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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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What exactly do you not understand in here ? –  Sasha Sep 27 '11 at 19:46
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It is true since $x \cdot 1 = x$ for all values of $x$. $\frac{5 \theta}{5 \theta}$ is just a fancy way to write $1$. The only thing you need to be careful about is that $5 \theta \neq 0$. But that only occurs when $\theta = 0$, which is precluded from the domain of the original expression anyways. –  JavaMan Sep 27 '11 at 19:47
    
If you simplify both expressions you get $\frac{\sin \theta}{\sin (5 \theta)}$. That means that they are equal. –  Beni Bogosel Sep 27 '11 at 19:50
    
I was told simplify $\frac{\sin(\theta)}{\sin(5\theta)}$ by multiplying it by $\frac{5\theta}{5\theta}$. What I didn't get it how it yielded that solution. –  Mike Gates Sep 27 '11 at 20:00
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I imagine you were told to simplify the limit of that as $\theta\to0$, right? –  anon Sep 27 '11 at 20:14
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2 Answers

up vote 4 down vote accepted

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. –  Mike Gates Sep 28 '11 at 0:25
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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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I still don't see where that $\frac{1}{5}$ comes from. –  Mike Gates Sep 27 '11 at 19:55
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@Mike: $\frac{1}{5\theta}$ gets split into $\frac{1}{5}\frac{1}{\theta}$ and the two parts are moved around separately. –  anon Sep 27 '11 at 20:14
    
Oh, I see it now! Thank you very much. –  Mike Gates Sep 28 '11 at 0:25
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