what are the phase shift and period for the function y = 4 cos 1/3 ( 0 + 45 degrees ) - 3?
+37170 I'll hazard a guess...first of all I will guess that 0+45 is really 'theta' +45 and I'll guess that 45 would be the phase shift.
The 1/3 in the equation lends me to believe the period will be 360 x 3 ..... these are all guesses....educate me...THANX !
+37170 I'll hazard a guess...first of all I will guess that 0+45 is really 'theta' +45 and I'll guess that 45 would be the phase shift.
The 1/3 in the equation lends me to believe the period will be 360 x 3 ..... these are all guesses....educate me...THANX !
+118723 Yes I am with you ElectricPavlov, we need a theta,
The phase shift is actually -45 degrees
I think of it like this
\(\theta+45=0\\ \theta=-45\\ \)
Maybe that "logic' won't work for you but it works for me :)
Here is a graph.
+118723 Hi ElectricPavlov and asker. Here is more of a look if you are interested :)
y = 4 cos 1/3 ( theta + 45 degrees ) - 3]
\(y = 4 cos[\frac{ 1}{3} (\theta + 45 ) ]- 3\)
I like to rewrite this as
\(y+3 = 4 cos[\frac{ 1}{3} (\theta + 45 ) ]\)
1) amplitude=4
2) Centre of wave y+3=0 y=-3
(there is probably a better name for this but I don't know it)
3) \(Period = \frac{360}{\frac{1}{3}} = 360\div\frac{1}{3} = 360*3 = 1080\; degrees\)
4) Phase shift
\(\theta+45=0 \qquad \theta=-45\\ \mbox{That is 45 degrees in the negative direction}\)
Here is a you tube clip if you want to watch it :)
