# Tricky Trig Function Find the eqn Help

Find the equation of a sine function that has a vertical displacement 2 units down, a horizontal phase shift pi/3 to the right, a period of pi/6, reflection in the y-axis and the amplitude of 3.

my answer $$y=3\sin [-12(x+\frac {\pi} {3})] -2$$

is this what you get?

mezarashi
Homework Helper
A sine wave function can be expressed generally as:

$$f(x)=A\sin (\frac{2\pi x}{T}+ \theta}) + C$$

where A is peak to peak amplitude, T is period (or 1/T = frequency f), theta is phase shift (horizontal), and C is the vertical displacement.

P.S. Your original question. Having the function be both even (reflected about the y-axis), and also forcing a phase-shift generally doesn't make sense (given you want a continuous function). Having both conditions satisfied would be purely coincidental.

Edit: I forgot to note, when your theta is positive, the phase shift wil be to the left. A negative theta will result in a shift to the right.

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I think it should be pi/36 instead of pi/3 although I could be wrong.

Anyone else get the same answer as me?

At first glance it looks OK. The only thing I found is that for a phase shift to the right, it would be x-pi/3 rather than x+pi/3.