Question

求详细步骤谢谢

Answer 1

三、解答题
（1）
f(7π/6)=cos^2 x+sin x-1
=cos^2(π+6/π)+sin(π+6/π)-1
=(-cos 6/π)^2+(-sin 6/π)-1
=(-√3/2)^2+(-1/2)-1
=3/4-1/2-1
=-3/4

(2)
f(x)=cos^2 x+sin x-1
=1-sin^2 x+sin x-1
=-sin^2 x+sin x
=-(sin^2 x-sin x+1/4-1/4)
=-(sin x-1/2)^2+1/4
函数f(x)为开口向下的二次函数，因此当sin x=1/2时，顶点即f(x)的最大值为1/4，此时x=π/6，属于[6/π,2π/3]；而最小值，就需要比较二者：
f(-6/π)=-[sin(-6/π)-1/2]^2+1/4
=-[-sin(6/π)-1/2]^2+1/4
=-1+1/4
=-3/4
f(2π/3)=-[sin(2π/3)-1/2]^2+1/4
=-[sin(π-π/3)-1/2]^2+1/4
=-[sin(π/3)-1/2]^2+1/4
=-(√3/2-1/2]^2+1/4
=-3/4+√3/2＞-3/4
所以f(x)的范围是[-3/4,1/4]

Answer 2

大哥大蝴蝶好看吗五

追答

大哥大蝴蝶好看吗五