Parametric equation of inward pointing half sin waves
I can create a circle in red and I can create a sin wave that goes around a circle in green.
Parametric equation:
x=7cos(t)
y=7sin(t)+sin(2pi*5t)
But how can I create a half sin wave pointing inward going around in a circle using parametric equations see image below.
First, to help work out how we might create something like this, we will think more with respect to polar coordinates. If you don't know what polar coordinates are, all you need to understand for this answer is that we will think about representing points with an angle and a radius, rather than an $x$ and $y$ coordinate. We do this because it will allow us to generate graphs that repeat around the origin more easily.
If we call the angle $t$, and we let the angle be any value around the origin (but only each angle once) then we shall say $0 \leq t < 2\pi$. Now we know that a circle can be parametrised by $(rcos(t), rsin(t))$ for some fixed $r>0$, so we could try and make this $r$ vary as $t$ varies to change the distance of the graph to the origin. As per your image, we want this $r$ to vary from some value $a>0$ to some value $b>a$. Then if we select $$r=\frac{(b-a)sin(kt)+a+b}{2}$$ for some $k \in \mathbb{N}$, we see that $a \leq r \leq b$ as desired. The reason I inserted this $k$ here is to indicate the number of turning points in the shape. $sin(t)$ only has two turning points for $0 \leq t < 2\pi$, but our graph we are trying to plot has $16$ turning points. So, if we select $k=8$ then we would have $16$ turning points as desired. This yields the following:
If you want your graph rotated as in your image then we rotate with $(u, v) \to (u cos(p)-vsin(p), usin(p)+vcos(p))$. Specifically, we want a turning point of our graph (one of which occurs when $8t=\pi/2 \iff t=\pi/16$) to be at $(0,1) = (cos(\pi/2), sin(\pi/2))$. Thus we need $p = \pi/2 - \pi/16 = 7 \pi/16$.
Final product:
Note: this clearly does not look exactly like the graph shown - I assumed that was a more general image. The image shown, to me, does not appear to be very trigonometric. It could likely be closer modelled with a series of straight lines and sections of circles, but this will yield a much less pleasant result.