已知函数y1=3sin(2x-π/3),y2=4sin(2x +π/3),那么函数y=y1+y2的振幅A的值是:

问题描述:

已知函数y1=3sin(2x-π/3),y2=4sin(2x +π/3),那么函数y=y1+y2的振幅A的值是:
A、5 B、7 C、根号下13 D、13

y1=3sin(2x-π/3),y2=4sin(2x +π/3),
y=3sin(2x-π/3)+4sin(2x +π/3)
=3(1/2*sin2x-√3/2*cos2x)+4(1/2*sin2x+√3/2*cos2x)
=(3/2+2)sin2x+(2√3-3√3/2)cos2x
=7/2*sin2x+√3/2*cos2x
=√13*[sin2x*(7/2√13)+cos2x*(√3/2√13)]
=√13*sin(2x+a),其中:cosa=7/2√13,sina=√3/2√13;所以振幅=√13.