# How to find the Period and Phase angle?

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

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.

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