Lesson Math OLYMPIAD level problems on Trigonometry

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Math OLYMPIAD level problems on Trigonometry



Problem 1

Let   f(x) = %28x+-+sqrt%283%29%29%2F%28x%2Asqrt%283%29+%2B+1%29.

What is   f^{2012}(x),  where the function is being applied/composed  2012  times?
This notation indicates repeated composition of functions,  not exponentiation of functions.
For example,   f^2(x) = f(f(x))  and not   f(x)*f(x).   Similarly,  f^3(x) = f(f(f(x))).

Solution


            This problem allows absolutely unexpected and elegant solution. 


Let  x = tan(a)  (which means "let a = arctan(x)").  Then


    f(x) = %28tan%28a%29+-+sqrt%283%29%29%2F%28tan%28a%29%2Asqrt%283%29%2B1%29 = %28tan%28a%29+-+tan%28pi%2F3%29%29%2F%281%2Btan%28a%29%2Atan%28pi%2F3%29%29 = tan%28a-pi%2F3%29.


It means that taking f(x) for x= tan(a) returns tan%28a-pi%2F3%29.


Obviously, that taking the composition (fof)(x) = f(f(x))      will return  tan%28a-2pi%2F3%29;

                taking the composition (fofof)(x) = f(f(f(x))) will return  tan%28a-3pi%2F3%29;


                . . . . and so on . . . 


                taking the composition f^{2012}(x) = f(f(f...f(x)))...)  will return  tan%28a-2012pi%2F3%29.


It easy to calculate:  2012pi%2F3 = 670%2Api + 2pi%2F3;   THEREFORE  


    tan%28a-2012pi%2F3%29 = tan%28a-670pi+-+2pi%2F3%29 = tan%28a-2pi%2F3%29 = %28tan%28a%29+-+tan%282pi%2F3%29%29%2F%281+%2B+tan%28a%29%2Atan%282pi%2F3%29%29 = %28tan%28a%29-%28-sqrt%283%29%29%29%2F%281%2Btan%28a%29%2A%28-sqrt%283%29%29%29 = %28x%2Bsqrt%283%29%29%2F%281-sqrt%283%29%2Ax%29.    ANSWER



ANSWER.  f^{2012}(x) = %28x%2Bsqrt%283%29%29%2F%281-sqrt%283%29%2Ax%29.


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