9. Proof:
$$\begin{aligned} \text { Left side } & =\sum_{k=1}^{n}[\cos (4 k-3) \alpha+\sin (4 k-1) \alpha] \\ & =\sum_{k=0}^{n-1} \cos (4 k+1) \alpha+\sum_{k=0}^{n-1} \sin (4 k+3) \alpha . \end{aligned}$$

In equation (37), let $\theta=\alpha, \varphi=4 \alpha$, we get
$$\sum_{k=0}^{n-1} \cos (4 k+1) \alpha=\frac{\sin 2 n \alpha}{\sin 2 \alpha} \cos (2 n-1) \alpha$$

In equation (38), let $\theta=3 \alpha, \varphi=4 \alpha$, we get
$$\sum_{k=0}^{n-1} \sin (4 k+3) \alpha=\frac{\sin 2 n \alpha}{\sin 2 \alpha} \sin (2 n+1) \alpha$$

The condition $\left\{\frac{\varphi}{2 \pi}\right\} \neq 0$ is equivalent to $\alpha \neq \frac{k}{2} \pi$, so
$$\begin{aligned} \cos \alpha & +\sin 3 \alpha+\cos 5 \alpha+\sin 7 \alpha+\cdots+\sin (4 n-1) \alpha \\ = & \frac{\sin 2 n \alpha}{\sin 2 \alpha}[\cos (2 n-1) \alpha+\sin (2 n+1) \alpha] \\ = & \frac{\sin 2 n \alpha}{\sin 2 \alpha}[\cos 2 n \alpha \cos \alpha+\sin 2 n \alpha \sin \alpha \\ & \quad+\sin 2 n \alpha \cos \alpha+\cos 2 n \alpha \sin \alpha] \\ = & \frac{\sin 2 n \alpha}{\sin 2 \alpha}(\cos 2 n \alpha+\sin 2 n \alpha)(\cos \alpha+\sin \alpha) \end{aligned}$$