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We first give an auxiliary-polynomial proof, rather than assuming a transcendence theorem for exponentials. Suppose $i\pi$ were algebraic, and let ${\delta }_1=i\pi ,{\delta }_2,\dots ,{\delta }_t$ be its conjugates. Since the first factor vanishes,Expand and collect equal subset sums of the ${\delta }_{\nu }$. This gives $k+{\sum }_{j=1}^s{b}_j{e}^{{\beta }_j}=0$, where $k\ge 1$ 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 $p$ defineAt 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 thusGalois invariance makes ${\sum }_j{b}_j{F}_p({\beta }_j)$ rational; it is therefore an integer, divisible by $p$. ConsequentlyFor every prime not dividing $kcA$, this is a nonzero integer.

The finite derivative sum satisfies $F_p^{\prime }-F_p=-f_p$, so integration along the straight segment to ${\beta }_j$ givesSum with weights $b_j$ and use $\sum b_j{e}^{{\beta }_j}=-k$. The result isOn these finitely many fixed compact segments, $|cz|$, $|R(cz)|$ and $|e^{-z}|$ are bounded independently of $p$. Hence $|S_p|\le C_1{C}_2^p/(p-1)!\to 0$. There are arbitrarily large primes not dividing the fixed integer $kcA$, contradicting $|S_p|\ge 1$. Therefore $\boxed{\pi \text{ is transcendental}}$. 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 givesAfter 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 $\alpha ,\beta \ne 0$. Its coefficient is a positive integer, even when two exponents coincide. The relation is therefore nontrivial and contradicts the theorem. Thus $\boxed{\cos \alpha \cos \beta \text{ is transcendental}}$, including the cases $\alpha =\pm \beta$.