q)²/[p²+(1-q)²]
sin²(α+β)=p²/[p²+(1-q)²]
sin(α+β)^2+psin(α+β)cos(α+β)+q(cos(α+β))^2
=p²/[p²+(1-q)²]+ptan(α+β)cos²(α+β)+q(1-q)²/[p²+(1-q)²]
=p²/[p²+(1-q)²]+p*[-p/(1-q)][(1-q)²/[p²+(1-q)²]+q(1-q)²/[p²+(1-q)²]
=[p²-p²(1-q)+q(1-q)²]/[p²+(1-q)²]
=q

Answer 2

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