How do you find the period of #y = (1/2) sin ((x/3)-π)#?

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Answer 1 · 2016-06-07T22:24:22

#6pi#

A sine function appears in the form

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

Where the period of the function can be found by taking #(2pi)/b#.

Here, we can slightly rewrite the given function to make the value of #b#, the coefficient of #x#, more clear:

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

Thus, #a=1/2,b=1/3,# and #c=-pi#.

So, the period of the function is #(2pi)/b=(2pi)/(1/3)#.

Note that dividing by #1/3# is the same as multiplying by #3#, so #(2pi)/(1/3)=2pi*3=6pi#.