We may write these equations as$$\cos \theta+\cos\phi=a,\qquad$$$$\cos\theta\sin\phi+\sin\theta\cos\phi=b\sin\theta\sin\phi,\qquad$$$$\qquad\sin\theta+\sin\phi=c\sin\theta\sin\phi.$$Let us write $\alpha:=\frac12(\theta+\phi)$ and $\beta=\frac12(\theta-\phi)$. Then the above equations may be written, using standard trigonometric formulae, as$$2\cos\alpha\cos\beta=a,\qquad\qquad\qquad\qquad(1)$$$$4\sin\alpha\cos\alpha=b(\cos2\beta-\cos2\alpha),\qquad$$$$4\sin\alpha\cos\beta=c(\cos2\beta-\cos2\alpha).\qquad$$Consider first the special case when $\cos2\beta=\cos2\alpha.$ This is when either $\theta$ or $\phi$ is a multiple of $\pi$. But then the original equations would be undefined; so we can rule this case out. Thus we may divide the last two displayed equations to obtain$$c\cos\alpha=b\cos\beta.\qquad\qquad\qquad\qquad(2)$$By subtracting the same equations, writing $\cos2\beta-\cos2\alpha=2(\cos\beta-\cos\alpha)(\cos\beta+\cos\alpha)$, and dividing through by $\cos\beta-\cos\alpha$ (which must be nonzero in the present case), we get$$2\sin\alpha=(c-b)(\cos\alpha+\cos\beta).\qquad(3)$$ Now substituting for $\cos\beta$ from eqn $2$ into eqns $1$ and $3$, using the fact that $\cos^2\alpha+\sin^2\alpha=1$, and some straightforward algebra yields$$a(c^2-b^2)^2=4b(2c-ab).$$
These are the parametric equations of an epicycloid, where the large circle has radius $R=1$, and the smaller exterior rolling circle has radius $r=\frac{1}{4}$.
A possible simplification is writing
$$4a=4(x+i y) = 5 z - z^ 5\\
4b=4(x-i y) = \frac{5}{z} - \frac{1}{z^5}$$
where $z = \cos \theta + i \sin \theta$
and get a relation between $a$, $b$, that implies a relation between $x$ and $y$, see link. Or, we can just give the result provided by WA :
$$-81 - 45 x^2 + 365 x^4 - 15 x^6 - 480 x^8 + 256 x^{10} - 45 y^2 - 2395 x^2 y^2 - 45 x^4 y^2 - 1920 x^6 y^2 + 1280 x^8 y^2 + 365 y^4 - 45 x^2 y^4 - 2880 x^4 y^4 + 2560 x^6 y^4 - 15 y^6 - 1920 x^2 y^6 + 2560 x^4 y^6 - 480 y^8 + 1280 x^2 y^8 + 256 y^{10}=0$$
a curve of degree $10$
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We can simplify the defining polynomial by writing it as a polynomial in $x^2 + y^2$ and $x^2 y^2$.
$\bf{Added:}$ Following the solution of @ACB: we have the polar representation ($x = r \cos \phi$, $y=r \sin \phi$)
$$\cos^2 2 \phi = \frac{(9 - 4 r^2)(16 r^4+ 3 r^2 + 6)^2}{3125 r^4}= f(r)$$
The function in $r>0$ on RHS is strictly decreasing on $(0, \infty)$. Indeed, we have
$$f'(r) = -24\, \frac{(-1 + r^2)^2 (9 + 16 r^2) (6 + 3 r^2 + 16 r^4)}{3125\, r^5}$$
Now, $f(1) = 1$, and $f(\frac{3}{2})=0$. We see that our curve is situated between the circles of radius $1$, and $\frac{3}{2} = 1 + 2\cdot \frac{1}{4}$, as it should.
So now we can write the curve in polar coordinates
$$r = f^{-1}(\cos^2 2\phi)$$
where $f^{-1}\colon [0,1]\to [1, \frac{3}{2}]$ is the inverse function of $f$.
Best Answer
You've done the hard part. Of A):
This is where the Pythagorean Identity (as you call it) comes in very handy, indeed:
$$\quad\quad\quad\sin^2(\theta) + \cos^2(\theta) = 1$$
$$\iff \left(\sin\theta \right)^2 + \left(\cos\theta \right)^2 = 1$$
$$x^2 + \left(\dfrac{y}{2x}\right)^2 = 1$$
Then just simplify!
Part B:
$$x = 3(1-2\sin^2\theta) \tag{$\cos(2\theta) = 1 - 2\sin^2\theta$}$$ $$x= \frac 32(2 - (1 - \cos^2\theta)) \tag{$\sin^2\theta = 1 - \cos^2 \theta$}$$ $$x = \frac 32 (\cos^2 \theta + 1) \iff \frac 23 x - 1 = \cos^2\theta$$ $$ $$ $$ y = 2 \sin \theta \iff \sin \theta = y/2$$ $$ $$ $$\sin^2 \theta + \cos ^2 \theta = 1\tag{Pythagorean identity again}$$ $$\left(\frac y2\right)^2 + \left(\frac 23 x - 1\right) = 1$$
Simplify.