Closed form of $\sum_{n=1}^{\infty}a^n \cos(nt)$

Question

I'm trying to find a closed form (if one exists) for

$$\sum_{n=1}^{\infty}a^n \cos(nt)$$

where $a \in (0, 1)$. I know that

$$\begin{matrix} \sum_{n=1}^{\infty} \frac{a^n}{n} \cos(nt) &= &\ln(a^2-2a\cos(t)+1) \\ \sum_{n=1}^{\infty} \frac{1}{n} \cos(nt) &= &2 \ln|2 \sin(\frac{t}{2})| \end{matrix}$$

It feels like these two facts combined should make it relatively easy to come up with the closed form for $\sum_{n=1}^{\infty}a^n \cos(nt)$ but I'm having difficulty seeing it.

Answer 1

$$ \begin{align} \sum_{n=1}^\infty a^n\cos(nt) &=\frac12\sum_{n=1}^\infty\left(a^ne^{int}+a^ne^{-int}\right)\tag1\\ &=\frac12\left(\frac{ae^{it}}{1-ae^{it}}+\frac{ae^{-it}}{1-ae^{-it}}\right)\tag2\\ &=\frac{a\cos(t)-a^2}{1-2a\cos(t)+a^2}\tag3 \end{align} $$ Explanation:
$(1)$: $\cos(x)=\frac{e^{ix}+e^{-ix}}2$
$(2)$: sum of a geometric series with ratios $ae^{it}$ and $ae^{-it}$
$(3)$: $\cos(x)=\frac{e^{ix}+e^{-ix}}2$

Answer 2

Using the fact that $\operatorname{Re} e^{int} = \cos nt$, your series is just geometric. Indeed, $$\sum_{n=1}^\infty a^n \cos nt = \operatorname{Re}\sum_{n=1}^\infty \left(ae^{it}\right)^n.$$ You should then get $$\operatorname{Re}\left(\frac{ae^{it}}{1-ae^{it}}\right) = \frac{a(a+\cos t)}{1+a^2-2a\cos t}.$$