Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Here is the complete problem but (c) is the part that I am having problems with, I have already solved (a) and (b):

(a) If $t=\tan\left(\frac{x}{2}\right)$,$-\pi<x<\pi$, sketch a right triangle or use trigonometric identities to show that $$\cos\left(\frac{x}{2}\right)=\frac{1}{\sqrt{1+t^2}}\qquad\sin\left(\frac{x}{2}\right)=\frac{t}{\sqrt{1+t^2}}$$

(b) Show that $$\cos x=\frac{1-t^2}{1+t^2}\qquad\sin x=\frac{2t}{1+t^2}$$

(c) Show that $$dx = \frac{2}{1+t^2}dt$$

I am aware that it is relatively simple to obtain the correct result by $x = 2\arctan t$ and if $y = \arctan x$ then $\frac{dy}{dx} = \frac{1}{1+x^2}$ so we obtain the result above. My problem is that I attempted to do it by $x = \arcsin \frac {2t}{1+t^2}$ and knowing that if $y = \arcsin x$ then $\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$ I obtained the following result $$ dx = -\frac{2}{1+t^2}dt$$ I have reviewed my solution several times and I cannot find an algebraic mistake. In the case that the result is algebraically correct, I am speculating that both results are equivalent because of something that has to do with the restrictions imposed when defining inverse trigonometric functions but I am lost and cannot figure out the connection.


I understand the mistake now, the restriction of $x \in (-\pi/2,\pi/2)$ needs to be made as dictated by the definition of $\arcsin$ and then $t=\tan(x/2)\in[\tan(-\pi/4),\tan(\pi/4)]=[-1,1]$. Now, my question is the following: when attempting to find $dx$ by $x = 2\arctan t$ we impose the restriction of $x \in (-\pi,\pi)$ because $\arctan t \in (-\pi/2,\pi/2)$ but, doesn't this contradict the restrictions we imposed on $x$ when finding $dx$ by $x = \arcsin \frac {2t}{1+t^2}$?

share|cite|improve this question
I noticed you didn't accept any of the answers for your questions so far. You're supposed to accept one answer to your question if you consider it to be satisfactory. – Git Gud Jan 14 '13 at 13:47
@LanceFerd One has to be carefull when taking root of a squared function: $$\sqrt{f^2(x)} = \big|f(x)\big|.$$ See my answer for details. – Pragabhava Jan 14 '13 at 14:19
If $x\in[-\pi/2,\pi/2]$, then $t=\tan(x/2)\in[\tan(-\pi/4),\tan(\pi/4)]=[-1,1]$. If $x$ is not in $[-\pi/2,\pi/2]$, then $|\tan(x/2)|>1$. – David Mitra Jan 14 '13 at 15:29
When $|t|>1$, you know that $x\in(\pi/2,\pi]\cup [-\pi ,-\pi/2)$. In this case $x=\pi-\arcsin{2t\over 1+t^2}$. – David Mitra Jan 14 '13 at 15:33
Setting $x=2\arctan t$ is fine: you have $-\pi< x <\pi\iff -\pi/2<x/2<\pi/2$. Then $2\arctan t$ will give you $x$ back, always. I'm not sure what you meant about the "above restrictions for $x$"; but, what I said in my earlier comments applied to $\arcsin{2t\over 1+t^2}$. – David Mitra Jan 15 '13 at 0:25
up vote 4 down vote accepted

If $$ t = \tan(x) $$ you can use impicit differentiation, i.e. $$ \frac{d}{dt}t = \frac{d}{dt}\tan\left(\frac{x(t)}{2}\right) $$ so $$ 1 = \frac{1}{2}\sec^2\left(\frac{x(t)}{2}\right) x'(t), $$ then $$ 2 \cos^2\left(\frac{x(t)}{2}\right) = x'(t) $$ and using (a) $$ 2 \cos^2\left(\frac{x(t)}{2}\right) =\frac{2}{1 + t^2} = x'(t) $$


The principal branch of $\arcsin \xi$ is defined only when $\xi \in (-1,1)$. Then $\arcsin \frac{2t}{1+t^2}$ is defined when $t \in (-1,1)$. Now, let $$ x = \arcsin \frac{2 t}{1 + t^2} $$ then \begin{align} \frac{d x}{d t} &= \frac{1}{\sqrt{1-\frac{4 t^2}{(1+t^2)^2}}} \frac{d}{d t}\left\{\frac{2 t}{1+t^2}\right\} = - \frac{2}{\sqrt{1-\frac{4 t^2}{(1+t^2)^2}}} \frac{t^2 -1}{(1 + t^2)^2}\\ &= -2 \frac{1+t^2}{\big|t^2 - 1\big|} \frac{t^2 -1}{(1 + t^2)^2} \end{align} and given that $t \in (-1,1)$, $$ \big|t^2 - 1\big| = 1 - t^2 $$ Finally $$ \frac{d x}{d t} = \frac{2}{1+t^2} $$

share|cite|improve this answer
@GitGud Edited. By the way, if you use \arcsin x, you produce $$\arcsin x.$$ – Pragabhava Jan 14 '13 at 14:05
I do not follow why $t \in (-1,1)$. – Lucas Alanis Jan 14 '13 at 15:20
@LanceFerd It's the domain of definition for $\arcsin t$, i.e. the range where $\sin x$ is invertible -in the principal branch, that is-. – Pragabhava Jan 14 '13 at 17:13
What would be restrictions on $x$ and $t$ if we were to do it by $x=2\arctan t$? – Lucas Alanis Jan 14 '13 at 20:18
Please, look at my last question to David Mitra above. – Lucas Alanis Jan 14 '13 at 21:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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