Question

Hi all,
Can somebody please explain how to calculate amplitude, period and phase angle for −4cos(x + π/2)? (Session1, Problem set1: 1A-7-b)

Answer 1

For a function of the form \[A \cos (Bx-C)\] The amplitude is \[A\] The period is \[\frac{ 2 \pi }{ B }\] The phase shift is \[\frac{C}{B}\]

Answer 2

Thanks Raizel.

Answer 3

If I may ask, in the same question, how is -4cos(x+pi/2) simplified to 4sinx?
Also, is it just me or there are quite a few questions in this problem set up until this point, the material for which hasn't been covered in the lectures (inverse functions and amplitude for e.g.). Have you found this problematic?

Answer 4

cos(x+pi/2) = cos(x+90deg) = cos x cos 90 - sin x sin 90 = 0 - sin x = -sinx
-4 cos(x+pi/2) = -4 * - sinx = 4 sin x

here if reference is A sin (Bx + C) then amp A = 4 and phase C = 0
whereas if reference is A cos (Bx + C) then amp A = -4 and phase C = pi/2 = 90 deg

all yield the same solution...it just a matter of what you want to see :D