\(y=a* cos[n(\theta+p) ]+ L\)
Amplitude =a
phase shift = p (units in the NEGATIVE direction - opposite direction to what most people expect)
wave length \(\lambda = \frac{2\pi}{n}\)
L is the vertical shift
SO CONSIDER
\(y=1.8cos(3\theta+\frac{\pi}{2})-1.5\\ rewrite\;\; as \\ y=1.8cos(3[\theta+\frac{\pi}{6}])-1.5\\\)
It has the basic \(y=cos(\theta) \)shape.
wavelength = \(\frac{2\pi}{3}\)
Phase (horizontal) shift =\( \frac{\pi}{6}\;\)units in the negative direction
Amplitude =1.8
Vertical shift is 1.5 units DOWN
check
Here is the graph.
You can play iwth the circles on the left to see how I 'developed' the graph
https://www.desmos.com/calculator/bluugr6nna