Sin and cos equation

Question

f(x) = 1 2 1 0 1 2 1 0 1

x = 0 pi/2 pi 3pi/2 2pi 5pi/2 3pi 7pi/2 4pi

What is the sin equation for this and what is the cos equation for this relationship?

Answer 1

Let f(x) = Asin[B(x-C)] + D or f(x) = Acos[B(x-C)] + D (the cosine graph is always a shift π/2 right of the sine graph)

Where A = amplitude [(Max - Min)/2]

B = frequency (number of cycles per 2π radians)

C = phase shift

D = vertical shift

Since f(0) = 1 and f(π/2) = 2 it appears that A = 1 (or max occurs at 2 and min at 0 and (2-0)/2=1)

Since f(0) = f(2π) we see that B = 1

Since the max = 2 and min = 0, the mid-line must be (2+0)/2 = 1 (the average)

So: f(x) = sinx + 1 the cosine starts at the max and therefore the shift is π/2 to the right

So: f(x) = cos(x-π/2) + 1

This method works for all sine/cosine graphs.