已知sin（A+π/4）=7根号2/10,A∈（π/4,π/2）.求cosA的值,求函数f（x）=cos2x+5/2sinAsinx的值域复制搜索
问题描述：
已知sin（A+π/4）=7根号2/10,A∈（π/4,π/2）.求cosA的值,
求函数f（x）=cos2x+5/2sinAsinx的值域
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答
sin(A+π/4)=7√2/10,
A属于(π/4,π/2)
cos(A+π/4)
=-√2/10
cosA=cos[(A+π/4)-π/4]
=cos(A+π/4)cosπ/4+sin(A+π/4)sinπ/4
=(-√2/10)*(√2/2)+(7√2/10)*(√2/2)=3/5
A∈（π/4，π/2），所以sinA=4/5，
f（x）=cos2x+5/2sinAsinx
= cos2x+2sinx
=1-2sin²x+2sinx
=-2(sinx-1/2)²+3/2
sinx的值域是（-1,1）
Sinx=1/2时，函数取到最大值3/2.
Sinx=-1时，函数取到最小值-3.
所以函数值域是[-3,3/2].
答
sin(A+π/4)=7√2/10,
A属于(π/4,π/2)
cos(A+π/4)
=-√2/10
cosA=cos[(A+π/4)-π/4]
=cos(A+π/4)cosπ/4+sin(A+π/4)sinπ/4
=(-√2/10)*(√2/2)+(7√2/10)*(√2/2)=3/5
A∈（π/4,π/2）,所以sinA=4/5,
f（x）=cos2x+5/2sinAsinx
= cos2x+2sinx
=1-2sin²x+2sinx
=-2(sinx-1/2)²+3/2
Sinx=1/2时,函数取到最大值3/2.
Sinx=-1时,函数取到最小值-3.
所以函数值域是[-3,3/2].