Question

求函数y=(3sinx-3)/(2cosx+10)的最值

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

Answer 2