# Prove $\tan(\frac{x}{2}) = \frac{\sin x}{1 + \cos x}$ using the quadratic formula

I am trying to prove the fact that $$\tan \frac{x}{2} = \frac{1-\cos x}{\sin x}$$ or alternatively $$\tan \frac{x}{2} = \frac{1- \cos x}{\sin x}$$. (I understand that it can be proved using the half-angle identities of $$\sin$$ and $$\cos$$ but I want to understand how to get to the solution from this specific method of derivation.)

\begin{align*} \tan(2x) &= \frac{2\tan(x)}{1-\tan(x)^2} \\ \tan(x) &= \frac{2\tan(\frac{x}{2})}{1-\tan(\frac{x}{2})^2} \\ \end{align*} I now let $$A=\tan x$$ and $$B=\tan \frac{x}{2}$$ \begin{align*} A\cdot(1-B^2) &= 2B\\ AB^2+2B-A &= 0 \\ \end{align*} Now I solve for B using the quadratic formula. \begin{align*} B &= \frac{-2\pm \sqrt{4+4A^2}}{2A} \\ B &= \frac{-1\pm \sqrt{1+A^2}}{A} \\ \tan(\frac{x}{2}) &= \frac{-1\pm \sqrt{1+\tan(x)^2}}{\tan(x)}\\ \tan(\frac{x}{2}) &= \frac{-1\pm \sqrt{(\sec x)^2}}{\tan(x)}\\ \tan(\frac{x}{2}) &= \frac{-1\pm |\sec x|}{\tan(x)} \end{align*} I am confused as to how to continue at this point (firstly, not sure how to deal with the absolute value, and secondly not sure how to deal with the plus-minus). Any help is greatly appreciated, as I feel like I do not fully understand how to manipulate absolute values and the meaning of the plus-minus.

The quadratic formula alone won't help you obviate the $$\pm$$ sign. It's better to note $$\sin x=\frac{2t}{1+t^2}$$ ($$t$$ is more common notation than $$B$$) while $$\cos x=\frac{1-t^2}{1+t^2}$$, so $$\frac{\sin x}{1+\cos x}=t$$. Alternatively, $$\frac{\sin x}{1+\cos x}=\frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^2\frac{x}{2}}=t.$$

• So is it not possible to simplify my last line to something else to solve it? I am more interested in how I could simplify expressions with absolute values and $\pm$ – Saran Wrap Oct 10 '19 at 21:37
• @SaranWrap Multiplying top and bottom by $\cos x$ gives $\frac{-\cos x\pm\operatorname{sgn}\cos x}{\sin x}$, but you can't verify the $\pm\operatorname{sgn}\cos x$ is $1$ with quadratics alone. – J.G. Oct 10 '19 at 21:43
• Alright thank you very much. – Saran Wrap Oct 10 '19 at 21:47

Hint. Note that $$\pm|x|=\pm x,$$ without loss of any generality. Then split into two cases.

Because of the $$\pm$$, the absolute value is superfluous.

\begin{align} B &= \frac{-1\pm |\sec x|}{\tan x} \\ B &= \frac{-1\pm \sec x}{\tan x} \\ B &= \frac{(-1\pm \sec x)(\cos x)}{(\tan x)(\cos x)} \\ B &= \frac{-\cos x\pm 1}{\sin x} \\ \end{align}

We know that $$\dfrac{1 - \cos x}{\sin x} = \tan \dfrac x2$$

Also

\begin{align} \dfrac{-1 - \cos x}{\sin x} &= -\dfrac{1 + \cos x}{\sin x}\\ &= -\dfrac{1 + (2 \cos^2 \frac x2 - 1))} {2 \sin \frac x2 \cdot \cos \frac x2} \\ &= - \cot \frac x2 \end{align}

So the roots of the quadratic $$AB^2 + 2B -A = 0$$ are $$B = \tan \frac x2$$ and $$B =-\cot \frac x2$$.

That is to say

$$\tan x \cdot \left(\tan \frac x2\right)^2 + 2\tan \frac x2 - \tan x = 0$$

and

$$\tan x \cdot \left(-\cot \frac x2\right)^2 - 2\cot \frac x2 - \tan x = 0$$

• Wow this is great thank you! – Saran Wrap Oct 15 '19 at 12:47