<s>[INST] Is there a way to simplify $\prod_{i=1}^n\cos(a^i\theta)$, where $a<1$? I recently came upon the following expression in an attempt at getting a closed-form solution for a recursive relation:
$$\prod_{i=1}^n \cos(a^i\theta)$$
where $a<1$. Is there a way to make this product into a sum or otherwise make it simpler, or approximate it? In particular for the problem I was looking at, $a$ was $\frac{3}{4}$ and $\theta$ was $\frac{\pi}{4}$. Thanks!
 [/INST] Note this is @saulspatz answer with the computations added (I couldn't resist).
With 
$$
f(\theta)=\prod_{k=1}^n\cos(a^k\theta)
$$ 
Then noting 
$$
-\int_0^{a^k\theta}\tan x\, \mathrm{dx}=\log\cos(a^k\theta)
$$
we have on taking the logarithm
\begin{align*}
\log f(\theta) &= \sum_{k=1}^n\log\cos(a^k\theta)=\log\cos(a\theta)+\dotsb+\log\cos(a^n\theta)\\
&=-\sum_{k=1}^n\int_0^{a^k\theta}\tan x \,\mathrm{dx}\\
&=-\sum_{k=1}^n\int_0^{a^k\theta}x + \frac{1}{3}x^3+\frac{2}{15}x^5+\frac{17}{315}x^7+\dots\, \mathrm{dx}\quad\text{(by the Maclaurlin series for $\tan{x}$)}\\
&=-\left(\sum_{k=1}^n\Big{[}\frac{x^2}{2}\Big{]}_0^{a^k\theta}
+\sum_{k=1}^n\Big{[}\frac{x^4}{12}\Big{]}_0^{a^k\theta}
+\sum_{k=1}^n\Big{[}\frac{x^6}{45}\Big{]}_0^{a^k\theta}
+\sum_{k=1}^n\Big{[}\frac{17x^8}{2520}\Big{]}_0^{a^k\theta}
+\dots\right)\\
\end{align*}
Note each of these sums are all finite geometric sums: For the first
\begin{align*}
\sum_{k=1}^n\Big{[}\frac{x^2}{2}\Big{]}_0^{a^k\theta} &=\frac{\theta^2}{2}(a^2+a^4+\dotsb+a^{2n})\\
&=\frac{a^2\theta^2}{2}(1+a^2+\dotsb+a^{2(n-1)})\\
&= \frac{a^2\theta^2}{2}\cdot\frac{1-a^{2n}}{1-a^2}=\frac{a^2\theta^2}{2}\cdot\frac{(1-a^{n})(1+a^{n})}{(1-a)(1+a)}\\
\end{align*}
Hence
\begin{align*}
\log f(\theta) &=-\left(\sum_{k=1}^n\Big{[}\frac{x^2}{2}\Big{]}_0^{a^k\theta}
+\sum_{k=1}^n\Big{[}\frac{x^4}{12}\Big{]}_0^{a^k\theta}
+\sum_{k=1}^n\Big{[}\frac{x^6}{45}\Big{]}_0^{a^k\theta}
+\sum_{k=1}^n\Big{[}\frac{17x^8}{2520}\Big{]}_0^{a^k\theta}
+\dots\right)\\
&=\frac{a^2\theta^2}{2}\cdot\frac{a^{2n}-1}{1-a^2}+\frac{a^4\theta^4}{12}\cdot\frac{a^{4n}-1}{1-a^4}+\frac{a^6\theta^6}{45}\cdot\frac{a^{6n}-1}{1-a^6}+\frac{17a^8\theta^8}{2520}\cdot\frac{a^{8n}-1}{1-a^8}+\dotsb
\end{align*}
Now take the exponential of the sum to compute $f(\theta)$:
\begin{align*}
f(\theta) &=
\exp\left(\frac{a^2\theta^2}{2}\cdot\frac{a^{2n}-1}{1-a^2}\right)\cdot
\exp\left(\frac{a^4\theta^4}{12}\cdot\frac{a^{4n}-1}{1-a^4}\right)\cdot
\exp\left(\frac{a^6\theta^6}{45}\cdot\frac{a^{6n}-1}{1-a^6}\right)\\
&\qquad\qquad\qquad\qquad\cdot\exp\left(\frac{17a^8\theta^8}{2520}
\cdot\frac{a^{8n}-1}{1-a^8}\right)\dotsb\\
&=\prod_{k=1}^{\infty}\exp\left(\frac{\tan^{(2k-1)}(0)\,a^{2k}\theta^{2k}}{2k(2k-1)!}\cdot\frac{a^{2kn}-1}{1-a^{2k}}\right)
\end{align*}
where $\tan^{(2k-1)}(0)$ is the $(2k-1)$th derivative of $\tan x$ evaluated at $0$.
</s>