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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:

$\sqrt{\tan^2\left(\frac{A}{2}\right)}$

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.

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    Yes, but I did$n$’t think of that approach. How would have been? cos(A + 0)? (Trying it right now)2011-09-30

5 Answers 5

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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|. $

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    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!2011-09-30
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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.

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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

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

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$^1$ 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.)

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    @AndrésBotero: Thanks!2011-09-30
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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.

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    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.2011-10-01
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$\tan(a/2)$ 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:

$\tan^2(A/2)=\frac{1-\cos(A)}{1+\cos(A)}$

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