I’ve got to proof that:

$$\tan\left(\frac{A}{2}\right) = \sqrt{\frac{1 -\cos(A)}{1 + \cos(A)}}$$

My attempt was (tried with right side):

$$= \pm \sqrt{\frac{1 -\cos(A)}{1 + \cos(A)} \cdot \frac{1 -\cos(A)}{1 - \cos(A)}}$$ $$= \pm \sqrt{\frac{\left(1 -\cos(A)\right)^2}{1 - \cos^2(A)}}$$ $$= \pm \frac{1 -\cos(A)}{\sin(A)}$$

They are not even similar.

I tried Wolfram Alpha online calculator and it showed one of the alternate answers as:


That’s the answer I suppose I should get to, but I’ve tried it many times and I can’t find (imagine) a possible route.

Please, if you could point me in the right direction, I’d greatly appreciate it.

Thank you very much in advance.

  • $\begingroup$ Do you already have the half-angle formulas for $\sin$ and $\cos$ at your disposal? Because then the problem is easy: just replace $\tan$ with $\sin/\cos$. $\endgroup$ – anon Sep 30 '11 at 20:53
  • $\begingroup$ It is not necessary that you show that the RHS is equal to the LHS, although you may do. You may transform your identity into an equivalent one and then prove it. $\endgroup$ – Américo Tavares Sep 30 '11 at 21:15
  • $\begingroup$ Excellent question! No, I did not have the half-angle formulas. Many answers already, but I didn’t mention anything of half-angle because I didn’t even know about them until now. $\endgroup$ – Andrés Botero Sep 30 '11 at 21:16
  • $\begingroup$ But do you know the addition/subtraction formulas? $\endgroup$ – Américo Tavares Sep 30 '11 at 21:17
  • $\begingroup$ Yes, but I didn’t think of that approach. How would have been? cos(A + 0)? (Trying it right now) $\endgroup$ – Andrés Botero Sep 30 '11 at 21:24

HINT You can go from $$ \frac{1-\cos A}{\sin A} $$ to $\tan (A/2)$ using the double-angle formulas: $$ \sin (2x) = 2 \sin x \cos x $$ and $$ \cos (2x) = \cos^2 x - \sin^2 x = 2 \cos^2 x - 1 = 1 - 2 \sin^2 x, $$ where, of course, $x = A/2$. (In the formula for $\cos (2x)$, all three formulas are really the same, but one of them will work "directly". I will leave it to you to figure which of the three formulas to use.)

Thanks to AD for pointing this out.

The expression $$ \sqrt{\frac{1-\cos A}{1+\cos A}} $$ is not equal to $$ \pm \sqrt{\frac{1-\cos A}{1+\cos A}} $$ that you have written down in the next line. There is no need for the $\pm$ sign here. Similarly, in the next line, you are manipulating the expression inside the radical, so again, the $\pm$ is unnecessary when you say $$ \pm \sqrt{\frac{1-\cos A}{1+\cos A} \cdot \frac{1-\cos A}{1-\cos A}}. $$ But in the next line, you are actually taking square roots, and you do not know the sign of $$ \frac{1-\cos A}{\sin A}. $$ In this case, it is best to say $$ \sqrt{\frac{1-\cos A}{1+\cos A} \cdot \frac{1-\cos A}{1-\cos A}} = \left| \frac{1-\cos A}{\sin A} \right|. $$ Although this is the correct way to write, it is, unfortunately, kind-of customary to be sloppy when people are doing trigonometry :). In particular, it is quite common to drop the absolute value signs here and just say $$ \sqrt{\frac{1-\cos A}{1+\cos A} \cdot \frac{1-\cos A}{1-\cos A}} = \frac{1-\cos A}{\sin A}. $$ As I showed in the hint, you can express the right hand side as $\tan(A/2)$.

If one is careful with her/his absolute value signs, one will end up with $$ \left| \tan \frac{A}{2} \right|. $$

  • $\begingroup$ +1 While you are at it you should also say something about $\pm\sqrt{\ldots}$ $\endgroup$ – AD. Sep 30 '11 at 20:56
  • $\begingroup$ Wait a minute, why 2x now? I mean, why the double angle now? How did you come to the conclusion that you should use those formulas in order to solve the problem? I mean, besides “experience”. $\endgroup$ – Andrés Botero Sep 30 '11 at 20:57
  • $\begingroup$ @Andrés I used $x$ simply as a shorthand. Here $x$ stands for $A/2$. Think of it this way: you learn a single formula $\sin (2x) = 2 \sin x \cos x$. In a particular question, you will need to figure out what your $x$ is. In this case, $x$ happens to be $A/2$. In some other question, it might be $\theta$ and so on. It's what is often called a "dummy variable". $\endgroup$ – Srivatsan Sep 30 '11 at 21:00
  • $\begingroup$ @Srivatsan I mean, why the double angle formulas? Simply because they could work? Sorry, perhaps I’m overcomplicating, but I’m trying to understand the background of that decision. $\endgroup$ – Andrés Botero Sep 30 '11 at 21:03
  • $\begingroup$ Yes, I solved it using the 1 - 2sin^2x variant, but I would really appreciate if you could please explain the above inquiry before checking this topic. Oh, by the way: thank you! $\endgroup$ – Andrés Botero Sep 30 '11 at 21:09

Hint: From the start, $$\sqrt{\frac{1-\cos (A)}{1+\cos(A)}},$$ try using the double angle identities for cosine. Specifically, use $$\cos(A)=2\cos^2\left(\frac{A}{2}\right)-1$$ for the denominator, and $$\cos(A)=1-2\sin^2\left(\frac{A}{2}\right)$$ for the numerator. You will see that things simplify very nicely.

Remark: Why should we expect to use double angle identities? The reason why is that I have $\tan \frac{A}{2}$ on the left hand side, and $\cos(A)$, $\sin(A)$ on the right hand side. This means that the only way to change the right hand side into the left hand side is to use the double angle identities to cut $A$ in half.


You can solve most trignometric identities using Euler's formula:

$$ e^{i \theta }=\cos (\theta )+i \sin (\theta ) $$

Solving for sin and cos, we see:

$$ \cos (\theta )=\frac{1}{2} \left(e^{-i \theta }+e^{i \theta }\right) $$

$$ \sin (\theta )=\frac{1}{2} i \left(e^{-i \theta }-e^{i \theta }\right) $$

Plugging in the formula above on the right side yields:

$$ \sqrt{\frac{1+\frac{1}{2} \left(-e^{-i \theta }-e^{i \theta }\right)}{1+\frac{1}{2} \left(e^{-i \theta }+e^{i \theta }\right)}} $$

Algebraic simplication yields:

$$ \sqrt{-\frac{\left(-1+e^{i \theta }\right)^2}{\left(1+e^{i \theta }\right)^2}} $$

Now, replacing the left side with sin/cos and substituting as above yields:

$$ \frac{i \left(e^{-\frac{i \theta }{2}}-e^{\frac{i \theta }{2}}\right)}{e^{-\frac{i \theta }{2}}+e^{\frac{i \theta }{2}}} $$

Simplifying yields:

$$ -\frac{i \left(-1+e^{i \theta }\right)}{1+e^{i \theta }} $$

which is identical to the right side after you take the square root.

  • $\begingroup$ In fact, you can prove all of the elementary trig identities by converting to exponentials. For more complicated identities we can save some painful algebra by simply reading off by inspection the degree $d$ of the polynomial identity in $e^{i\theta}$ the trig identity implies, and verifying the identity indeed holds for $d+1$ values and hence identically. $\endgroup$ – Ragib Zaman Oct 1 '11 at 4:01

$\tan(a/2)$ is not uniquely determined by $\cos(a)$. For the formula to hold, one needs to restrict the angles to an interval, or take absolute values of both sides.

A correct statement without any qualification is that the squares of the two sides of the equation are equal:


which follows from the half- or double-angle formulas for $\sin$ and $\cos$.


Knowing that all [direct] trigonometric functions of the simple angle (half-angle) can be expressed rationally as a function of the $\tan$ of the half-angle (double angle), we can think of transforming the given identity (edited: with the LHS replaced by its absolute value, as pointed by others)

$$\left|\tan \frac{A}{2}\right|=\sqrt{\frac{1-\cos A}{1+\cos A}}\tag{1}$$

into this equivalent one, by simple algebraic manipulation, i.e. solving for $\cos A$

$$\cos A=\frac{1-\tan ^{2}\frac{A}{2}}{1+\tan ^{2}\frac{A}{2}}.\tag{2}$$

And now prove it$^1$, e.g. as follows

$$\cos A=\cos \left(\frac{A}{2}+\frac{A}{2}\right)=\cos ^{2}\frac{A}{2}-\sin ^{2}\frac{A}{2}=\frac{\cos ^{2}\frac{A}{2} -\sin ^{2}\frac{A}{2}}{\cos ^{2}\frac{A}{2}+\sin ^{2}\frac{A}{2}}=\frac{ 1-\tan ^{2}\frac{A}{2}}{1+\tan ^{2}\frac{A}{2}}.$$ $$\tag{3}$$


$^1$ Assuming one knows the addition formula for $\cos$. From

$$\cos (\alpha +\beta )=\cos \alpha \cdot \cos \beta -\sin \alpha \cdot \sin \beta$$

for $\alpha =\beta =\frac{A}{2}$, we have

$$\cos A=\cos ^{2}\frac{A}{2}-\sin ^{2}\frac{A}{2}.$$

(Also in this answer of mine.)

  • $\begingroup$ Nice, keeping-it simple approach (plus the addition in the edit). I will choose Srivatsan Narayanan answer mostly because he took off from there I left. In any case, thank you very much! PS: I’m reading your other answer, and some linked in that question. $\endgroup$ – Andrés Botero Sep 30 '11 at 21:41
  • $\begingroup$ @AndrésBotero: Thanks! $\endgroup$ – Américo Tavares Sep 30 '11 at 22:03

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