Given that $0 < x < \pi/2$, it can be shown that $$\arcsin(\cos(x)) = \pi/2-x,$$ however, I was unable to make Mathematica to simplify the LHS into the RHS. I was guessing that,

Simplify[ArcSin[Cos[x]],Assumptions -> {0 < x < Pi/2}]

would have done the trick, but Mathematica seems to be reluctant on manipulating this expression at all, so I was starting to wonder if it knows about these kinds of identities in the first place?

Since the relation is proven from the general $$\arcsin(x)+\arccos(x) = \pi/2,$$ I was thinking at first that the issue might be evaluating $\arccos(\cos(x))$. However, this can be done using PowerExpand, which does not help in the case of $\arcsin(\cos(x))$ though?

  • $\begingroup$ Probably need FullSimplify with a custom ComplexityFunction. ArcSin[Cos[x]] has a LeafCount of only 3, so it's about as simple as you can get. $\endgroup$
    – John Doty
    Commented Jan 14, 2020 at 23:31
  • $\begingroup$ This is valid for 0<x<Pi. Maples gets it: simplify(arcsin(cos(x))) assuming 0<x,x<Pi gives Pi/2 - x !Mathematica graphics As mentioned in comment above, you probably need to use custom complexityfunction for this in Mathematica. $\endgroup$
    – Nasser
    Commented Jan 14, 2020 at 23:33
  • 5
    $\begingroup$ Strongly related, if not a duplicate. Alternative form of ArcSin[Sin[x]] $\endgroup$
    – Carl Woll
    Commented Jan 14, 2020 at 23:58

1 Answer 1


Use PowerExpand:

PowerExpand[ArcSin[Cos[x]], Assumptions -> 0 < x < π/2]

π/2 - x

  • 1
    $\begingroup$ This works. I am confused as I am certain I did try this beforehand... Anyway, thank you. $\endgroup$
    – Patrick.B
    Commented Jan 15, 2020 at 10:17
  • $\begingroup$ Didn't work with $arccos(-cos(x))$ $\endgroup$
    – dtn
    Commented Dec 28, 2023 at 11:30

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