Disproving an "almost true" trigonometric identity

First note that $[\mathbb{Q}(\sin\tfrac{\pi}{51}):\mathbb{Q}]=32$ so $[\mathbb{Q}(\sin\tfrac{\pi}{51}, \sqrt{2}):\mathbb{Q}]\in\{32,64\}.$ Now $[\mathbb{Q}(\cos\tfrac{\pi}{74}):\mathbb{Q}]=36$ and since $36\not\mid 64$ we conclude that $\cos\tfrac{\pi}{74}\not\in\mathbb{Q}(\sin\tfrac{\pi}{51}, \sqrt{2})$ and in particular $\cos\tfrac{\pi}{74}\neq \frac{3}{2\sqrt{2}}-\sin\tfrac{\pi}{51}$.

Well, algebraic number theory handles it without difficulty, but my argument may seem purely mumbojumbo if you’re not familiar with that subject.

For a positive integer $n$, let $\zeta_n$ be a primitive $n$-th root of unity, such as $\cos(2\pi/n)+i\sin(2\pi/n)$. We have $\sin(\pi/51)=\frac1{2i}(\zeta_{102}-\zeta_{102}^{-1})$, and we expect that this will be an irrationality of degree $64=2\phi(3)\phi(17)$ over the rationals $\mathbb Q$. In the same way, $\cos(\pi/74)=\frac12(\zeta_{148}+\zeta_{148}^{-1})$. Here the degree over $\mathbb Q$ is equal to $36=\frac12\phi(4)\phi(37)$. The first irrationality involves ramification at $3$ and $17$, the second at $2$ and $37$. The sum must involve ramification at all of these, which your expression in $\sqrt2$ does not.

EDIT: At the prompting of both @pew and @WimC, I have developed a more transparent explanation of why the two sides of the relation are unequal. It still depends on ramification theory, but in a way that most people are willing to accept. I’ll point out where.

I need to establish some facts about primitive roots of unity and the minimal $\mathbb Q$-polynomials they satisfy. First, in $\mathbb C$, any finite multiplicative subgroup $C$ is cyclic. If of order $N$, then $C$ has $\phi(N)$ generators, where $\phi$ is the “totient” function that counts the number of positive integers up to $N$ that are relatively prime to $N$. So, $\phi(1)=\phi(2)=1$, and more generally for a prime $p$, $\phi(p^k)=p^k-p^{k-1}$. One sees, by counting which roots of unity generate which cyclic subgroups, that $N=\sum_{d|N}\phi(d)$, and from the Möbius Inversion Formula, $$ \Phi(N)=\sum_{d|N}\mu(d)\frac Nd\,. $$ In the same way, if we take the polynomial $\Phi_N(X)$ (the $N$-th cyclotomic polynomial) whose roots are the primitive $N$-th roots of unity, we get \begin{align} X^n-1&=\prod_{d|N}\Phi_d(X)\\ \Phi_N(X)&=\prod_{d|N}(X^{N/d}-1)^{\mu(d)}\,, \end{align} which shows that $\Phi_N(X)$ is not only a polynomial (which it is from the definition), but has $\mathbb Z$-coefficients. And you see by the parallelism of the $\phi(N)$ with $\Phi_N(X)$ that $\deg(\Phi_N)=\phi(N)$.

Now it’s easy, by a trick using the Eisenstein Criterion, to show that when $N$ is a prime power, $\Phi_N$ is $\mathbb Q$-irreducible. It requires ramification theory, as far as I know, to show that all the cyclotomic polynomials are $\mathbb Q$-irreducible.

Next I want to point out that if we adjoin a primitive $n$-th root of unity, say $\zeta_n$, to $\mathbb Q$, we get a field of degree $\phi(n)$ over $\mathbb Q$: its irreducible polynomial is $\Phi_n$. And then I argue that if $m$ and $n$ are relatively prime (so that $\phi(mn)=\phi(m)\phi(n)$), then if we call $K=\mathbb Q(\zeta_m,\zeta_n)$, it turns out that $K$ is also equal to $\mathbb Q(\zeta_{mn})$. Certainly a field containing the $m$-th roots and the $n$-th roots of unity contains the $mn$-th roots (remember that $m$ and $n$ are relatively prime), but might at first glance be bigger, giving the inequality $[K\colon\mathbb Q]\ge\phi(mn)=\phi(m)\phi(n)$. But you get the $n$-th roots of unity by adjoining $\zeta_n$ to $\mathbb Q(\zeta_m)$, so a root of the polynomial $\Phi_n$, which might not be irreducible over the larger field, but at least the total degree of this doubly-adjoined field over $\mathbb Q(\zeta_m)$ is less than or equal to $\phi(n)$. But the $\phi$’s multiply, showing that $$ [K\colon\mathbb Q]=[K\colon\mathbb Q(\zeta_m)]\cdot\phi(m)\le\phi(n)\phi(m)\,. $$ the two degrees are equal, so the two fields are equal.

Now let’s go back to the original problem: we’re looking at the quantity $$ \frac1{2i}(\zeta_{102}-\zeta_{102}^{-1}) + \frac12(\zeta{148}+\zeta_{148}^{-1})\,. $$ Of course $i=\zeta_4$, and you see that everything is taking place in the field $L=\mathbb Q(\zeta_{4\cdot3\cdot17\cdot37})$. And now I claim that this field does not contain the eighth roots of unity. Its degree over $\mathbb Q$ is $\phi(4)\phi(3)\phi(17)\phi(37)$, but with the eighth roots of unity there, the total degree would be the same product but with $\phi(4)=2$ replaced by $\phi(8)=4$. So the eighth roots of unity are not in $L$.

It follows that $\sqrt2\notin L$, because $\zeta_8=(1+i)/\sqrt2$. Since $\sqrt2$ is not in the (presumably) larger field $L$, it’s not in the field generated by $\sin(\pi/51)$ and $\cos(\pi/74)$. And that’s it.