MHB Can You Prove $\sin(2^{25})^\circ = -\cos(2^\circ)$?

[anemone]

Here is this week's POTW:

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Prove that $\sin(2^{25})^{\circ}=-\cos 2^{\circ}$

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to the kaliprasad for his correct solution, which you can find below::)

Spoiler

Let us first find $2^{25}\,\pmod {360}$
we have $360 = 2^3 * 45$
$2^{25} \equiv 0 \pmod {2^3}\cdots(1)$
to evaluate $2^{25}\, \pmod \, {45} $ we proceed as below
$\phi(45) = \phi(5 * 3^2) = 45 * ( 1- \frac{1}{5})( 1- \frac{1}{3})$
as 2 and 45 are coprimes So as per Eluer's theorem
$2^{\phi(45)} \equiv 1 \pmod{45}$
or $2^{24} \equiv 1 \pmod{45}$
hence $2^{25} \equiv 2 \pmod{45}\cdots(2)$
using (1) and (2) we should able to find $2^{25}\,\pmod {360}$
we have from (2) the value should be $45k+2$ ( k from 0 to 7) and from (1) $45k+2 \equiv 0 \pmod 8$
k cannot be odd as 45k+2 need to be even
k cannot be multiple of 4 (that is neither 0 nor 4) then 45k+2 shall not be divisible by 8
checking for k = 2 and 6 we get k = 6
so $2^{25} \equiv 272 \pmod{360}$
hence
$\sin(2^{25})^\circ = \sin(272^\circ) = \sin(360-272)^\circ = \sin\,- 88^\circ = - \sin\, 88^\circ = - \cos \,2^\circ$