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?

  • $\begingroup$ $1 - \cos x $ and $1 + \cos x $ are nonnegative, but $\sin x $ can be negative. $\endgroup$ – Will Jagy Aug 1 '12 at 4:14
  • 2
    $\begingroup$ There shouldn't be an absolute value in there... $\endgroup$ – J. M. is a poor mathematician Aug 1 '12 at 4:17
  • $\begingroup$ I'm sorry for the confusion here ... $sin(x)$ came from $\sqrt {sin^2(x)}$. I'll edit the question. $\endgroup$ – Daniel B. Aug 1 '12 at 4:23
  • $\begingroup$ The source of the confusion has been clarified in Daniel's comments on lab's answer: math.stackexchange.com/a/177448 $\endgroup$ – Jonas Meyer Aug 1 '12 at 5:25

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

  • $\begingroup$ 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. $\endgroup$ – Daniel B. Aug 1 '12 at 4:21
  • $\begingroup$ 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 $\tan x/2$. $\endgroup$ – anon Aug 1 '12 at 4:22
  • $\begingroup$ @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 $\tan(x/2)$. $\endgroup$ – Jonas Meyer Aug 1 '12 at 4:24
  • $\begingroup$ "$x^2$ 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. $\endgroup$ – Jonas Meyer Aug 1 '12 at 4:33
  • $\begingroup$ Edited ... hope that clarifies. $\endgroup$ – Daniel B. Aug 1 '12 at 4:37

Avoid unnecessary squaring wherever possible.

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


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


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.

  • $\begingroup$ I appreciate that, though it doesn't explain the absolute value in the answer book (they didn't use your approach) $\endgroup$ – Daniel B. Aug 1 '12 at 4:52
  • 1
    $\begingroup$ @DanielBall It appears the absolute value in the answer book is simply wrong. As you note, it covers up a sign ambiguity inherent in $\pm$. $\endgroup$ – anon Aug 1 '12 at 5:06
  • 1
    $\begingroup$ ... 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. $\endgroup$ – Daniel B. Aug 1 '12 at 5:13
  • 1
    $\begingroup$ And on the next page they use the proof that lab provided. I ... wow I just wasted a lot of people's time. $\endgroup$ – Daniel B. Aug 1 '12 at 5:13
  • $\begingroup$ @Daniel: Don't worry, many of us have been there. Thanks for sticking with it and clearing it up in the end! $\endgroup$ – Jonas Meyer Aug 1 '12 at 5:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.