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. Jan 14, 2013 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. Jan 14, 2013 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 $|\tan(x/2)|>1$. Jan 14, 2013 at 15:29
• When $|t|>1$, you know that $x\in(\pi/2,\pi]\cup [-\pi ,-\pi/2)$. In this case $x=\pi-\arcsin{2t\over 1+t^2}$. Jan 14, 2013 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 $\arcsin{2t\over 1+t^2}$. Jan 15, 2013 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 $$\arcsin x.$$ Jan 14, 2013 at 14:05
• I do not follow why $t \in (-1,1)$. Jan 14, 2013 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-. Jan 14, 2013 at 17:13
• What would be restrictions on $x$ and $t$ if we were to do it by $x=2\arctan t$? Jan 14, 2013 at 20:18