= Solution
The fourth <Chebyshev polynomial> eliminates the parameter:
$$
y=\cos4\theta=8x^4-8x^2+1,\qquad -1<x<1.
$$
Its derivatives are $y'=16x(2x^2-1)$ and $y''=16(6x^2-1)$. The interior <stationary points> are a local maximum $(0,1)$ at $\theta=\pi/2$, and two minima $(\pm1/\sqrt2,-1)$ at $\theta=\pi/4,3\pi/4$. At the maximum $y''=-16<0$, while at either minimum $y''=32>0$.
The two <inflection points> are
$$
\boxed{(x,y)=\left(\pm\frac1{\sqrt6},-\frac19\right),\qquad \theta=\arccos\frac1{\sqrt6},\ \pi-\arccos\frac1{\sqrt6}.}
$$
The curvature changes sign there. The endpoint limits are $(\pm1,1)$, at $\theta=0,\pi$; they are excluded from the open curve. The central maximum and the two minima are also the attained global extremal values on that open curve.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2004/paper-2-parametric-curves.png]
{title=Parametric curves after eliminating the cosine parameter}
{description=Original plots of the parabola and fourth Chebyshev polynomial on their open parameter domains. Arrows show traversal from right to left; the quartic's extrema and inflection points are marked.}
Back to article page
