How do you find the period of $y = 1 + 2sin [((pi x)/2) - (pi/4)]$ ?

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Answer 1 · 2016-12-07T03:47:57

Please see the explanation.

For a sine function of the general form:

$y = Asin(Bx + C) + D$

The period, $tau$ , is found, using the equation:

$tau = (2pi)/B$

We observe that $B = pi/2$ in the given equation:

$tau = (2pi)/(pi/2)$

$tau = 4$

How do you find the period of y = 1 + 2sin [((pi x)/2) - (pi/4)]y = 1 + 2sin [((pi x)/2) - (pi/4)]?