Question

f(x)=sin(ωx+φ)(ω大于0，φ大于等于0 小于等于π)为偶函数,其图像上相邻的两个最高点之间的距离为2π

已知函数f(x)=sin(ωx+φ)(ω大于0，0≤φ≤π)为偶函数,其图像上相邻的两个最高点之间的距离为2π
1.求f(x)的解析式
2.若a属于(-π/3,π/2)，f(a+(π/3))=1/3,求sin(2a+(5π/3))的值

Answer 1

T=2π，w=1,因为偶函数，φ=π/2，所以sin(x+π/2)
f(a+(π/3))=sin（a+π/3+π/2）
sin（a+5π/6）=
=cos（π/2-a-5π/6）
=cos（-a-π/3）=1/3
cos（a+π/3）=1/3 所以sin（a+π/3）=根号8除以9
a属于(-π/3,π/2)，
sin(2a+(5π/3))=2cos（a+π/3）sin（a+π/3）=-根号8除以9

Answer 2

2。可知sin(a+π/3+π/2)=1/3
cos(a+5π/6)=-2根号2/3
sin(2a+5π/3)=sin(2(a+5π/6))=2sin(a+5π/6)cos(a+5π/6)=-4根号2/9

Answer 3

1，f（x）=sin（x+∏/2)