题目

求下列三角函数值域
1.y=-2sinx=1
2.y=sin^2(x)-2sinxcosx=3cos^2(x)
3.y=1/2sin2x=4sin^2(x)
4.f(x)=sinx+根号3(cosx) x属于[-90度,90度]
1.y=-2sinx+1
2.y=sin^2(x)-2sinxcosx+3cos^2(x)
3.y=1/2sin2x+4sin^2(x)
4.f(x)=sinx+根号3(cosx) x属于[-п/2,п/2]

提问时间：2020-12-25

答案

1、
-1<=sinx<=1
所以-2<=-2sinx<=2
-2+1<=-2sinx+2<=2+1
所以值域[-1,3]
2、
y=(sin²x+cos²x)-sin2x+(2cos²x-1)+1
=1-sin2x+cos2x+1
=-(sin2x-cos2x)+2
=-√2sin(2x-π/4)+2
-1<=sin(2x-π/4)<=1
所以-√2<=-√2sin(2x-π/4)<=√2
-√2+2<=-√2sin(2x-π/4)+2<=√2+2
所以值域[-√2+2,√2+2]
3、
y=1/2*sin2x+4(1-cos2x)/2
=1/2*sin2x-2cos2x+2
=√[(1/2)²+2²]sin(2x-z)+2
=(√17/2)sin(2x-z)+2
其中tanz=2/(1/2)=4
-1<=sin(2x-z)<=1
-(√17/2)<=(√17/2)sin(2x-z)<=(√17/2)
-(√17/2)+2<=(√17/2)sin(2x-z)+2<=(√17/2)+2
所以值域[-(√17/2)+2,(√17/2)+2]
4、
f(x)=√[(1²+(√3)²]sin(x+z)
其中tanz=√3/1=√3
z=π/3
所以f(x)=2sin(x+π/3)
-π/2<=x<=π/2
-π/6<=x+π/3<=5π/6
所以x+π/3=-π/6时最小
x+π/3=π/2时最大
所以
sin(-π/6)<=sin(x+π/3)<=sin(π/2)
-1/2<=sin(x+π/3)<=1
再乘以2
所以值域[-1,2]