0
$\begingroup$

I want to determine the domain of the equation below. When I solve the equation algebraically, I get that all the terms cancel out and the equation is defined for all x values. However, I know that the inverse of the trigonometric functions are defined within a certain intervall. For example arcus sine is defined for $|x|\leq 1$ and arcus tangent is defined for all $x\in\mathbb{R}$. I also know that the function inside arctan is defined for $|x|\leq 1$. But how would I determine the domain of an equation with one inverse trig function at one side and another inverse trig function at the other side?
$$\arcsin(x) = 2\arctan\left(\frac{x}{1+\sqrt{1-x^2}}\right)$$

Thanks in advance.

$\endgroup$
18
  • $\begingroup$ I’m not sure what you mean by “domain of an equation”. If you mean “what are the $x$’s for which both sides are (independently) defined), then you’ve solved the problem on the left side ($|x|\leq 1$), and on the right you need to make sure you can do all the operations (which also require $|x|\leq 1$). But if you mean solutions (not domain), then that’s a different question altogether. Functions have domains. Equations don’t. $\endgroup$ Sep 22, 2018 at 21:43
  • $\begingroup$ There are some nice relations between the arcus functions and the trigonometric functions of different kinds. For example $\tan(\arcsin x)=\frac{x}{\sqrt{1-x^2}}$ which can be derived by using the identities $\tan x=\frac{\sin x}{\cos x}$ and $\cos x = \sqrt{1-\sin^2 x}$ which could be helpful for your equation. $\endgroup$
    – mrtaurho
    Sep 22, 2018 at 21:44
  • $\begingroup$ If you solve the equation then, like a mentioned in the post, all the terms will cancel out and the solutions will be $x\in\mathbb{R}$. The task is to determine all x, 'the domain', which satisfies the equation above. How would I combine the domain of the left hand side with the domain of the right hand side. $\endgroup$ Sep 22, 2018 at 21:51
  • $\begingroup$ If you correctly solve the equation, then you will only get the $x$s for which the original equation is sensible. It is possible that you are taking some steps that are not as reversible or as correct as you think they are, but that is difficult to know without knowing exactly what it is you did to solve the equation. But that is not “domain”; that is solution set of the equation. $\endgroup$ Sep 22, 2018 at 22:03
  • $\begingroup$ For example, if you took the equation $\sqrt{x}=-x$, squared both sides to get $x=x^2$, and then from here you got to $x^2-x=0$, hence $x(x-1) = 0$, hence $x=0$ or $x=1$, then the first step is not reversible (you can go from $\sqrt{x}=-x$ to $x=x^2$, but you can’t go from $x=x^2$ to $\sqrt{x}=-x$), which is why you get answers that are not really solutions to the original equation (namely, $x=1$ does not solve the original equation). You are likely doing something similar when solving, which means you are getting a superset of the solution set, rather than the solution set itself. $\endgroup$ Sep 22, 2018 at 22:06

1 Answer 1

1
$\begingroup$

Recall that

  • $\arcsin(x)$ is defined for $|x|\le 1$
  • $\arctan(y)$ is defined on $\mathbb{R}$ then we need $1-x^2 \ge 0 \implies|x|\le 1$

then by $x=\sin u$ and $-\frac{\pi}2 \le u \le \frac{\pi}2$, such that $|x|\le 1$, we have

$$\arcsin(\sin u) = 2\arctan\left(\frac{\sin u}{1+\sqrt{1-\sin^2 u}}\right)$$

$$ u = 2\arctan\left(\frac{\sin u}{1+\cos u}\right)$$

$$ \frac u2 = \arctan\left(\frac{\sin u}{1+\cos u}\right)$$

$$ \tan \left(\frac u2\right) = \frac{\sin u}{1+\cos u}$$

which always holds, therefore the given equality holds for any $x$ such that $|x|\le 1$.

$\endgroup$
7
  • $\begingroup$ How do you get to your claim about the left hand side? $\endgroup$ Sep 22, 2018 at 21:44
  • $\begingroup$ @ArturoMagidin I've used $-1\le \arcsin x\le 1$...something wrong? $\endgroup$
    – user
    Sep 22, 2018 at 21:46
  • $\begingroup$ That tells you something about when the equation can be true, rather than tell you something about what you need for “the existence of the left hand side”. The left hand side by itself is sensible whenever $|x|\leq 1$. I understand the question itself is unclear, but then talking about LHS and RHS separately does not help clarify things. $\endgroup$ Sep 22, 2018 at 21:49
  • $\begingroup$ @ArturoMagidin Thanks for your kind suggestion. I try to revise that! $\endgroup$
    – user
    Sep 22, 2018 at 21:51
  • $\begingroup$ @ArturoMagidin I've updated the discussion. $\endgroup$
    – user
    Sep 22, 2018 at 22:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.