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) $$−4\cos(x + π/2)$$
$\endgroup$ 51 Answer
$\begingroup$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
