Suppose you want cos(nx) and sin(nx) in terms of sin(x).  You'll need cos(x) as well which can be found using 
cos(x)=±1(sin(x))2

First note that eix=cos(x)+isin(x)
Raising both sides to the nth power:
enix=(cos(x)+isin(x))n
And note that 
enix=cos(nx)+isin(nx)
Therefore, by the transitive property
(cos(x)+isin(x))n=cos(nx)+isin(nx)
So 
cos(nx)=Re(cos(x)+isin(x))n)
and 
sin(nx)=Im(cos(x)+isin(x))n)
Expanding using the binomial theorem yields:
(cos(x)+isin(x))n=k=0n(nk)(cos(x))nk(isin(x))k

Splitting it into real and imaginary parts:
cos(nx)=k=0n2(n2k)(1)k(sin(x))2k(cos(x))n2k
And
sin(nx)=k=0n12(n2k+1) 

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