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Are the functions $\arcsin x$ and $\arccos x$ equal up to a constant?

When I was solving the indefinite integral $\int\frac{\mathrm dx}{\sqrt{1-x^2}}$ I got two different results depending on the kind of the trigonometric substitution I make:

$\displaystyle \int\frac{\mathrm dx}{\sqrt{1-x^2}}=-\arcsin x$, if $x=\sin \theta$

$\displaystyle \int\frac{\mathrm dx}{\sqrt{1-x^2}}=\arccos x$, if $x=\cos\theta$

Where's my mistake?

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    $\begingroup$ You mean like $\arcsin x + \arccos x = \pi/2$ ? $\endgroup$ – A---B Sep 10 '17 at 16:47
  • $\begingroup$ @A---B I didn't understand, if $-\arcsin x=\arccos x$, then $\arcsin x+ \arccos x=0$, no? $\endgroup$ – user42912 Sep 10 '17 at 16:52
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    $\begingroup$ $-\arcsin x \ne \arccos x$ but $\arccos x =\pi/2 - \arcsin x$. So, $\arccos x + C=C_2 - \arcsin x$, where $C, C_2 \in \Bbb R$ and $C_2 = C + \pi/2$. $\endgroup$ – A---B Sep 10 '17 at 16:53
  • $\begingroup$ @user42912 No! What you can deduce for your computations is that $\int\frac{\mathrm dx}{\sqrt{1+x^2}}=\arcsin x+C_1$ and that $\int\frac{\mathrm dx}{\sqrt{1+x^2}}=-\arccos x+C_2$, where $C_1$ and $C_2$ are constants. Note, by the way, that there is an error in your signs. $\endgroup$ – José Carlos Santos Sep 10 '17 at 16:54
  • $\begingroup$ @A---B ha ok, I got it. Thank you! $\endgroup$ – user42912 Sep 10 '17 at 16:58
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No mistake. It turns out that$$\left(\forall x\in\left[-1,1\right]\right):\arcsin(x)+\arccos(x)=\frac\pi2.$$It is easy to justify this geometrically.

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  • $\begingroup$ I didn't understand, if $-\arcsin x=\arccos x$, then $\arcsin x+ \arccos x=0$, no? $\endgroup$ – user42912 Sep 10 '17 at 16:53
  • $\begingroup$ @user42912 Read my comment to your question. $\endgroup$ – José Carlos Santos Sep 10 '17 at 16:55
  • $\begingroup$ I got it! Thank you very much! Muito obrigado :) $\endgroup$ – user42912 Sep 10 '17 at 16:59
  • $\begingroup$ @user42912 De nada. ;) $\endgroup$ – José Carlos Santos Sep 10 '17 at 17:00
  • $\begingroup$ @José Carlos Santos: In the question there are no bounds given for the defining integrals. This identity also appears to hold outside the normal geometric bounds of [-1,1] for x, i.e. when both arcsin(x) and arccos(x) give values in the complex plane. $\endgroup$ – James Arathoon Sep 10 '17 at 20:05

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