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I was tasked with proving the identity $\tan(\frac x 2) = \dfrac {\sin(x)}{1+\cos(x)}$

I used the quotient identity for tangent and the half angle identities for sine and cosine to get $ \pm \dfrac {\sqrt{\dfrac {1-\cos(x)}{2}}}{\sqrt{\dfrac {1-\cos(x)}{2}}}$

which I reduced to $\pm \sqrt{\dfrac {1-\cos(x)}{1+\cos(x)}}$

I multiplied the fraction (within the square root) by $ \dfrac {1+ \cos(x)}{1+\cos(x)}$

Resulting in $\pm \sqrt{\dfrac {1-\cos^{2}(x)}{(1+\cos(x))^2}}$

Using the Pythagorean identity, I get $\pm\sqrt{\dfrac {\sin^{2}(x)}{(1+\cos(x))^2}}$

Taking the square root of the numerator and denominator I further reduced to $\pm \dfrac {\sin(x)}{1+\cos(x)}$

I thought I was done but when I checked my work in the answer book, it showed $ \left|\dfrac {\sin(x)}{1+\cos(x)}\right|$

Where do they get the absolute value from?

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    $1 - \cos x $ and $1 + \cos x $ are nonnegative, but $\sin x $ can be negative.2012-08-01
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    There shouldn't be an absolute value in there...2012-08-01
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    I'm sorry for the confusion here ... $sin(x)$ came from $\sqrt {sin^2(x)}$. I'll edit the question.2012-08-01
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    The source of the confusion has been clarified in Daniel's comments on lab's answer: http://math.stackexchange.com/a/177448/2012-08-01

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You are tasked with proving that $\tan(\frac x 2) = \dfrac {\sin(x)}{1+\cos(x)}$. If we suppose that this is not an impossible task (i.e. that the identity is correct) then neither $\tan(\frac x 2) =\pm\dfrac {\sin(x)}{1+\cos(x)}$ nor $\tan(\frac x 2) =\left|\dfrac {\sin(x)}{1+\cos(x)}\right|$ are satisfactory places to end up. The first version, which you arrived at, is not exactly incorrect, but rather incomplete, because it doesn't say which sign applies. The second version, from the answer book, is incorrect when $\tan(\frac x2)<0$.

In general, $a=\pm b$ gives you the same information as $|a|=|b|$. However, in this case we can determine which sign applies. Your work essentially shows that $\left|\tan(\frac x 2)\right| =\left|\dfrac {\sin(x)}{1+\cos(x)}\right|$. But note that $1+\cos(x)\geq 0$ so it does not affect the sign, and $\sin(x)$ always has the same sign as $\tan(x/2)$. To see that the last point is true, it is enough to work in the interval $(-\pi,\pi)$ by periodicity. Both $\tan(x/2)$ and $\sin(x)$ are positive when $x$ is in $(0,\pi)$, and both are negative when $x$ is in $(-\pi,0)$.

That said, I agree with lab bhattacharjee that avoiding methods that require later working out signs is a good idea.


Here is a side issue, which in the original version of my answer was all I posted:

$\sqrt{x^2}=|x|$ is an identiy for real numbers $x$. The reason is that for a nonegative real number $a$, $\sqrt{a}$ is defined to be the unique nonnegative squareroot of $a$. Since $|x|^2=x^2$, it follows that $|x|$ is the nonnegative real number whose square is $x^2$.

Therefore $\sqrt{\left(\dfrac{\sin x}{1-\cos x}\right)^2}=\left|\dfrac{\sin x}{1-\cos x}\right|$.

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    I see what I was missing now. $x^2$ is always positive, so the square root has to be positive. That makes it clear, thanks Jonas.2012-08-01
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    But $\tan x/2$ can be negative, hence the original $\pm$, but the $\pm$ got discarded without explanation. If we have absolute values, then the expression is *not* $\tan x/2$.2012-08-01
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    @Daniel: You're welcome. Could you please clarify the question? What exactly is the relationship to $\tan(x/2)$ that you are trying to work on? As J.M. and anon have pointed out, if you're starting with something that can be negative, you should end with something that can be negative. Neither your nor the book's answer are correct for $\tan(x/2)$.2012-08-01
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    "$x^2$ is always positive, so the square root has to be positive." That reason is incorrect. The fact that $x^2$ is always positive (or zero) only tells you that a *real* square root of $x^2$ always exists. It is the definition of the function $\sqrt{a}$ for $a\geq 0$ that tells us that $\sqrt{a}\geq 0$ for all $a\geq 0$. Squares of reals are nonnegative (a fact that can be proved), but nonnegative square roots are nonnegative *by definition* of nonnegative square root.2012-08-01
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    Edited ... hope that clarifies.2012-08-01
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    @Daniel: Thank you. Yes, that does clarify the intent of the problem (and I will edit my answer to match).2012-08-01
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    @Jonas Huhm. OK, I see where you determine that the signs are always the same, and thus prove the identity. I'm not sure I follow the answer for the original response now, though. If I take $\sqrt{x^2}$ can't the answer be ±x? Why did you restrict a to non-negative numbers?2012-08-01
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    @Daniel: In a sense it is true that $\sqrt{x^2}$ is $\pm x$, if that is shorthand for saying that $\sqrt{x^2}$ equals $x$ or $-x$. However, $\sqrt{x^2}=|x|$ says this and more. It points out specifically that we take whichever of $x$ or $-x$ is nonnegative. It is also true that $|x|$ equals $\pm x$ in this sense, but $\pm x$ is ambiguous, whereas $|x|$ is specified by $|x|=x$ if $x\geq 0$, and $|x|=-x$ if $x\leq 0$, which is the same value as $\sqrt{x^2}$.2012-08-01
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    How does this relate to the question? You were asked to prove that $a=b$, but ended up with $a=\pm b$, while the answer book says $a=|b|$. Since $a$ (which in this case is $\tan(\frac x2)$) can be negative, the answer book is wrong. Your work showing $a=\pm b$ isn't wrong, just not done yet. Once you know that $a=b$ or $a=-b$, you can rule out the latter by the method indicated above: show that $a$ and $b$ have the same sign.2012-08-01
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    I suppose it doesn't, really. I was just thrown by the absolute value ... I'd never seen that square root identity before, or at least had forgotten it. Perhaps I should post it as a separate question before I get yelled at for having a conversation here.2012-08-01
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    @DanielBall: Whatever suits you. (Sorry, I was asking that question somewhat rhetorically before attempting to answer it, probably unclearly.)2012-08-01
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Avoid unnecessary squaring wherever possible.

If we use the following approach, no such confusion arises.

$$\tan\frac{x}{2}=\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}$$

Now multiply numerator & denominator by $2\cos\frac{x}{2}$

$$\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}=\frac{2\cos\frac{x}{2}\sin\frac{x}{2}}{2\cos^2\frac{x}{2}}.$$

Now $\sin2A=2\sin A\cos A$ and $\cos2A=2\cos^2A-1$

So, $\tan\frac{x}{2}=\frac{\sin x}{1+\cos x}$

If the multiplier is $2\sin\frac{x}{2}$, $\tan\frac{x}{2}$ will be $\frac{1-\cos x}{\sin x}$ which is same as $\frac{\sin x}{1+\cos x}$


If we follow your approach also, just observe that $1+\cos x$ can not be negative as $-1≤\cos A≤1$ and $\frac{\tan\frac{x}{2}}{\sin x}=\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}2\cos\frac{x}{2}\sin\frac{x}{2}}=\frac{1}{2\cos^2\frac{x}{2}}$ which also $>0$.

So the sign of $\frac{\tan\frac{x}{2}}{\frac{\sin x}{1+\cos x}}$ is positive.

So the sign of $\tan\frac{x}{2}$ and $\frac{\sin x}{1+\cos x}$ are same.

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    I appreciate that, though it doesn't explain the absolute value in the answer book (they didn't use your approach)2012-08-01
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    @DanielBall It appears the absolute value in the answer book is simply wrong. As you note, it covers up a sign ambiguity inherent in $\pm$.2012-08-01
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    ... Oh my, I feel sheepish. The book isn't wrong, I didn't see the problem was continued and they explained the absolute value being removed because sine and tan have the same signs.2012-08-01
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    And on the next page they use the proof that lab provided. I ... wow I just wasted a lot of people's time.2012-08-01
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    @Daniel: Don't worry, many of us have been there. Thanks for sticking with it and clearing it up in the end!2012-08-01