= Solution
We first give an auxiliary-<polynomial> proof, rather than assuming a transcendence theorem for exponentials. Suppose $i\pi$ were algebraic, and let $\delta_1=i\pi,\delta_2,\ldots,\delta_t$ be its conjugates. Since the first factor vanishes,
$$
\prod_{\nu=1}^t(1+e^{\delta_\nu})=0.
$$
Expand and collect equal subset sums of the $\delta_\nu$. This gives $k+\sum_{j=1}^s b_je^{\beta_j}=0$, where $k\geq1$ counts the zero sums, the $\beta_j$ are distinct nonzero <algebraic numbers>, and each $b_j$ is a positive <integer>. The exponents and their weights are invariant under every Galois permutation.
Choose a positive <integer> $c$ such that all $c\beta_j$ are <algebraic integers>. Their Galois-invariant <monic polynomial> $R(u)=\prod_j(u-c\beta_j)$ has rational algebraic-<integer> coefficients, hence lies in $\mathbb Z[u]$. Its constant term $A=R(0)$ is a nonzero <integer>. For a <prime> $p$ define
$$
f_p(z)=\frac{(cz)^{p-1}R(cz)^p}{(p-1)!},\qquad
F_p(z)=\sum_{h=0}^{(s+1)p-1}f_p^{(h)}(z).
$$
At each $\beta_j$, derivatives below order $p$ vanish. Every higher derivative is $c^h h!/(p-1)!$ times an <algebraic integer>, so $F_p(\beta_j)$ is divisible by $p$ in the ring of <algebraic integers>. At zero the only derivative not visibly divisible by $p$ is the one of order $p-1$, and thus
$$
F_p(0)\in\mathbb Z,\qquad F_p(0)\equiv c^{p-1}A^p\pmod p.
$$
Galois invariance makes $\sum_jb_jF_p(\beta_j)$ rational; it is therefore an <integer>, divisible by $p$. Consequently
$$
S_p=kF_p(0)+\sum_jb_jF_p(\beta_j)\in\mathbb Z,\qquad
S_p\equiv kc^{p-1}A^p\pmod p.
$$
For every <prime> not dividing $kcA$, this is a nonzero <integer>.
The finite derivative sum satisfies $F_p'-F_p=-f_p$, so integration along the straight segment to $\beta_j$ gives
$$
e^{\beta_j}F_p(0)-F_p(\beta_j)
=e^{\beta_j}\int_0^{\beta_j}e^{-z}f_p(z)\,dz.
$$
Sum with weights $b_j$ and use $\sum b_je^{\beta_j}=-k$. The result is
$$
S_p=-\sum_jb_je^{\beta_j}\int_0^{\beta_j}e^{-z}f_p(z)\,dz.
$$
On these finitely many fixed compact segments, $|cz|$, $|R(cz)|$ and $|e^{-z}|$ are bounded independently of $p$. Hence $|S_p|\leq C_1C_2^p/(p-1)!\to0$. There are arbitrarily large <primes> not dividing the fixed <integer> $kcA$, contradicting $|S_p|\geq1$. Therefore \b[$\boxed{\pi\text{ is transcendental}}$]. This is the <Hermite auxiliary-polynomial proof for pi>.
The more general <Lindemann–Weierstrass theorem> states that exponentials of distinct <algebraic numbers> are linearly independent over $\overline{\mathbb Q}$. Equivalently, exponentials of rationally independent <algebraic numbers> are algebraically independent over $\overline{\mathbb Q}$. In particular, the exponential of a nonzero <algebraic number> is transcendental.
Suppose $C=\cos\alpha\cos\beta$ were algebraic. The exponential formula for <cosine> gives
$$
e^{i(\alpha+\beta)}+e^{i(\alpha-\beta)}+e^{i(-\alpha+\beta)}+e^{-i(\alpha+\beta)}-4C e^0=0.
$$
After combining equal exponents this is an algebraic-coefficient linear relation among exponentials of distinct <algebraic numbers>. At least one exponent is nonzero, since $\alpha+\beta$ and $\alpha-\beta$ cannot both vanish when $\alpha,\beta\ne0$. Its coefficient is a positive <integer>, even when two exponents coincide. The relation is therefore nontrivial and contradicts the theorem. Thus \b[$\boxed{\cos\alpha\cos\beta\text{ is transcendental}}$], including the cases $\alpha=\pm\beta$.
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