Prove the identity... $$\frac{\cos{2x}-\sin{4x}-\cos{6x}}{\cos{2x}+\sin{4x}-\cos{6x}}=\tan{(x-15^{\circ})}cot{(x+15^{\circ})}$$
So, here's what I've done so far, but don't know what do do next:
$$\frac{\cos{2x}-2\sin{2x}\cos{2x}-\cos{6x}}{\cos{2x}+2\sin{2x}\cos{2x}-\cos{6x}}=$$
$$\frac{\cos{2x}-4\sin{x}\cos{x}\cos{2x}-\cos{(4x+2x)}}{\cos{2x}+4\sin{x}\cos{x}\cos{2x}-\cos{(4x+2x)}}=$$
$$\frac{\cos{2x}-4\sin{x}\cos{x}\cos{2x}-\cos{4x}\cos{2x}+\sin{4x}\sin{2x}}{\cos{2x}+4\sin{x}\cos{x}\cos{2x}-\cos{4x}\cos{2x}+\sin{4x}\sin{2x}}=...$$
I have no idea what to do next. I have a big feeling that I'm going in a totally wrong direction. Is there anything I can do with the expression in the beginning?
 A: Well, don't expand to $\sin(x)$ or $\cos(x)$...
Consider complex numbers...
Let $z = \exp( 2 \textbf{i} x )$
Then
$
\begin{eqnarray}
2 \cos(2x) &=& z + \bar{z}\\
2 \sin(2x) &=& - \textbf{i} \Big( z - \bar{z} \Big)\\
2 \sin(4x) &=& - \textbf{i} \Big( z^2 - \bar{z}^2 \Big)\\
&=& - \Big( z + \bar{z} \Big) \textbf{i} \Big( z - \bar{z} \Big)\\
2 \cos(6x) &=& z^3 + \bar{z}^3\\
&=& \Big( z + \bar{z} \Big) \Big( z^2 - z\bar{z} + \bar{z}^2 \Big)
\end{eqnarray}
$
Then we can write
$
\begin{eqnarray}
2\cos(2x) + 2\sin(4x) - 2\cos(6x)
&=& \Big(z + \bar{z}\Big)
\left[ 1 - \textbf{i} \Big( z - \bar{z} \Big) - z^2 + z\bar{z} - \bar{z}^2 \right]\\
&=& \Big(z + \bar{z}\Big)
\left[ - \textbf{i} \Big( z - \bar{z} \Big) - z^2 + 2 z\bar{z} - \bar{z}^2 \right]\\
&=& \Big(z + \bar{z}\Big)
\left[ \textbf{i}^2 \Big( z - \bar{z} \Big)^2 - \textbf{i} \Big( z - \bar{z} \Big) \right]\\
&=& - \textbf{i} \Big( z - \bar{z} \Big) \Big(z + \bar{z}\Big) \left[ - \textbf{i} \Big( z - \bar{z} \Big) + 1 \right]\\
\end{eqnarray}
$
and
$
\begin{eqnarray}
2\cos(2x) - 2\sin(4x) - 2\cos(6x)
&=& - \textbf{i} \Big( z - \bar{z} \Big) \Big(z + \bar{z}\Big) \left[ - \textbf{i} \Big( z - \bar{z} \Big) - 1 \right]\\
\end{eqnarray}
$
So
$
\begin{eqnarray}
\frac{\cos(2x) - \sin(4x) - \cos(6x)}{\cos(2x) + \sin(4x) - \cos(6x)}
&=& \frac{ - \textbf{i} \Big( z - \bar{z} \Big) - 1 }{ - \textbf{i} \Big( z - \bar{z} \Big) + 1 }\\
\end{eqnarray}
$
But $\sin(2x) = - \textbf{i} \Big( z - \bar{z} \Big)$, whence
$
\begin{eqnarray}
\frac{\cos(2x) - \sin(4x) - \cos(6x)}{\cos(2x) + \sin(4x) - \cos(6x)}
&=& \frac{ 2 \sin(2x) - 1 }{ 2 \sin(2x) + 1 }\\
\end{eqnarray}
$

We can write
$
\begin{eqnarray}
\frac{ 2 \sin(2x) - 1 }{ 2 \sin(2x) + 1 }
&=& \frac{ 2 \sin(2x) - 2 \sin(2y) }{ 2 \sin(2x) + 2 \sin(2y) }\\
\end{eqnarray}
$
where $2 \sin(2y) = 1$, thus $\sin(2 y) = \frac{1}{2}$, so $2 y = 30^o$,
whence $y = 15^o$.
But
$
\begin{eqnarray}
\sin(2x) \pm 2 \sin(2y) &=&
2 \sin\big( x \pm y \big) \cos\big( x \mp y \big)
\end{eqnarray}
$
so
$
\begin{eqnarray}
\frac{\cos(2x) - \sin(4x) - \cos(6x)}{\cos(2x) + \sin(4x) - \cos(6x)}
&=& \frac{ \sin\big( x - 30^o \big) \cos\big( x + 30^o \big) }{ \sin\big( x - 30^o \big) \cos\big( x - 30^o \big) }\\
\end{eqnarray}
$
So we obtain
$
\begin{eqnarray}
\frac{\cos(2x) - \sin(4x) - \cos(6x)}{\cos(2x) + \sin(4x) - \cos(6x)}
&=& \tan\big( x - 30^o \big) \cot\big( x + 30^o\big)
\end{eqnarray}
$
A: I'd start by using $2x=4x-2x$ and $6x=4x+2x$ to get to
$$
\cos(2x)-\cos(6x)=2\sin(2x)\sin(4x)
$$
A: This is an exercise in the sum-to-product and product-to-sum identities.
On the LHS, use
$$ \cos a - \cos b = -2\sin(\tfrac12(a+b))\sin(\tfrac12(a-b)) $$
on the top and bottom (as suggested by Lutzl).  For the RHS, write
$$ \tan(x-15^\circ)\cot(x+15^\circ)
= \frac{\sin(x-15^\circ)\cos(x+15^\circ)}{\sin(x+15^\circ)\cos(x-15^\circ)} $$
and then use
$$ \sin a \cos b = \tfrac12(\sin(a+b) + \sin(a-b)) $$
on the top and bottom.
A: Using Prosthaphaeresis Formula, $$\frac{\cos2x-\cos6x}{\sin4x}=\frac{2\sin4x\sin2x}{2\sin4x}=\frac{\sin2x}{\dfrac12}$$
$$\implies \frac{\cos2x-\cos6x}{\sin4x}=\frac{\sin2x}{\sin30^\circ}$$
Now apply Componendo and dividendo and again apply Prosthaphaeresis formulae $$\sin C\pm\sin D$$
