Expressing $x = \tan{\frac{u}{2}}$ into a form using $\arctan$ -- Spivak

Question

The question that is being asked is: If $x = \tan{\frac{u}{2}}$, express $\sin(u)$ and $\cos(u)$ in terms of $x$. But my question isn't about how to do this, but how they treated the $\arctan(x)$ operation. In particular how did the solution manage to express $u = 2\arctan(x)$ ? The solution was presented as such:

Because either way you want to interpret

$$x = \tan{\frac{u}{2}} \text{or}\ \frac{\tan{u}}{2}$$

you still have to take the operation over the whole thing, unless there is something I'm missing about the treatment of $\arctan(x)$ that is idiosyncratic. Also from all I've read it is supposed to be $\tan{\frac{u}{2}}$, which makes things more confusing.

Answer 1

It is $x=\tan\left(\frac u2\right)$. So$$x=\tan\left(\frac u2\right)\iff\arctan(x)=\frac u2\iff2\arctan(x)=u.$$

Answer 2

Since we have that

$$u=2\arctan x$$

then your interpretation for $x=\tan{\left(\frac{u}{2}\right)}$ is correct indeed

$$x=\tan{\left(\frac{u}{2}\right)} \implies \arctan x=\arctan\left(\tan{\left(\frac{u}{2}\right)}\right)=\frac u 2 \implies u=2\arctan x$$

Answer 3

Let's consider only principal values for inverse trigonometric ratios that follow. $\tag{1}$
Note that if $y=\tan \theta, -\pi/2\lt\theta\lt\pi/2$, then we have $\theta=\arctan y$

It is $\tan (u/2)$ and not $(\tan u)/2$ in your post.

So $x=\tan (u/2)\implies (u/2)=\arctan x\implies u=2\arctan x=2\phi$, where $\phi=\arctan x $ and of course as per our convention in $(1): -\pi/2\lt\phi\lt\pi/2$ and hence $\tan \phi=x$.

Now, $\sin u=\sin (2\phi)=\frac{2\sin \phi \cos \phi}{1}=\frac{2\sin \phi \cos \phi}{\sin^2\phi+\cos^2\phi}=\frac{2\cos^2\phi\tan\phi }{\cos^2\phi(\tan^2\phi+1)}=\frac{2\tan\phi }{\tan^2\phi+1}=\frac{ 2x}{x^2+1}$ etc.

Answer 4

If $x=\tan(u/2)$ and $-\pi<u<\pi$, then $u=2\arctan x$.

For instance, if $u=3\pi/2$, then $\tan(u/2)=-1$ and $2\arctan(-1)=-\pi/2\ne3\pi/2$.

By definition, $\arctan x$ returns the unique number $v\in(-\pi/2,\pi/2)$ such that $\tan v=x$.

So, if $x=\tan(u/2)$, then you can say that, when $-\pi/2<u/2<\pi/2$, $$ \frac{u}{2}=\arctan x $$ and therefore $u=2\arctan x$.

The limitation $-\pi<u<\pi$ cannot be raised. Given $x=\tan v$ you can at most say that there exists an integer $k$ such that $$ v=k\pi+\arctan x $$