 Solution

The fourth Chebyshev polynomial eliminates the parameter:Its derivatives are $y^{\prime }=16x(2x^2-1)$ and $y^{\prime \prime }=16(6x^2-1)$. The interior stationary points are a local maximum $(0,1)$ at $\theta =\pi /2$, and two minima $(\pm 1/\sqrt{2},-1)$ at $\theta =\pi /4,3\pi /4$. At the maximum $y^{\prime \prime }=-16<0$, while at either minimum $y^{\prime \prime }=32>0$.

The two inflection points areThe curvature changes sign there. The endpoint limits are $(\pm 1,1)$, 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.