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Why is this true?

$\frac{\sin(\theta)}{\sin(5\theta)} \cdot \frac{5\theta}{5\theta} = \frac{1}{5} \cdot \frac{\sin(\theta)}{\theta} \cdot \frac{5\theta}{\sin(5\theta)}$

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    I imagine you were told to simplify the limit of that as $\theta\to0$, right?2011-09-27

2 Answers 2

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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?

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    That's *exactly* what I was supposed to do. Thank you.2011-09-28
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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.

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    Oh, I see it now! Thank you very much.2011-09-28