已知-π/2<α<π/2,-π/2<β<π/2,且tanα,tanβ是方程x²+6x+7=0的两个根求α+β的值
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已知-π/2<α<π/2,-π/2<β<π/2,且tanα,tanβ是方程x²+6x+7=0的两个根求α+β的值
已知-π/2<α<π/2,-π/2<β<π/2,且tanα,tanβ是方程x²+6x+7=0的两个根
求α+β的值
已知-π/2<α<π/2,-π/2<β<π/2,且tanα,tanβ是方程x²+6x+7=0的两个根求α+β的值
∵tanα,tanβ是方程x²+6x+7=0的两个根
∴tanα+tanβ=-60
∴tanα,tanβ同为负值
∵-π/2<α<π/2,-π/2<β<π/2
∴-π/2<α
由韦达定理知tanα+tanβ=-6 tanα×tanβ=7
所以tan(α+β)=(tanα+tanβ)/(1-tanα×tanβ)=1
又因为-π<α+β<0,所以α+β=-3π/4