<s>[INST] Why does the equation with $2 \arctan(x)$ and other Inverse Trigonometric functions have weird conditions? I have these three equations:
$$2\arctan(x) = \arcsin(\frac{2x}{1+x^2}),\left |x\right|\leq1$$
$$2\arctan(x) = \arccos(\frac{1-x^2}{1+x^2}),x\geq0$$
$$2\arctan(x) = \arctan(\frac{2x}{1-x^2}),\left |x\right|<1 $$
I can verify these by simple substitution taking $x = \tan\theta$ for some $\theta$ but I could not understand why these conditions for $x$ are given.
I can see their graphs makes sense but is there an algebraic way of proving/verifying these conditions?
Also the graph is:

 [/INST] An exercise I found helpful.
Consider the intersection of the line $y = tx$ and the circle $x^2 + y^2 = 2x$
The circle has radius 1 and is centered at $(1,0)$
$x^2 + (tx)^2 = 2x\\
x = \frac {2}{1+t^2}\\
y = \frac {2t}{1+t^2}$
Now lets translate everything one unit to the left.  Yes, we could have calculated the intersection of $x^2+ y^2 = 1$ and the line $y = t(x+1)$ but the algebra is simpler this way.
$x = \frac {2}{1+t^2}-1\\
x = \frac {1-t^2}{1+t^2}\\
y = \frac {2t}{1+t^2}$
Do these equations look familiar?  What are $x,y,t$ geometrically / trigonometrically?
The line and the circle and the x axis form an angle on the circle, which has half the measure of the central angle to that same arc.
$t = \tan \frac 12 \theta$
And of course,
$x = \cos \theta\\
y = \sin \theta$
Regarding the domain restrictions. If $t> 1$  the line $y = t(x+1)$ will be intersecting the circle in QII.
$\arcsin (\frac {2t}{1+t^2})$ will return a value corresponding to a point in QI
Similarly if $t < 0$
$\arccos (\frac {1-t^2}{1+t^2})$ will return a value for a point in QI or QII while our reference point is actually in QIII or QIV
I found this exercise to more helpful to build intuition than mechanically chug through the trig identities.
Here would be the mechanical approach:
$\sin (\arctan \frac {a}{b}) = \frac {a}{\sqrt {a^2 + b^2}}\\
\cos (\arctan \frac {a}{b}) = \frac {b}{\sqrt {a^2 + b^2}}\\
\theta = 2\arctan t\\
\sin \theta = 2\sin(\arctan t)\cos(\arctan t) = 2\frac {1}{\sqrt {1+t^2}}\frac {t}{\sqrt {1+t^2}} = \frac {2t}{1+t^2}\\
\cos \theta = \cos^2(\arctan t)-\sin^2(\arctan t) = \frac {t^2}{1+t^2} - \frac {1}{1+t^2}= \frac {1-t^2}{1+t^2}$
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