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I'm currently brushing up my trig and found these two problems. I'm totally clueless on how to start. Please help.

Find the period , amplitude , and phase angle, and use these to sketch

a) $3\sin(2x − π)$

b) $−4\cos(x + π/2)$

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    well what jm meant was to use these values and the addition formulae to see if u can arrive at the answer2011-09-06

1 Answers 1

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In simple words. When talking about a periodic function:

  • Amplitude is its highest absolute value. Well it's actually a subject to convention/definition what is called the amplitude, but at least for sin/cos functions this is so.
  • Period is the minimal value that you may add to the argument without the function change.
  • Phase is a matter of the convention. You may define one function as one with phase=0. Then if another function may be brought to this one by "shifting" (i.e. subtracting the shift from argument) - its phase is said to be equal to this shift.

Whereas the period has a strict absolute definition, the amplitude and the phase are subject for the convention. There is however a strict definition for relative amplitude and phase.

Now, about your exercise.

If you have a function of the form f(x) = |a| sin (bx + c) then:

  • |a| is the amplitude
  • 2π/b is the period
  • c is the phase

Note: we actually defined a convention here. The amplitude is the maximum value of the function, and the phase=0 is defined for the point where the function is 0 with positive derivative.

f(x) = 3sin(2x−π) = 3sin(2x+π)

  • Amplitude = 3
  • Period = π
  • Phase = π

f(x) = −4cos(x+π/2) = 4sin(x) [trigonometry equality]

  • Amplitude = 4
  • Period = 2π
  • Phase = 0
  • 0
    My sincere appologies for $t$he delayed response.The in$t$ernet was down in my locality!!2011-10-30