Question

高一三角函数问题，谢谢！

Answer 1

y=sinx==>y=2sin(2x-π/3)
(1)将Y轴坐标扩大到原来的2倍，得y=2sinx
(2)再将X轴坐标缩小到原来的1/2，得y=2sin2x
(3)将y=2sin2x图像右移π/6，得y=2sin(2x-π/3)图像
单调递增区：2kπ-π/2<=2x-π/3<=2kπ+π/2==>kπ-π/12<=x<=kπ+5π/12
单调递减区：kπ+5π/12<=x<=kπ+11π/12
最大值：在x= kπ+5π/12处取最大值2
最小值：在x= kπ-π/12处取最小值-2

追问

谢谢！