三角函数的最值4

1.
设t=sinx+cosx=根号2*sin(x+45)
那么t的范围就是[-根号2,根号2]
y=1+sinx+cosx+sinxcosx
=1+t+(t^2-1)/2
=t^2/2+t+1/2
=(1/2)*(t+1)^2
所以当t=-1时,y的最小值是0
当t=根号2时,y的最大值是根号2+(3/2)
2.
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
3.
设t=1-sinx
那么t的范围就是[0,2]
y=(1-sinx)/[2-2sinx+(sinx)^2]
=(1-sinx)/[1+(1-sinx)^2]
=t/(1+t^2)
=1/[t+(1/t)]
因为有t+1/t≥2,
此时t=1时,,就是y取最大值1/2
t=0时,y取最小值0

1.y=1/2*(sinx+cosx)^2+sinx+cosx+1/2=1/2*(sinx+cosx+1)^2
(sinx+c