Question

函数y=cos^2(x)-sinx的值域是？

Answer 1

函数y=cos²(x)-sinx
=-sin²(x)-sinx+1
=5/4-(sinx+1/2)²

∵-1≤sinx≤1
∴-1/2≤sinx+1/2≤3/2
∴ 0≤(sinx+1/2)²≤9/4
-1≤y≤5/4

y值域(-1, 5/4)

Answer 2

y=cos²x-sinx
=(cos²x-1+1)-sinx
= -sin²x-sinx+1
=5/4-(sinx+1/2)²

因为-1≤sinx≤1
所以-1/2≤sinx+1/2≤3/2
所以0≤(sinx+1/2)²≤9/4
所以-1≤5/4-(sinx+1/2)²≤5/4

所以值域为[-1,5/4]

Answer 3

y=cos^2(x)-sinx
=1-sin^2(x)-sinx
=-(sinx+1/2)^2+5/4

-1<=sinx<=1
-1/2<=(sinx+1/2)<=3/2

0<=(sinx+1/2)^2<=9/4

-9/4<=-(sinx+1/2)^2<=0

-1<=-(sinx+1/2)^2+5/4<=5/4

-1<=y<=5/4

Answer 4

1、y=cos^2(x)-sinx=1-sin^2(x)-sinx=-(sin^2(x)+sinx-1)=-(sinx+1/2)^2+5/4
2、sinx+1/2的值域是（-1/2,3/2),
3、(sinx+1/2)^2的值域是（1/4,9/4)，-(sinx+1/2)^2的值域是（-9/4,0)
4、-(sinx+1/2)^2+5/4的值域是（-1,5/4)
5、注以上均为闭区间

Answer 5

y=cos(x)*cos(x)-sin(x)
=1-sin(x)^2-sin(x)
=1-(sin(x)^2+sin(x))
=1-(sin(x)^2+sin(x)+(1/2)^2-(1/2)^2)
=1-(sin(x)+1/2)^2+1/4
=5/4-(sin(x)+1/2)^2
sin(x)=-1/2时，是y最大值5/4
sin(x)=1是，是y的最小值-1

Answer 6

我提供哈思路吧把COS(X)2用1-SIN（X)2代替 然后再配方即可