Question

高中高一数学问题

高中高一数学问题第一问答案中距离d怎么得那个的？

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Answer 1

设t=tan(x/2) y=(3sinx-3)/(2cosx+10) =-3(1-sinx)/2(cosx+5) =-3[sin(x/2)-cos(x/2)]^2/2[2(cos(x/2))^2-1+5] =-3[sin(x/2)-cos(x/2)]^2/4[(cos(x/2))^2+2] =-3[sin(x/2)-cos(x/2)]^2/4[(3cos(x/2))^2+2(sin(x/2))^2] =(-3/4)*(t-1)^2/(2t^2+3) 就是得到:y=(-3/4)*(t-1)^2/(2t^2+3) 再化为方程: (8y+3)t^2-6t+(3+12y)=0 那么就要有判断式: 6^2-4(8y+3)(3+12y)≥0 也就是: 36-12(8y+3)(1+4y)=36-12(8y+32y^2+3+12y) =-12(32y^2+20y) =-12*4y(8y+5)≥0 就得到:-5/8≤y≤0 也就是,,最大值是0;;最小值是-5/8