Question

已知函数fx=sin(x+兀/3)+sinx(1)求函数fx的最小正周期和最小值

已知函数fx=sin(x+兀/3)+sinx(1)求函数fx的最小正周期和最小值。(2)若fx=-4根号3/5，且-兀/2<x<0.求cos(x+兀/6)的值 求过程，不要思路

Answer 1

f(x)=sin(x+π/3)+sinx
=sinxcosπ/3+cosxsinπ/3+sinx
=(1/2+1)sinx+√3/2cosx
=3/2sinx+√3/2cosx
=√3(√3/2sinx+1/2cosx)
=√3sin(x+π/6)
(1) T=2π
当sin(x+π/6)=-1时，f(x)取得最小值：f(x)min=-√3
(2) f(x)=-4√3/5
√3sin(x+π/6)=-4√3/5
sin(x+π/6)=-4/5
∵-π/2<x<0
∴-π/3<x+π/6<π/6
cos(x+π/6)>0
cos(x+π/6)=√(1-sin^2(x+π/6))
=√(1-(-4/5)^2)
=3/5