设θ是三角形的最小内角,且acos²θ/2+sin²θ/2-cos²θ/2-asin²θ/2=a+1,则a的取值范
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![设θ是三角形的最小内角,且acos²θ/2+sin²θ/2-cos²θ/2-asin²θ/2=a+1,则a的取值范](/uploads/image/z/2981961-9-1.jpg?t=%E8%AE%BE%CE%B8%E6%98%AF%E4%B8%89%E8%A7%92%E5%BD%A2%E7%9A%84%E6%9C%80%E5%B0%8F%E5%86%85%E8%A7%92%2C%E4%B8%94acos%26%23178%3B%CE%B8%2F2%2Bsin%26%23178%3B%CE%B8%2F2-cos%26%23178%3B%CE%B8%2F2-asin%26%23178%3B%CE%B8%2F2%3Da%2B1%2C%E5%88%99a%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83)
设θ是三角形的最小内角,且acos²θ/2+sin²θ/2-cos²θ/2-asin²θ/2=a+1,则a的取值范
设θ是三角形的最小内角,且acos²θ/2+sin²θ/2-cos²θ/2-asin²θ/2=a+1,则a的取值范
设θ是三角形的最小内角,且acos²θ/2+sin²θ/2-cos²θ/2-asin²θ/2=a+1,则a的取值范
cos²θ/2-sin²θ/2=cosθ
所以(a-1)cosθ=a+1
cosθ=(a+1)/(a-1)
θ是三角形的最小内角
所以0
难
1
令x=sinθ/2 y=cosθ/2 x^2+y^2=1 则原式为ay^2+x^2-y^2-ax^2=a(y^2-x^2)+(x^2-y^2)=a(1-x^2-x^2)+(x^2-1-x^2)=a(1-2x^2)-1=a+1所以a=-1/x^2将分子的1化成x^2+y^2则a=-(x^2+y^2)/x^2=-(1+y^2/x^2)=-(1+tan^2θ/2) 0≤θ≤60 所以0≤tan^2θ/2≤1/3 1≤tan^2θ/2+1≤4/3 所以 -4/3 ≤a ≤-1