query about the cosine of an irrational multiple of an angle? de Moivre's identity
$$
(\cos \theta + i \sin \theta)^n = \cos n\theta + i \sin n\theta
$$
only applies as written when $n \in \mathbb{Z}$. if the exponent is a fraction $\frac{m}{n}$ then there will be $n$ values of $(\cos \theta + i \sin \theta)^{\frac{m}{n}}$, whilst for irrational $\alpha$ the set $\{e^{i\alpha (\theta +2\pi k)} \}$ is a dense subset of the unit circle.

nevertheless $e^{i \alpha \theta}$
is distinguished amongst the values of $(e^{i\theta})^\alpha$ even when $\alpha$ is not an integer. so can we proceed as follows?

when $0 \lt \theta \lt \frac{\pi}4$
$$
(\cos \theta + i \sin \theta)^\alpha = \cos^{\alpha}\theta(1+i \tan \theta)^{\alpha} = \cos \alpha \theta + i \sin \alpha \theta \tag{?}
$$
using the binomial theorem, which converges for the specified range of values for $\theta$, we have:

$$
\cos \alpha \theta = \cos^{\alpha}\theta \sum_{m=0}^\infty (-1)^m \binom{\alpha}{2m} \tan^{2m}\theta \tag{1}
$$
if (1) is true, then if $0 \lt \alpha, \beta \lt \frac{\pi}4$ we would have
$$
\cos^{\alpha}\beta \sum_{m=0}^\infty (-1)^m \binom{\alpha}{2m} \tan^{2m}\beta = \cos^{\beta}\alpha \sum_{m=0}^\infty (-1)^m \binom{\beta}{2m} \tan^{2m}\alpha
$$
Since your question is something along "how to prove step (?)" I will solve that part.  I can't do anything beyond that.

$$\cos^{\alpha}\theta(1+i\tan\theta)^{\alpha}=[\cos\theta(1+i\tan\theta)]^{\alpha}=(\cos\theta+i\sin\theta)^{\alpha}$$$$\cos*\tan=\cos*\frac{\sin}{\cos}=\sin$$

Simple exponent rule of "factoring" out the exponent and distributing the terms.
Since your question is something along "how to prove step (?)" I will solve that part.  I can't do anything beyond that.

$$\cos^{\alpha}\theta(1+i\tan\theta)^{\alpha}=[\cos\theta(1+i\tan\theta)]^{\alpha}=(\cos\theta+i\sin\theta)^{\alpha}$$$$\cos*\tan=\cos*\frac{\sin}{\cos}=\sin$$

Simple exponent rule of "factoring" out the exponent and distributing the terms.