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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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First: do you know the addition formulae for $\sin$ and $\cos$? – J. M. Sep 6 '11 at 8:05
you mean sin(A+B) = sinAcosB + cosAsinB and cos(A+B) = cosAcosB - sinAsinB right? – alok Sep 6 '11 at 8:10
Yes, precisely. You also know the special values, e.g. $\sin\,\pi$ and $\cos\,\frac{\pi}{2}$ right? – J. M. Sep 6 '11 at 8:11
sorry for the delayed response.$$sin{\pi} = 0$$ and $$cos{\pi/2}= 0$$ – alok Sep 6 '11 at 8:33
well what jm meant was to use these values and the addition formulae to see if u can arrive at the answer – Bhargav Sep 6 '11 at 8:45
up vote 3 down vote accepted

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
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My sincere appologies for the delayed response.The internet was down in my locality!! – alok Oct 30 '11 at 9:12

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