函数y=cos^2(x)-sinx的值域是? 函数y=cos(sinx)的值域是???

\u51fd\u6570y=cos^2x-sinx\u7684\u503c\u57df\u662f

y=cos^2x-sinx=1-\uff08sinx\uff09^2-sinx=5/4-(sinx+1/2)^2\uff1b
sinx\u2208[-1,1],(sinx+1/2)^2\u2208[0,9/4]\uff1b
\u6240\u4ee5\u503c\u57dfy\u2208[-1,5/4]

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\u4ee4t=sinx,\u5219
-1=<t<=1
cos1=<cost<=1
\u6240\u4ee5y\u7684\u503c\u57df\u4e3a[cos1, 1]

\u5982\u679c\u6709\u5e2e\u5230\u60a8 \u8bf7\u7ed9\u4e88\u597d\u8bc4 \u8c22\u8c22\u62c9#^_^#\u795d\u60a8\u6109\u5feb

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)

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]

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

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

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

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

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