Question

已知函数f(x)=根号3sinx/4cosx/4+cosx/4的平方(1)若f(x)=1,求cos(2pai/3-x）的值

Answer 1

已知函数f(x)=√3sin(x/4)cos(x/4)+cos^2(x/4)
(1)若f(x)=1,求cos(2π/3-x)的值
f(x)=√3sin(x/4)cos(x/4)+cos^2(x/4)
=√3/2sin(x/2)+1/2(2cos^(x/4)-1)+1/2
=sin(x/2)cos(π/6)+cos(x/2)sin(π/6)+1/2
=sin(x/2+π/6)+1/2
f(x)=1
sin(x/2+π/6)+1/2=1
sin(x/2+π/6)=1/2
x/2+π/6=π/6+2kπ x/2+π/6=5π/6+2kπ
x/2=2kπ x/2=2π/3+2kπ
x=4kπ x=4π/3+4kπ
2π/3-x=2π/3+4kπ 2π/3-x=-2π/3+4kπ
cos(2π/3-x)=-√3/2 cos(2π/3-x)=-√3/2

追问

最后两步是什么意思呀

追答

当2π/3-x=2π/3+4kπ
cos(2π/3-x)
=cos(2π/3+4kπ)
=cos(2π/3)
=-√3/2

当2π/3-x=-2π/3+4kπ
cos(2π/3-x)
=cos(-2π/3+4kπ)
=cos(-2π/3)
=cos(-2π/3)
=-√3/2

cos(2π/3-x)=-√3/2

追问

不好意思啊 可是你上面算出来一个是x=4kpai 另一个是4pai/3+4kpai嘛 那然后不是应该用2pai/3-x=4kpai 和2pai/3-x=2pai/3+4kpai 然后算出来结果一个等于-1/2一个等于1吗

追答

x=4kπ x=4π/3+4kπ

-x=4kπ -x=-4π/3+4kπ

2π/3-x=2π/3+4kπ 2π/3-x= 2π/3-4π/3+4kπ

2π/3-x=2π/3+4kπ 2π/3-x=-2π/3+4kπ

应都=-1/2

Answer 2

cos(2pai/3)=-1/2,,ok?