How do you find the period of #y=1/2sin[(pix)/3]#?

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Answer 1 · 2018-04-05T13:04:08

#color(blue)(6)#

If we express the the sine function in the form:

#y=asin(bx+c)+d#

Where:

#bba \ \ \ \ \ ="amplitude"#

#bb((2pi)/b)="period"#

#bb((-c)/b)="phase shift"#

# \ \ \ \ bbd \ \ ="vertical shift"#

#y=1/2sin(pi/3x)#

#:.#

#b=pi/3#

Period is:

#(2pi)/b=(2pi)/(pi/3)=(6pi)/pi=color(blue)(6)#

The graph confirms this:

enter image source here