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

y=sinx最小正周期是2π
水平移动和上下伸缩不影响周期,只有在水平伸缩时影响周期,这里是缩小到1/3所以f(x)=Asin(3x+a)最小正周期是2π/3
f(x)=sinx最小正周期是2π
水平移动和上下伸缩不影响周期,只有在水平伸缩时影响周期,这里是缩小到1/3所以f(x)=Asin(3x+a)最小正周期是2π/3
0Asin(π/4+a)=4
A=4
π/4+a=π/2+π*n
0a=π/4
f(x)=4sin(3x+π/4)
f(2a/3+π /12)
=4sin(2a+π/2)
=-4cos2a
=12/5
cos2a=-3/5

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