Which graph corresponds to which equation? The figure below represents the graphs of parametric curve $x(t) = \sin(t)$, $y(t) = \frac{1}{2}\sin(2t)$, $0 \le t \le 2\pi$ and polar curve $r^2 = \cos(2\theta)$, $0\le \theta \le 2\pi$. Which curve corresponds to which equation?
I simply have no idea how to approach this, as it seems that $t$ doesn't have a direct relationship to $\theta$.
 A: The point $(x,y)=(1,0)$ belongs to both curves, at $t=\pi/2$ for the parametric curve and at $\theta=0$ for the polar curve, so let us compare the behaviour of the two curves in a neighborhood of this point.
Our only prerequisite shall be the one term Taylor expansion $$\sqrt{1-2z}=1-z+o(z^2)$$ when $z\to0$, which (if need be) follows from the double inequality, with an elementary proof, stating that, for every $|z|\leqslant\frac12$, $$1-z-2z^2\leqslant\sqrt{1-2z}\leqslant1-z$$


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*On the polar curve, $r^4=r^2(\cos^2\theta-\sin^2\theta)$ hence $(x^2+y^2)^2=x^2-y^2$. Solving this quadratic identity in $x^2$ yields $2x^2=1-2y^2\pm\sqrt{1-8y^2}$. 
Thus, in a neighborhood of $(x,y)=(1,0)$, on the polar curve, $$2x^2=1-2y^2+\sqrt{1-8y^2}$$
The limited expansion when $y\to0$ is $$2x^2=2-6y^2+o(y^2)$$

*On the parametric curve, $y^2=\sin^2t\cos^2t=x^2(1-x^2)$ hence $x^4-x^2+y^2=0$. Solving this quadratic identity in $x^2$ yields $2x^2=1\pm\sqrt{1-4y^2}$. 
Thus, in a neighborhood of $(r,\theta)=(1,0)$, which is also $(x,y)=(1,0)$, on the parametric curve, $$2x^2=1+\sqrt{1-4y^2}$$
The limited expansion when $y\to0$ is $$2x^2=2-2y^2+o(y^2)$$
This proves that the parametric curve leaves $(1,0)$ by staying slightly farther from $(0,0)$ than the polar curve, thus the parametric curve is drawn in red and the polar one in green.
Edit: For a non asymptotic approach, one should study the sign of $$u(y)=\left(1+\sqrt{1-4y^2}\right)-\left(1-2y^2+\sqrt{1-8y^2}\right)$$ which is also $$u(y)=2y^2+\left(\sqrt{1-4y^2}-\sqrt{1-8y^2}\right)$$ The $2y^2$ term and the term in the parenthesis are both nonnegative hence $u(y)\geqslant0$ everywhere. This shows that the polar curve is always "inside" the parametric curve, extending the validity of the conclusion above from a neighborhood of $(1,0)$ to the whole curves.
A: From $r(\theta)=\sqrt{\cos(2\theta)}$ we obtain $y(\theta)=\sqrt{1-2\sin^2\theta}\sin\theta$. Writing $\sin^2\theta=:u$ we therefore have
$$y^2(\theta)=(1-2u)u=2\> u\left({1\over2}-u\right)\ .$$
The RHS is maximal for  $u={1\over4}$, with maximal value  ${1\over8}$. It follows that the maximal $y$-value along the "polar curve" is ${1\over\sqrt{8}}$, whereas the maximal $y$-value along the other curve is obviously ${1\over2}>{1\over\sqrt{8}}$.
