Tristian Velazquez

2022-06-22

How can I prove that :
$2\mathrm{arctan}\left(x\right)+\mathrm{arcsin}\left(\frac{2x}{1+{x}^{2}}\right)=\pi$ , x>1
What is the best way to do this ?

upornompe

An easy way is to differentiate the function
$f\left(x\right)=2\mathrm{arctan}\left(x\right)+\mathrm{arcsin}\left(\frac{2x}{1+{x}^{2}}\right)$
and show that f′(x)=0 so that f is a constant. Then calculate the constant by letting x be a well-chosen number.

deceptie3j

You can prove it without differentiating: set $t=2\mathrm{arctan}x$. This means
$\mathrm{tan}\frac{t}{2}=x\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}-\pi
Furthermore, as x>1, we have $\phantom{\rule{thickmathspace}{0ex}}\frac{\pi }{2}
Now, by the half-angle formulae (actually this is where the formula we seek to prove comes from),
$\mathrm{sin}t=\frac{2x}{1+{x}^{2}},\phantom{\rule{1em}{0ex}}\text{whence}\phantom{\rule{1em}{0ex}}t\equiv \left\{\begin{array}{l}\mathrm{arcsin}\frac{2x}{1+{x}^{2}}\phantom{\rule{1em}{0ex}}\text{or}\\ \pi -\mathrm{arcsin}\frac{2x}{1+{x}^{2}}\end{array}\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2\pi .$
Knowing $\phantom{\rule{thickmathspace}{0ex}}\frac{\pi }{2}, we necessarily have
$t=2\mathrm{arctan}x=\pi -\mathrm{arcsin}\frac{2x}{1+{x}^{2}}.$

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