- #1

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

is this what you get?

- Thread starter aisha
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- #1

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

is this what you get?

- #2

mezarashi

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A sine wave function can be expressed generally as:

[tex] f(x)=A\sin (\frac{2\pi x}{T}+ \theta}) + C [/tex]

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.

[tex] f(x)=A\sin (\frac{2\pi x}{T}+ \theta}) + C [/tex]

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.

Last edited:

- #3

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

- #4

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Anyone else get the same answer as me?

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