So we start with the following tangent half angle formula: $$ \tan\left(\frac \theta2\right) = \pm\sqrt{\frac {1 - \cos \theta}{1 + \cos \theta}} $$
If I do some algebraic manipulation I end up with the following below: $$ \tan \left(\frac \theta2\right)= \pm\frac {1 - \cos \theta} {\sin \theta}$$
Now according to Michael Corral's Trigonometry the minus sign is not possible so I only end up with:
$$ \tan \left(\frac \theta2\right)= \frac {1 - \cos \theta} {\sin \theta} $$
Can you please explain why that is true?
$\endgroup$ 43 Answers
$\begingroup$$$\tan { \left( \frac { \theta }{ 2 } \right) } =\sqrt { \frac { 1-\cos { \theta } }{ 1+\cos { \theta } } } =\sqrt { \frac { { \left( 1-\cos { \theta } \right) }^{ 2 } }{ \left( 1+\cos { \theta } \right) \left( 1-\cos { \theta } \right) } } =\sqrt { \frac { { \left( 1-\cos { \theta } \right) }^{ 2 } }{ \sin ^{ 2 }{ \theta } } } =\frac { 1-\cos { \theta } }{ \sin { \theta } } $$
$\endgroup$ 8 $\begingroup$The fussiness over $\pm$ arises because $\tan \frac\theta2$ can be either positive or negative, whereas the square root is always considered positive. Your algebraic manipulation of $\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$ is OK except in the final step: the result should be $\left|\frac{1-\cos\theta}{\sin\theta}\right|$, as in @haqnatural's derivation, which leads to $$ \tan\frac\theta2 = \pm\left|\frac{1-\cos\theta}{\sin\theta}\right|.\tag1 $$ Equation (1) is true, but doesn't get us any closer to resolving the question of which sign to choose! One way to decide this is to consider different ranges of $\theta$, as in @imranfat's answer.
If this is not satisfying, you can try an alternative derivation that doesn't fuss with plus and minus signs: $$ \tan\frac\theta2=\frac{\sin\frac\theta2}{\cos\frac\theta2}=\frac{2\sin\frac\theta2\sin\frac\theta2}{2\sin\frac\theta2\cos\frac\theta2}=\frac{1-\cos\theta}{\sin\theta}, $$ using the double-angle identities $$ \cos2t=1-2\sin^2t $$ and $$\sin2t = 2\sin t\cos t.$$
$\endgroup$ $\begingroup$One way to tackle this problem is to look at the sign outcome of $tan\theta/2$. This happens if $\theta$ is on interval $(\pi,2\pi)$. In other words, when $\theta$ (not $\theta/2$!!) is in Quadrant 3 or 4, $tan\theta/2<0$. Now in $\frac{1-cos\theta}{sin\theta}$, the numerator is positive, so consider the denominator. The sine graph produces negative values when $\theta$ is in Q3 or Q4 and so the outcome is negative.Similarly for angles given in Quadrant 1 and 2, the sine outcome is positive. Now for the expression that contains a square root, the author puts the plus/minus in front but the reader need to determine the quadrant in question and then chose the correct sign. Otherwise, the correct identity is $tan\theta/2=\frac{1-cos\theta}{sin\theta}$ without the absolute values.
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