How do you find the amplitude, period, phase shift given #y=-5sin3x+12cos3x#?

1 Answer
Aug 1, 2018

Amplitude = 13, period = #(2pi)/3# and phase shift #= - 0.6552#, nearly.

Explanation:

#y = -5 sin 3x + 12 cos 3x#

# = 13 ( (-5/13 )sin 3x + (12/13) cos 3x#

#13 ( sin 3x cos alpha + cos 3x sin alpha )#,

#alpha = pi - sin^(-1)(12/13)#

#= 13 sin ( 3x + alpha )#. Now, easily

amplitude = 13, period = #(2pi)/3# and

phase shift# = - alpha/3 = 1/3 ( sin^(12/13) - pi )#

# =1/3 ( 1. 1760 - 3.1416 ) = - 0.6552#, nearly.

See ( not-to-uniform-scale ) graph, with all these aspects:
graph{(y-13 sin ( 3x -1.9656))(y -13+0x )(y+13+0x)( x - 1/3pi+0.0001y)( x + 1/3pi+0.0001y)=0[-2 2 -16 16]}