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The following is part of a question and the solution from my textbook:

Question

Given that $2 \sin{2\theta} = \cos{2\theta}$, show that $\tan{2\theta} = 0.5$

Solution

$2 \sin{2\theta} = \cos{2\theta}$
$\Rightarrow \frac{2 \sin{2\theta}}{\cos{2\theta}} = 1$
$\Rightarrow 2 \tan{2\theta} = 1~~~\tan{2\theta} = \frac{\sin{2\theta}}{\cos{2\theta}}$

I am unsure about this part: Since it is possible that $\cos(2\theta) = 0$, isn't it bad form to divide by $\cos(2\theta)$?

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1  
When dividing you have to assume that $\cos (2\theta) \neq 0$. You can check the case $\cos (2\theta) = 0$ for itself. –  please delete me May 25 '11 at 10:17
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You're absolutely right. It is indeed very bad form to divide by zero and it is a very good thing that you noticed that problem. However, as pointed out in the comments and answers, the damage is not too bad in the present case. –  t.b. May 25 '11 at 10:23
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This is why $\tan(2\theta)$ is undefined wherever $\cos(2\theta) = 0$ –  Dancrumb May 25 '11 at 14:26
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As a quick summary (there are many good answers and comments), $\sin()$ and $\cos()$ of the same value are never both zero and in your given equation $\cos()$ being zero would imply $\sin()$ is zero, so it's impossible for $\cos()$ to be zero and have that equation be true, so you can divide by $\cos()$ without having to worry about it being zero. –  Isaac May 25 '11 at 20:16
    
@Issac very clearly put thank you! –  Danny King Jun 3 '11 at 12:28

3 Answers 3

up vote 9 down vote accepted

If $\cos(2\theta) = 0$ then $\sin(2\theta)$ should be $1$ or $-1$. So $2\sin(2\theta)$ could not be equal to $\cos(2\theta)$.

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2  
Use $\backslash\cos$ and $\backslash\sin$. –  Did May 25 '11 at 12:24

You're absolutely right. It is very bad form to divide by zero and it is a very good thing that you noticed that problem.

As was already pointed out in the other answers and comments, in the present situation this isn't too bad, as it doesn't lead to a wrong conclusion, but this is rather an accident than good and careful reasoning. Were I to grade this solution, the textbook author wouldn't get full marks for that (assuming that what you display is the entire explanation).

It would have been much better to start with AlbertH's observation, so that's how I would have written the solution:

We would like to divide by $\cos{(2\theta)}$ in order to get the tangent on the left hand side. In order to do this, we should exclude the possibility that $\cos{(2\theta)} = 0$. Since $\sin^{2}{(2\theta)} + \cos^{2}{(2\theta)} = 1$, we conclude from $\cos{(2\theta)} = 0$ that $\sin{(2\theta)} = \pm 1$. Thus, we would have $\pm 2 = 2 \sin{(2\theta)} \neq \cos{(2\theta)} = 0$, so the given equation is not satisfied, and we can exclude $\cos{(2\theta)} = 0$. Now we may divide by $\cos{(2\theta)}$ and we get $1 = \frac{2 \sin{(2\theta)}}{\cos{(2\theta)}} = 2 \tan{2\theta}$ and thus $\tan{(2\theta)} = \frac{1}{2}$ as desired.

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If $\cos(2\theta) = 0$, then solving for $\theta$ we get one value of $\theta =\frac{\pi}{4}$, which says that the value of $$2 \sin{2\theta} = 2 \sin\frac{\pi}{2} = 2 \neq 0$$ which doesn't satisfy the given hypothesis $2 \sin{2\theta} = \cos{2 \theta}$.

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But there are many more values of $\theta$ you should consider... –  t.b. May 25 '11 at 10:56
    
@Theo: Yes, theo I know. This was just an illustration of why his claim wouldn't work. That's all. –  user9413 May 25 '11 at 10:57

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