Question

已知向量m=（sinA,cosA),n=(根号3，-1），m*n=1，且a为锐角，（1）求角A大小，（2）f(x)=cos2x+4cosAsinx

求函数的值域，X属于R。

Answer 1

因为m*n=1
所以√3sinA-cosA=1
√3/2sinA-1/2cosA=1/2
sinAcosπ/6-cosAsinπ/6=1/2
sin(A-π/6)=1/2
则有
A-π/6=2kπ+π/6
A=2kπ+π/3
因为A为锐角
所以A=π/3

(2)
f(x)=cos2x+4cosπ/3sinx
=1-2sin^2 x +2sinx
=-2(sin^2 x -sinx+1/4)+1+1/2
=-2(sinx-1/2)^2+3/2
当sinx=1/2时有最大值 f(x)=3/2
当sinx=-1时有最小值f(x)=-9/2+3/2=-3

Answer 2

(1)
m.n = 1
(sinA,cosA).(√3,-1) =1
√3sinA-cosA = 1
(√3/2) sinA - (1/2)cosA = 1/2
sin(A-π/6) = sin(π/6)
A-π/6 = π/6
A = π/3
(2)
f(x)=cos2x+4cosAsinx
= cos2x + 2sinx
= (1-(2sinx)^2) + 2sinx
= - 2(sinx - 1/2)^2 + 3/2
max f(x) = 3/2
minf(x) = -2(-3/2)^2 + 3/2
= -9/2 + 3/2
= -3
值域 = [-3, 3/2]

Answer 3

1，角A=43
2，fx=13