Can $\sin(\pi/25)$ be expressed in radicals

The answer to this question depends on exactly what you mean by expressed in radicals. In the sense which is usually meant in Galois theory courses, $\cos \pi/25$ is expressible in radicals, but in a dumb sense: We have $\cos (\pi/25) = \frac{1^{1/50}+1^{-1/50}}{2}$ for one of the $50$-th roots of $1$. You may argue that this is a useless expression, but it is an expression in radicals in the sense which is used in Galois's result that a polynomial is solvable in radicals if and only if its Galois group is solvable. (And, indeed, the Galois group in this case is abelian.)

It is also solvable in radicals in a slightly less dumb sense. We have $e^{(\pi i)/5} = \frac{1 + \sqrt{5}}{4} + \sqrt{\frac{5-\sqrt{5}}{8}} i$. So, if you allow complex numbers, then we have $e^{(\pi i)/25} = \sqrt[5]{\frac{1 + \sqrt{5}}{4} + \sqrt{\frac{5-\sqrt{5}}{8}} i}$ and thus $2 \cos(\pi/25) = \sqrt[5]{\frac{1 + \sqrt{5}}{4} + \sqrt{\frac{5-\sqrt{5}}{8}} i} + \sqrt[5]{\frac{1 + \sqrt{5}}{4} - \sqrt{\frac{5-\sqrt{5}}{8}} i}$. (I consider this a less dumb sense because now the standard branch cut for $\sqrt[5]{z}$ does the job, instead of using a highly nonstandard $50$-th root of $1$.)

However, perhaps you don't want to permit complex numbers. In this case, the answer is no. Isaacs proves the following result:

Suppose that $f \in \mathbb{Q}[x]$ is an irreducible polynomial all of whose roots are real. If any of the roots of $f$ are expressible by real radicals, then $\deg(f)$ is a power of $2$.

The minimial polynomial of $\cos (\pi/25)$ is $$f(x) = -1 - 10 x + 100 x^2 + 40 x^3 - 800 x^4 - 32 x^5 + 2240 x^6 - 2560 x^8 + 1024 x^{10}.$$ The degree of $f$ is not a power of $2$, and the roots of $f$, namely $\cos (j \pi/25)$ for $j \in \{ 1,2,3,4,6,7,8,9,11,12 \}$, are all real.

Here's an expression for $\xi$ in terms of radicals, according to Maple:

$$1/10\,\sqrt {20\, \left( -1/4-1/4\,\sqrt {5}-1/4\,\sqrt {-10+2\,\sqrt {5}} \right) \sqrt [5]{-{\frac {-3125\,i\sqrt {5}\sqrt {2}+12500\, \sqrt {5}\sqrt {{\frac {\sqrt {5}}{\sqrt {5}-1}}}+15625\,i\sqrt {2}- 25000\,\sqrt {{\frac {\sqrt {5}}{\sqrt {5}-1}}}}{2048\, \left( \sqrt { 5}-1 \right) ^{2}\sqrt {{\frac {\sqrt {5}}{\sqrt {5}-1}}}}}}+50+20\,{ \frac { \left( -1/4-1/4\,\sqrt {5}+1/4\,\sqrt {-10+2\,\sqrt {5}} \right) \left( {\frac {25}{32}}+1/2\,{\frac {{\frac {3125}{32768}}-{ \frac {3125\,\sqrt {5}}{32768}}}{-{\frac {125\,\sqrt {5}}{2048}}+{ \frac {125}{2048}}}} \right) }{\sqrt [5]{-{\frac {-3125\,i\sqrt {5} \sqrt {2}+12500\,\sqrt {5}\sqrt {{\frac {\sqrt {5}}{\sqrt {5}-1}}}+ 15625\,i\sqrt {2}-25000\,\sqrt {{\frac {\sqrt {5}}{\sqrt {5}-1}}}}{ 2048\, \left( \sqrt {5}-1 \right) ^{2}\sqrt {{\frac {\sqrt {5}}{\sqrt {5}-1}}}}}}}}} $$ This doesn't look very nice in a small window... can anybody make it look prettier? The raw Maple expression is

(1/10)*sqrt((20*(-1/4-(1/4)*sqrt(5)-(1/4)*sqrt(-10+2*sqrt(5))))*(-(3125/2048)*(-I*sqrt(5)*sqrt(2)+4*sqrt(5)*sqrt(sqrt(5)/(sqrt(5)-1))+(5*I)*sqrt(2)-8*sqrt(sqrt(5)/(sqrt(5)-1)))/((sqrt(5)-1)^2*sqrt(sqrt(5)/(sqrt(5)-1))))^(1/5)+50+(20*(-1/4-(1/4)*sqrt(5)+(1/4)*sqrt(-10+2*sqrt(5))))*(25/32+(1/2)*(3125/32768-(3125/32768)*sqrt(5))/(-(125/2048)*sqrt(5)+125/2048))/(-(3125/2048)*(-I*sqrt(5)*sqrt(2)+4*sqrt(5)*sqrt(sqrt(5)/(sqrt(5)-1))+(5*I)*sqrt(2)-8*sqrt(sqrt(5)/(sqrt(5)-1)))/((sqrt(5)-1)^2*sqrt(sqrt(5)/(sqrt(5)-1))))^(1/5))

As requested, R. Israel's Maple answer can be simplified. That complicated expression,

$$u = -\frac{-3125\sqrt{-10}+12500\sqrt{5}\sqrt{\frac{\sqrt{5}}{\sqrt{5}-1}}+15625\sqrt{-2}-25000\sqrt{\frac{\sqrt{5}}{\sqrt{5}-1}}}{2048(\sqrt{5}-1)^2\sqrt{\frac{\sqrt{5}}{\sqrt{5}-1}}}$$

is just equal to,

$$u=\big(\tfrac{5}{4}\big)^5\,\Big(\tfrac{1-\sqrt{5}}{4}-\sqrt{\tfrac{5+\sqrt{5}}{8}}i\Big)$$

Hence, after some simplification, his answer reduces to the ratio of arithmetic mean (AM) and geometric mean(GM),

$$\sin\big(\tfrac{\pi}{25}\big) = \frac{AM(a,b)}{GM(a,b)} = \frac{(a+b)/2}{\sqrt{ab}} = 0.1253332\dots$$

where,

$$a = e^{4\pi i/5} = \tfrac{-1-\sqrt{5}}{4}+\sqrt{\tfrac{5-\sqrt{5}}{8}}\,i$$

$$b = (e^{7\pi i/5})^{1/5} = e^{47\pi i/25} = e^{-3\pi i/25} = \left(\tfrac{1-\sqrt{5}}{4}-\sqrt{\tfrac{5+\sqrt{5}}{8}}\,i\right)^{1/5}$$

which should be prettier enough.