Question

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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Answer 1

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].

Answer 2

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].