# Tangent half angle formula

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?

• When using trig functions make sure you do \tan instead of tan Apr 19, 2017 at 21:16
• Sorry didn't know that just fixed it just now Apr 19, 2017 at 21:18
• No problem all good Apr 19, 2017 at 21:20
• @Omicron. Realize that $tan\theta/2$ is negative for exactly the same value of $\theta$ as when the sine is negative. So that's how the plus/minus signs are taken care off. You do not need the absolute values. Consider $\theta$ in Quadrant 3 and 4 Apr 19, 2017 at 21:38

## 3 Answers

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

• I actually forget the plus or minus sign infront of the radical sign in the first half angle formula. What about that? Apr 19, 2017 at 21:22
• @haqnatural Is your statement also valid for $\theta=225$ degrees? After all, $tan112.5<0$ but the absolute value on your result makes it positive. Apr 19, 2017 at 21:28
• @imranfat,it seems you forgot the "nature " of the absolute value.$\left| x \right| =\begin{cases} x\quad ,\quad x>0 \\ 0\quad ,\quad x=0 \\ -x\quad ,\quad x<0 \end{cases}$ Apr 19, 2017 at 21:33
• Your derivation suggests that the tangent function is never negative. This is the point of imranfat's comment. Apr 19, 2017 at 21:36
• @haqnatural. No, LHS of your given identity is not always positive, Look at interval $(\pi,2\pi)$ for $\theta$. What makes you think that $tan\theta/2$ would be just positive? You may graph the $tan\theta/2$ and your abs value function together and see... Apr 19, 2017 at 21:41

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 $$\sin2t = 2\sin t\cos t.$$

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.