 # Math Problem

Ok OG geniuses, here is a problem I can;t figure out:

solve for x:

sin(sin(sin(sin(x))))=cos(cos(cos(cos(x))))

Now i'll tell you from the get-go that their is no REAL solution, If someone really wants I'll do the proof. I suspect that there is no imaginary solution either, but can;t dissprove it.

Anywhoo, I gave up on it but would still like to see a solution.

Interesting problem! I just did a few numerical investigations:

I graphed the two sides of the equation and it indeed appears they don't cross anywhere on the real line.

I plugged the equation into Mathematica, as well. It won't solve it algebraically; however, it will solve it numerically, and given an initial guess of i, it gives a solution:

FindRoot[Sin[Sin[Sin[Sin[x]]]] == Cos[Cos[Cos[Cos[x]]]], {x, [ImaginaryI]}]

0.756887 + 0.610155*i

So there do seem to be imaginary roots. But I have no idea how to solve this analytically, truly no clue! Are you sure that it is even possible?

^^^^ hmmmmmmmmm I don't know if its solvable analyticly.

But that gives me some hope. I'm gonna work a little on it tomorrow, I'll let ya know of any insights.

Did you ever figure out how to answer this problem. I'd be very interested to know how you did it.

A little late but...

Sin[u] = Cos[u] => u = Pi/4 where u = Sin[Sin[Sin[x]]]]

So solve for Sin[u] = Pi/4

(Exp[iu] - Exp[-iu])/2i = Pi/4

Exp[2iu] - (i pi/2) Exp[iu] - 1 = 0
which has the solution

u = ArcTan[Pi/Sqrt[16 - Pi^2]]

So Sin[Sin[x]] = ArcTan[Pi/Sqrt[16 - Pi^2]]

Carry out the identical procedure twice more substituting the result from the previous step each time and you get

x = Pi/2 - i Ln[b + Sqrt[b^2 - 1]] where

b = ArcTan[(ArcTan[Pi/Sqrt[16 - Pi^2])/Sqrt[1 - (ArcTan[Pi/Sqrt[16 - Pi^2])^2 ]

according to Mathematica the result is approximately

1.5708 - 0.499742 I

Taking the Sine of this iteratively 4 times gives back approximately Pi/4