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) $1 + \cos x$
 A: $$
2^6 \cos^6 x = (e^{ix}+e^{-ix})^6=e^{6ix}+6e^{4ix}+15e^{2ix} +20+15e^{-2ix}+6e^{-4x}+e^{-6ix} \\
= 2(\cos 6x + 6 \cos 4x +15 \cos 2x +10)
$$
and so forth
A: Hint
Use multiple angles identities as they are given here and you will find a simple result (since $\cos(nx)$ can be expressed as a polynomial in $\cos(x)$).
A: 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 $2 \cos x$.
A: My answer is (B)
because when I threw it into Wolfy
and subtracted 
$2\cos(x)$
it came out zero.
Lazy in my old age.
(added later)
Well,
at 0 it is 2,
so only B and D are possible.
and at $\pi$
it is
$\frac{32}{-16}
=-2
$
so it can't be D,
so it must be B.
A: Use this identity and the work is done $cos(x)+cos(y)=2cos\frac{(x+y)}{2}.\frac{(x-y)}{2}$ and write $10=10cos0$ then create factors and nultiples to use this identity and after that multiple use of this identity you will get answer. As an example i give first step . $\frac{(cos(6x)+cos(4x)+5cos(4x)+5cos(2x)+10cos(2x)+10cos0)}{(cos(5x)+cos(3x)+4cos(3x)+4cos(x)+6cos(x))}=\frac{(2cos(5x).cos(x)+10cos(3x).cos(x)+20cos(x).cos(x))}{2cos(4x).cos(x)+8cos(2x).cos(x)+6cos(x)}$ now cancel $2,cos(x)$ then ........$=2cos(x)$ multiple times repeat the same process and its done. Bit long but thats necesarry. Hope its clear.
