$\frac{{\sin \theta \cos \theta}}{1!}+\frac{{\sin 2\theta \cos ^2\theta}}{2!}+\frac{{\sin 3\theta \cos ^3\theta}}{3!}+...\infty$

Question

I have tried replacing $\ x$ with $\cos x$ in the maclaurin series of $\ e^x$, but stuck with the $\sin$ terms. Is there any other method to solve this question.

Answer 1

Try $x\rightarrow \cos \theta \cdot e^{i\theta}$ in the same Maclaurin series. Then the imaginary part is what you are looking for.

$$e^{\cos\theta(\cos\theta + i\sin\theta)}=1+\frac{\cos\theta(\cos\theta + i\sin\theta)}{1!}+\frac{\cos^2\theta(\cos2\theta + i\sin2\theta)}{2!}+\frac{\cos^3\theta(\cos3\theta + i\sin3\theta)}{3!}+...$$

Answer 2

$$e^{ix} = \sum_{n=0}^{\infty} \frac{i^{n} \, x^{n}}{n!} = \cos(x) + i \, \sin(x)$$ $$e^{x \, e^{i \theta}} = \sum_{n=0}^{\infty} \frac{x^{n}}{n!} \, e^{i n \theta}$$ $$e^{x \, e^{-i \theta}} = \sum_{n=0}^{\infty} \frac{x^{n}}{n!} \, e^{-i n \theta}$$

\begin{align} \sum_{n=0}^{\infty} \frac{x^{n} \, \sin(n \theta)}{n!} &= \frac{1}{2i} \, \left( e^{x \, e^{i \theta}} - e^{x \, e^{-i \theta}} \right) = \frac{1}{2i} \, e^{x \, \cos(\theta)} \, \left( e^{i \, x \sin(\theta)} - e^{-i \, x \sin(\theta)} \right)\\ &= e^{x \, \cos(\theta)} \, \sin(x \, \sin(\theta)) \end{align}

$$\sum_{n=0}^{\infty} \frac{\cos^{n}(\theta) \, \sin(n \theta)}{n!} = e^{ \cos^2(\theta)} \, \sin(\cos(\theta) \, \sin(\theta)) $$