# Does Mathematica know about identity involving ArcSin(Cos(x))? [duplicate]

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?

• 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. Commented Jan 14, 2020 at 23:31
• 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. Commented Jan 14, 2020 at 23:33
• Strongly related, if not a duplicate. Alternative form of ArcSin[Sin[x]] Commented Jan 14, 2020 at 23:58

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

• Didn't work with $arccos(-cos(x))$