Luc's Trig problem
Matthew Daly -- Tan[ArcCos[Tan[ArcCos[Sin[ArcTan[sqrt[x]]]]]] = sqrt[x-1], which gives you 1 and 0 by repeated iterations.
Roel Huisman -- http://www.homepages.hetnet.nl/~eomer/PForms/start.html On your site there was a problem about converting '3' into other numbers, using just trig functions and square roots. When I was playing around with Maple I found something else that intrigued me a bit...although it may be nothing for I am not a star in mathematics but here it goes: tan(arcsin(cos(arctan(cos(arctan(sqrt(x))))))) gives 1 for x=0, 2 for x=3 (the one on your site),3 for x=8.....and so on. So the row 0 3 8 15 24 35 48 etc (+3 +5 +7 +9 +11 +13 etc) gives us all integers. I checked it in Maple with a simple 'for loop' and it worked for large numbers, although Maple started to make mistakes rounding off the larger numbers. I guess the above should be very obvious but I can't see it, can you help me with this?
--A little while ago you posted a piece from Roel Huisman wondering why tan(arcsin(cos(arctan(cos(arctan(sqrt(x)))))))
turns out integral when x = 0, 3, 8, 15, 24, ...
In case no-one's explained it yet, here goes:
Consider this right-angled
/ | sqrt(x)
1. Angle A = arctan(sqrt(x))
2. cos(A) = 1 / sqrt(1 + x)
Now consider this right-angled
/ | sqrt(1+x)
3. Angle E = arctan(1 / sqrt(1
+ x)) = arctan(cos(arctan(sqrt(x))))
4. Angle D = arcsin(cos(D) = arcsin(cos(arctan(cos(arctan(sqrt(x))))))
5. tan(D) = sqrt(1 + x), so the original expression
tan(arcsin(cos(arctan(cos(arctan(sqrt(x))))))) = sqrt(1 + x)
which is clearly an integer
(n) when x = n^2 - 1, ie 0, 3, 8, 15, 24, ...
Colin Backhurst -- Use the formula sec(arctan(n))=sqrt(n*n + 1) which is derived from the right angled triangle with sides (1, n, sqrt(n*n+1) ). So sec(arctan(3))=sqrt(10) sec(arctan(sec(arctan(3)))) = sqrt(11). Apply this five more times and you reach sqrt(16) which is 4. Apply it eleven more times and you reach sqrt(25) which is 5