# Is this proof of $\tan \frac{x}{2} = \frac{1-\cos x}{\sin x}$ incomplete?

So, for any angle $$\alpha$$ : $$\cos(2\alpha) = \cos^2\alpha - \sin^2\alpha = \dfrac{\cos^2\alpha-\sin^2\alpha}{\cos^2\alpha+\sin^2\alpha} = \dfrac{\dfrac{\cos^2\alpha-\sin^2\alpha}{\cos^2\alpha}}{\dfrac{\cos^2\alpha+\sin^2\alpha}{\cos^2\alpha}}= \dfrac{1-\tan^2\alpha}{1+\tan^2\alpha}$$ Now, $$\cos\alpha = \cos\Big(2\cdot\dfrac{\alpha}{2}\Big) = \dfrac{1-\tan^2\dfrac{\alpha}{2}}{1+\tan^2\dfrac{\alpha}{2}}$$

Now, using the componendo and dividendo rule, we get : $$\dfrac{\cos\alpha+1}{\cos\alpha-1} = \dfrac{2}{-2\tan^2\dfrac{\alpha}{2}} = \dfrac{-1}{\tan^2\dfrac{\alpha}{2}} \implies \tan^2\dfrac{\alpha}{2} = \dfrac{1-\cos\alpha}{1+\cos\alpha}$$ $$\implies \tan^2\dfrac{\alpha}{2} = \dfrac{(1-\cos\alpha)(1-\cos\alpha)}{(1+\cos\alpha)(1-\cos\alpha)} = \Big(\dfrac{1-\cos\alpha}{\sin\alpha}\Big)^2$$ $$\implies \Bigg|\tan\Big(\dfrac{\alpha}{2}\Big)\Bigg| = \Bigg|\dfrac{1-\cos\alpha}{\sin\alpha}\Bigg|$$ Now, only if $$\mathrm{sign}\Big(\tan\dfrac{\alpha}{2}\Big) = \mathrm{sign}\Big(\dfrac{1-\cos\alpha}{\sin\alpha}\Big)$$ is true, we can say that $$\tan\dfrac{\alpha}{2} = \dfrac{1-\cos\alpha}{\sin\alpha}$$

So, I think that without proving that, the proof will be incomplete but my Math textbook doesn't prove it.

So, is it necessary to prove it? If not, why not?

Thanks!

Your comment is correct. You can only get the final equality by proving that $$\tan\left(\frac{\alpha}{2}\right)$$ and $$\frac{1-\cos \alpha}{\sin\alpha}$$ have the same sign.

But this is not complicated to prove. $$\tan\left(\frac{\alpha}{2}\right)$$ is positive if and only if $$\frac{\alpha}{2} \in (k\pi, k\pi +\frac{\pi}{2})$$. Like $$\sin \alpha$$ while $$1- \cos \alpha$$ is always non negative.

Here's a simple way of proving the identity: $$\frac{1-\cos x}{\sin x}=\frac{1-(1-2\sin^2\frac{x}{2})}{2\sin\frac{x}{2}\cos\frac{x}{2}}=\frac{2\sin^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}}=\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}=\tan\frac{x}{2}$$ as required. I hope that was helpful:)

• $+1$. Nice way to derive it. I didn't accept your answer because that wasn't exactly what my question asked but a great answer, nonetheless :-) – Rajdeep Sindhu Aug 25 '20 at 15:43
• @RajdeepSindhu thanks for the compliment, it;s a pleasure to help! :) You might also be interested in the following: $$\frac{\sin2x}{1+\cos2x}=\frac{1-\cos2x}{\sin 2x}=\tan x$$ which can be proved in a very similar way. Don't worry about not accepting the answer,it's helping you that counts :) – A-Level Student Aug 25 '20 at 16:53
• It's helping you that counts. People like you are the reason I love the Stack Exchange community. :) – Rajdeep Sindhu Aug 26 '20 at 16:56
• @RajdeepSindhu thank you for the compliment, I appreciate it :)) – A-Level Student Aug 27 '20 at 18:33

Yes, it is necessary to show.

As $$\tan$$ has a periodicity $$\pi$$, it's enough to check the signs for $$\dfrac{\alpha}{2}$$ in the ranges, $$\left[0,\dfrac{\pi}{4}\right], \left[\dfrac{\pi}{4},\dfrac{\pi}{2}\right),\left(\dfrac{\pi}{2},\dfrac{3\pi}{4}\right]$$ and $$\left[\dfrac{3\pi}{4},\pi\right]$$