What's the expression $( \cos 6x + 6 \cos 4x + 15 \cos 2x + 10 ) / ( \cos 5x + 5 \cos 3x + 10 \cos x ) $ equal to? $$\frac{ \cos 6x + 6 \cos 4x + 15 \cos 2x + 10 }{ \cos 5x + 5 \cos 3x + 10 \cos x }$$
My approach so far : Tried to represent the denominator as a factor of numerator by manipulating numerator's $\cos 6x = \cos (5x+x)$ , $\cos 4x = \cos (3x+x)$ , so on .. but then $\sin x$ come up which make it more complex to solve . 
The options for the answer are: 
A) $\cos 2x$.
B) $2 \cos x$.
C) $\cos^2 x$.
D) $1 + \cos x$

In a concept similar to the Chebyshev Polynomial,
For a fixed angle $x$, We let $T(n)=\cos nx$. Note the relation of $T(n+1)=2T(1)T(n)-T(n-1)$.
Now we have the numerator as $$T(6)+6T(4)+15T(2)+10=2T(1)T(5)-T(4)+6T(4)+15T(2)+10 = 2T(1)T(5)+5T(4)+15T(2)+10 = 2T(1)T(5)+5(2T(1)T(3)-T(2))+15T(2)+10=2T(1)T(5)+10T(1)T(3)+10T(2)+10=2T(1)T(5)+10T(1)T(3)+20T(1)T(1)-10T(0)+10=2T(1)T(5)+10T(1)T(3)+20T(1)T(1)=2T(1)(T(5)+5T(3)+10T(1))$$
Since the denominator is just $T(5)+5T(3)+10T(1)$, we have the answer as $2T(1)$, or $2 \cos x$.
