# How to find the following derivative?

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.

EDIT

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 $x = \arcsin \frac {2t}{1+t^2}$?

• 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

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 $$\frac{d x}{d t} = \frac{2}{1+t^2}$$
• @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