Polynomial satisfied by $\cos^n(t)$ and $\sin^n(t)$ For any even $n$, there should be a polynomial $f(x,y)$ that vanishes on the points $(\cos^n(t),\sin^n(t))$ for all $t$ (since it is the image of the projective variety $x^2+y^2 = 1$ under the $n/2$th-power-in-each-coordinate map).  Is it possible to give an explicit form for $f$?
 A: You can use polynomial elimination, which is implemented in Macaulay2.
In fact, the parametric equations of your affine curve are $$x=\frac{(2t)^n}{(1+t^2)^n}, \quad y=\frac{(1-t^2)^n}{(1+t^2)^n},$$
so that an implicit equation $f(x, \,y)=0$ for it is obtained by eliminating the variable $t$ among these, i.e. eliminating $t$ in the ideal $$I=(x(1+t^2)^n-(2t)^n, \, y(1+t^2)^n-(1-t^2)^n) \subset \mathbb{Q}[x, \, y, \, t].$$ We can do this for all the values of $n$, non only the even ones. For instance, in the case $n=5$ the script is the following:
R=QQ[x, y, t];
n=5;
I=ideal(x*(1+t^2)^n-(2*t)^n, y*(1+t^2)^n-(1-t^2)^n);
J=eliminate(t, I);

Now the command "gens J" gives a list of generators for the elimination ideal, that consists only of the polynomial we are looking for:
gens J
| x10+5x8y2+10x6y4+10x4y6+5x2y8+y10-5x8+605x6y2-1905x4y4+605x2y6-5y8+10x6+1905x4y2+1905x2y4+
  ----------------------------------------------------------------------------------------------
  10y6-10x4+605x2y2-10y4+5x2+5y2-1 |

For even $n$, the generator of $J$ has degree $n/2$, as expected. For example, when $n=6$ the script produces the following polynomial of degree $3$:
gens J
| x3+3x2y+3xy2+y3-3x2+21xy-3y2+3x+3y-1 |

A: I don't know if this is irreducible, but it gives an answer: Let $\zeta$ be a primitive $n/2$-root of unity. Multiply out
$$\prod_{a=1}^{n/2} \prod_{b=1}^{n/2} (\zeta^a x^{2/n} + \zeta^b y^{2/n} - 1 ).$$
The result is a polynomial in $x$ and $y$, and one of the factors is $x^{2/n} + y^{2/n}-1$, so it vanishes when $(x,y) = (\cos^n t, \sin^n t)$.
A: Here are some small explicit solutions and general comments. 
Since the polynomials are symmetric, one would expect them to be more nicely expressed using $$S=x+y \\ M=x-y\\ P=xy.$$
Here $M$ should only appear to even powers and $M^2=-(x-y)(y-x).$ 
It seems convenient to use all three although either $P$ or $M^2$ suffices with $S$ since
$4P=S^2-M^2.$
Also, I will bend notations and introduce a symbol $T$ with $T^k$ interpreted as $M^k$ or $S^k$ according as $k$ is even or odd. 


*

*For $n=2$ the desired polynomial is , of course, $$S-1=0.$$

*For $n=4$ the desired polynomial is $$(S-1)^2-4P= \\ (T-1)^2=0$$

*For $n=6$ the desired polynomial is $$(S-1)^3+27P= \\ (T-1)^3+15P=0$$

*For $n=8$ the desired polynomial is $$(S-1)^4-8P\left(S^2-2P+14S+17\right)= \\(T-1)^4-112P(S+1)=0$$

*For $n=10$ I was unable to derive the polynomial (see comments below) however the degree $10$ polynomial given by Francesco satisfied by the points $(x,y)=(\cos^5(t),\sin^5(t))$ is even. If we halve the degrees it is degree $5$ and  satisfied by all the points $(x,y)=(\cos^{10}(t),\sin^{10}(t)).$
${x}^{5}+5\,{x}^{4}y+10\,{x}^{3}{y}^{2}+10\,{x}^{2}{y}^{3}+5\,x{y}^{4}+
{y}^{5}-5\,{x}^{4}+605\,{x}^{3}y-1905\,{x}^{2}{y}^{2}+605\,x{y}^{3}-5
\,{y}^{4}$
$+10\,{x}^{3}+1905\,{x}^{2}y+1905\,x{y}^{2}+10\,{y}^{3}-10\,{x
}^{2}+605\,xy-10\,{y}^{2}+5\,x+5\,y-1
$
$$(S-1)^5+625P(M^2+3S+1-P)=\\(T-1)^5+15P(39(T-1)^2+203S-47P))=0$$
The cases of $n=10,12,16$ can be worked out and continue to increase in complexity at about the same rate. 

The curve we want is $x^{1/m}+y^{1/m}=1$ where $n=2m.$ So we could as well think of this as an image of the` curve $x+y=1.$ Really we only want the restriction to the positive quadrant. For $m$ even the equation is only defined in the positive quadrant. This is not the case for $m$ odd.  Perhaps the right thing to do is to think of it as the unit ball in the $L_p$ metric for $p=\frac1m$: $$|u|^{1/m}+|v|^{1/m}=1.$$ 
For $n=4,6,8$ it isn't to bad to start from  $x^{1/m}+y^{1/m}=1$ and manipulate be rearranging and raising to powers to get a polynomial. I'm not sure that continues.
Instead I used David Speyer's method. The primitive $m$th roots of unity have reasonably simple expressions for $m=2,3,4,6,8.$ Also $m=5$ is nice but I couldn't quite get that to work. 

What about a polynomial satisfied by the points $(\sin^n(t),\cos^n(t))$ for $n$ odd? Just take the polynomial for $(\sin^{2n}(t),\cos^{2n}(t))$ and double all the exponents.
