Why is this true?
$\frac{\sin(\theta)}{\sin(5\theta)} \cdot \frac{5\theta}{5\theta} = \frac{1}{5} \cdot \frac{\sin(\theta)}{\theta} \cdot \frac{5\theta}{\sin(5\theta)}$
Why is this true?
$\frac{\sin(\theta)}{\sin(5\theta)} \cdot \frac{5\theta}{5\theta} = \frac{1}{5} \cdot \frac{\sin(\theta)}{\theta} \cdot \frac{5\theta}{\sin(5\theta)}$
You can switch the denominators and then pull out a $5$: $\frac{\sin\theta}{\sin 5\theta}\begin{matrix}| \\ \leftrightarrow\end{matrix}\frac{5\theta}{5\theta}=\frac{\sin\theta}{5\theta}\frac{5\theta}{\sin5\theta}=\frac{1}{5}\begin{matrix}|\\ \leftarrow\end{matrix}\frac{\sin\theta}{\theta}\frac{5\theta}{\sin5\theta}$ I imagine your task was to evaluate $\lim_{x\to0}\frac{\sin(x)}{\sin(5x)}$ given knowledge of $\lim_{x\to0}\frac{\sin(x)}{x}=1$; assuming this is true, do you see how to reason from here?
The terms have just been rearranged:
$ \begin{align*} \frac{\sin(\theta)}{\sin(5\theta)}\cdot \frac{5\theta}{5\theta} &= \sin(\theta)\cdot\frac{1}{\sin(5\theta)}\cdot \frac{1}{5}\cdot 5\theta\cdot \frac{1}{\theta}\\ &=\frac{1}{5}\cdot \sin(\theta)\cdot\frac{1}{\theta}\cdot 5\theta\cdot\frac{1}{\sin(5\theta)}\\ \end{align*} $
The last expression is what you have on the right side of your equation.