4
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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, $$\frac{\sin x}{1+\cos x}=\frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^2\frac{x}{2}}=t.$$

$\endgroup$
3
  • $\begingroup$ 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 $\pm$ $\endgroup$ – Saran Wrap Oct 10 '19 at 21:37
  • $\begingroup$ @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. $\endgroup$ – J.G. Oct 10 '19 at 21:43
  • $\begingroup$ Alright thank you very much. $\endgroup$ – Saran Wrap Oct 10 '19 at 21:47
2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

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

$\endgroup$
1
  • $\begingroup$ Wow this is great thank you! $\endgroup$ – Saran Wrap Oct 15 '19 at 12:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.