# Trigonometric Phase Lines

• April 8th 2010, 05:28 PM
isp_of_doom
Trigonometric Phase Lines
Hi,
As part of a recent assignment I have been asked to draw a phase line for the function $\frac{dy}{dt} = sin(y)$. I can find the main equilibrium point easily ( $y = 0$) but the problem I have is that $y = 180, y = 360, y = 540$ are also solutions. So in a way I have a phase line that continues infinitely. Is there any formal way of setting this out. The main idea that comes to mind is including an algebraic multiplier of both 180 and -180 on the phase line the stating $a \in Z$ when $a$ is the multiplier.

Thoughts?
• April 8th 2010, 05:36 PM
dwsmith
Usually with trig functions, one will write $k\pi$. That way for any integer you will hit all multiples of $\pi$ or $180$
• April 8th 2010, 05:39 PM
isp_of_doom
thanks dwsmith I figured it would be something like that.