Tristian Velazquez

2022-06-22

How can I prove that :

$2\mathrm{arctan}(x)+\mathrm{arcsin}{\textstyle (}\frac{2x}{1+{x}^{2}}{\textstyle )}=\pi $ , x>1

What is the best way to do this ?

$2\mathrm{arctan}(x)+\mathrm{arcsin}{\textstyle (}\frac{2x}{1+{x}^{2}}{\textstyle )}=\pi $ , x>1

What is the best way to do this ?

upornompe

Beginner2022-06-23Added 20 answers

An easy way is to differentiate the function

$f(x)=2\mathrm{arctan}(x)+\mathrm{arcsin}\left({\displaystyle \frac{2x}{1+{x}^{2}}}\right)$

and show that f′(x)=0 so that f is a constant. Then calculate the constant by letting x be a well-chosen number.

$f(x)=2\mathrm{arctan}(x)+\mathrm{arcsin}\left({\displaystyle \frac{2x}{1+{x}^{2}}}\right)$

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

Beginner2022-06-24Added 8 answers

You can prove it without differentiating: set $t=2\mathrm{arctan}x$. This means

$\mathrm{tan}\frac{t}{2}=x\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}-\pi <t<\pi .$

Furthermore, as x>1, we have $\phantom{\rule{thickmathspace}{0ex}}{\displaystyle \frac{\pi}{2}}<t<\pi $

Now, by the half-angle formulae (actually this is where the formula we seek to prove comes from),

$\mathrm{sin}t=\frac{2x}{1+{x}^{2}},\phantom{\rule{1em}{0ex}}\text{whence}\phantom{\rule{1em}{0ex}}t\equiv \{\begin{array}{l}\mathrm{arcsin}\frac{2x}{1+{x}^{2}}\phantom{\rule{1em}{0ex}}\text{or}\\ \pi -\mathrm{arcsin}\frac{2x}{1+{x}^{2}}\end{array}\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2\pi .$

Knowing $\phantom{\rule{thickmathspace}{0ex}}{\displaystyle \frac{\pi}{2}}<t<\pi $, we necessarily have

$t=2\mathrm{arctan}x=\pi -\mathrm{arcsin}\frac{2x}{1+{x}^{2}}.$

$\mathrm{tan}\frac{t}{2}=x\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}-\pi <t<\pi .$

Furthermore, as x>1, we have $\phantom{\rule{thickmathspace}{0ex}}{\displaystyle \frac{\pi}{2}}<t<\pi $

Now, by the half-angle formulae (actually this is where the formula we seek to prove comes from),

$\mathrm{sin}t=\frac{2x}{1+{x}^{2}},\phantom{\rule{1em}{0ex}}\text{whence}\phantom{\rule{1em}{0ex}}t\equiv \{\begin{array}{l}\mathrm{arcsin}\frac{2x}{1+{x}^{2}}\phantom{\rule{1em}{0ex}}\text{or}\\ \pi -\mathrm{arcsin}\frac{2x}{1+{x}^{2}}\end{array}\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2\pi .$

Knowing $\phantom{\rule{thickmathspace}{0ex}}{\displaystyle \frac{\pi}{2}}<t<\pi $, we necessarily have

$t=2\mathrm{arctan}x=\pi -\mathrm{arcsin}\frac{2x}{1+{x}^{2}}.$