The problem I'm looking at asks to prove that the following expression is an integer:

(tan(3pi/11) + 4sin(2pi/11))^2

I know that the expression is equal to 11, and I assume that isn't a coincidence, but I dont know how to relate tan(pi/11) to 11.

I'm not trying to get credit for this problem, I'm just curious about how to approach it.

Is this the reason that closed form expressions are never given for sin(pi/d), when d is composed only of prime numbers greater than 5?

I do not immediately see how to prove that but one thing I am certain of is that you do not have to look for a relationship between tan(pi/11) and 11. In all likelihood, it's a matter of using persistently some trig identities, like say

(sin(3pi/11) + 4sin(2pi/11)cos(3pi/11))^2 =
(sin(3pi/11) + 2sin(5pi/11) - 2sin(pi/11))^2

Then you square and again use either sin(a)sin(b) or sin(a)cos(b) or the third one. Probably, along the way, you may find that using

sin(5pi/11) = sin(6pi/11)

or something in this spirit will be of advantage. If you do not err, such a straightforward thing will likely lead to significant simplifications.

https://mathworld.wolfram.com/TrigonometryAnglesPi11.html

where they say:

Trigonometric functions of n*pi/11 for n an integer cannot be expressed in terms of sums, products, and finite root extractions on real rational numbers because 11 is not a Fermat prime.
However, exact expressions involving roots of complex numbers can still be derived using the multiple-angle formula

(1) sin(n*alpha) = (-1)^((n-1)/2) * T_n(sin(alpha))

where T_n(x) is a Chebyshev polynomial of the first kind.

Setting n=11 gives

(2) sin(11*alpha) = sin(alpha)*(11-220*sin^2(alpha)+1232*sin^4(alpha)-2816*sin^6(alpha)+2816*sin^8(alpha)-1024*sin^10(alpha)).

Letting alpha=pi/11 and x=sin^2(alpha) then gives

sin(pi)=0=11-220x+1232x^2-2816x^3+2816x^4-1024x^5.

etc.. (more in the reference)

But there must be another way to solve the problem of the week.