If I let $f(\theta)=((\mathrm{cos} \theta)X+(\mathrm{sin} \theta)Y)^{n1}$ and view the range of this curve as a subset of the space of homogeneous polynomials of degree $n1$ in two variables viewed as an $n$dimensional Euclidean space, then I can show that there are $n$ linearly independent points on the curve, because if I pick $n$ distinct values of $\theta$ in the range $(0,\frac{\pi}{2})$ then we can show that the determinant of the matrix which has the corresponding values of $f(\theta)$ as columns is a nonzero multiple of a Vandermonde determinant. But I would like to show that this curve has $n+1$ affinely independent points. What would be the simplest way of doing that?
This isn't true if $n$ is odd. For example, if $n=3$, then your formula is $(a,b,c) = (\cos^2 \theta, 2 \sin \theta \cos \theta, \sin^2 \theta)$ and it always lies in the hyperplane $a+c=1$. More generally, whenever $n$ is odd, the equality $1 = (\sin^2 \theta+ \cos^2 \theta)^{(n1)/2} = \sum \binom{(n1)/2}{k} \sin^{2k} \theta \cos^{n12k} \theta$ gives a hyperplane containing your points.
When $n$ is even, this is true. We want to prove that the following list of $n+1$ functions is linearly independent: $\cos^{n1k} \theta \sin^k \theta$, for $0 \leq k \leq n1$, and the constant function $1$. If $k$ is odd, then $\cos^{n1k} \theta \sin^k \theta$ is an odd function; if $k$ is even, then $\cos^{n1k} \theta \sin^k \theta$ is an even function, as is the constant function $1$. An odd function and an even function can never be proportional unless they are both zero, so it suffices to show that the odd and the even functions are linearly independent.
For the odd functions, your Vandermonde determinant argument works (as do many others). For the even functions, note that $\cos^{n12j} \theta \sin^{2j} \theta = \cos^{n12j} \theta (1\cos^2 \theta)^j$. This is a polynomial in odd powers of $\cos \theta$ with terms in degrees between $n1$ and $n1  2j$. Therefore, the matrix which transforms the even functions in your list into the functions $\cos^{2m+1} \theta$ is upper triangular with nonzero entries on the diagonal, and is thus invertible. For example, $$\begin{pmatrix} \cos^3 \theta \\ \cos \theta \sin^2 \theta \\ 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} \cos^3 \theta \\ \cos \theta \\ 1 \end{pmatrix} $$

$\begingroup$ Okay, great, thanks for pointing that out. I'd still be interested to know a proof of the case $n$ is even if there is one. $\endgroup$ – Rupert Mar 9 '15 at 12:14
