What is the period of #y=2-3sin(pi/4)(x-1)#?

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Answer 1 · 2016-11-08T03:36:09

The period, #tau = 8#

Given the general form, #y = Asin(Bx + C) + D#

#B = (2pi)/tau# where #tau# is the period

In this case, #B = pi/4#

#pi/4 = (2pi)/tau#

#1/4 = (2)/tau#

#tau = 2/(1/4)#

#tau = 8#