设函数f(x)=Asin(3x+φ)(A>0,x∈(-∞,+∞),0<φ<π)当x=π/12时取得最大值为4(1)求f(x)的最小正周期(2)若f(2/3,a+π/12)=12/5,求cos2a的值
问题描述:
设函数f(x)=Asin(3x+φ)(A>0,x∈(-∞,+∞),0<φ<π)当x=π/12时取得最大值为4
(1)求f(x)的最小正周期(2)若f(2/3,a+π/12)=12/5,求cos2a的值
答
1.T=2π/3
2.A=4 f(π/12)=4*sin(π/4+φ)=4
π/4+φ=2kπ+π/2 k为整数
因为0<0<φ<π φ=π/4
f(2/3*a+π/12)=4sin(3*(2/3*a+π/12)+π/4)=4sin(2a+π/2)=4cos2a=12/5
cos2a=3/5