TITLE: Finding the maximum value of $\sin\left((2^n)^\circ\right)$ where $n \in \Bbb N$
QUESTION [5 upvotes]: Found this question in my math textbook where it asked us to find $\max\sin\left((2^n)^\circ\right)$ where $n \in \Bbb N$ and $2^n$ expressed in god forbid degrees.
It's in the arithmetic section but the problem seems analytic at first.
I tried finding if $n$ exists such that $2^n=90+360k$ for some $k\in \Bbb N$ but I can't quite get some progress with it, then I tried using functions where I minimize the difference between $2^n-90-360k$ to get close to the max of the $\sin$ function that we all know is $1$ but not luck there either.
Any help would be appreciated!
PS: I know how to study functions using basic calculus, I just graduated high school. I wouldn't mind a little advanced math to explore and learn.
EDIT : just found out that $\sin(2^{96})=0.99..\gt \sin(2^6)$ which contradicts Ben's answer, maybe $2^n\mod360$ isn't periodic ?

REPLY [6 votes]: Credit to Henry for the inspiration on this solution.
I started by listing out the sequence generated by $2^n$ mod $360$: $2, 4, 8, 16, 32, 64, 128, 256, 152, 304, 248, 136, 272, 184, 8...$ The sequence produced by taking the sine of each term of this sequence is equal to the sequence produced by taking the sine of each term of the raw $2^n$ sequence, because subtracting (or adding) multiples of $360$ from the input of a sine function doesn't change its output.
As you can see, the sequence repeats itself (eventually looping back to 8). Therefore, there are finitely many values of $sin(2^n)$, and you can easily work out the maximum.