# is there a solution for equation $\arcsin((1-x)^{1/2})=\arccos(x^{1/2})$ in which $x$ is rational number

is there a solution for equation $\arcsin((1-x)^{1/2})=\arccos(x^{1/2})$ in which $x$ is rational number without using $\sin$ and $\cos$ functions in which?

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@Hashem: Please learn the elementary trig identities outside MO. – Wadim Zudilin Jul 3 '10 at 9:41
I can't see exactly what you are asking here. I suspect though that it would be a bit elementary for MO though. – Robin Chapman Jul 3 '10 at 9:42
@Wadim: see my answer. This is not elementary at all. – Jacques Carette Jul 3 '10 at 12:46
I don't understand. For every value of $x$ between $0$ and $1$, we have $\sin^{-1}((1-x)^{1/2}) = \cos^{-1}(x^{1/2})$. So take $x$ to be any rational number in this range. – David Speyer Jul 3 '10 at 12:59
@ David: a example about this equation is:$\sin^{-1}((1-{8/9})^{1/2})=cos^{-1}({8/9}^{1/2}) – Hashem sazegar Jul 3 '10 at 14:50 ## 1 Answer This may look elementary -- but it is most definitely not. This is because there are some nasty branch cuts involved, and making sure that the identity actually holds at all is not easy. The first thing to do is to look at what happens if one expands these functions into their simpler representation using logarithms: $$\arcsin \left( \sqrt {1-x} \right) =-i\ln \left( \sqrt {x}+i\sqrt {1-x} \right)$$ and $$\arccos \left( \sqrt {x} \right) =1/2\pi +i\ln \left( \sqrt {1-x}+i\sqrt {x} \right)$$ For essentially all real$x$outside$(0,1)$, these two quantities are complex. But, as it turns out, there are solutions. The simplest next step is to figure out 'where', and this is best done by splitting into cases. It can be written as $$\cases{i \left( -\ln \left( \sqrt {-x}+\sqrt {1-x} \right) -\ln \left( \sqrt {1-x}-\sqrt {-x} \right) \right) & x\leq 0 \cr -1/2\pi +\arctan \left( {\frac {\sqrt {1-x}}{\sqrt {x}}} \right) +\arctan \left( {\frac {\sqrt {x}}{\sqrt {1-x}}} \right) & x\in(0,1) \cr i \left( -\ln \left( \sqrt {x}-\sqrt {-1+x} \right) -\ln \left( \sqrt {-1+x}+\sqrt {x} \right) \right) & x \ge 1}$$ (the point being that all arguments of the square roots are now positive). With some extra work, it is then possible to in fact show that this is indeed$0$everywhere (with a minor quibble about$x=0$itself). From here the manipulations are indeed relatively straightforward. - Dear Jacques Carette: David Speyer wrote above: "I don't understand. For every value of$x$between$0$and$1$, we have$\sin^{-1}((1-x)^{1/2}) = \cos^{-1}(x^{1/2})$. So take$x\$ to be any rational number in this range." What's wrong with this argument? – Pierre-Yves Gaillard Jul 3 '10 at 18:05
Nothing wrong. But it was more interesting to look outside that range, to see globally (on the real line anyways) what was going on. More than what the OP asked for, sure, but this was more fun! – Jacques Carette Jul 3 '10 at 19:14
Jacques, I am very surprised to see the question open (= not closed). What I meant yesterday is said by David. If you wish to get fun from extending the OP, I would suggest you extend it in an interesting way. The fight with branches is pretty standard (although educationally important) but boring. – Wadim Zudilin Jul 4 '10 at 14:08
this question looks elementary,and i want to thanks of jacques carette,but i need the solution of this question without using log,so i need a reference of complating solution of this ,i prefer elementary proof – Hashem sazegar Jul 20 '10 at 20:49