## Tuesday, November 10, 2015

### Second Method: Find $k$ if $k\sin 6x=\sin 2x$ given $6\cos 6x=\cos 2x$.

Find $k$ if $k\sin 6x=\sin 2x$ given $\dfrac{\cos 6x}{\cos 2x}=\dfrac{1}{6}$.

Second method:

Note that we can rewrite the given equality $6\cos 6x=\cos 2x$ as $\dfrac{\cos 6x}{\cos 2x}=\dfrac{1}{6}$, also, our target expression as $\dfrac{\sin 6x}{\sin 2x}=\dfrac{1}{k}$.

If we are to add these the expressions on the LHS of both equations up, we get:

\begin{align*}\dfrac{\cos 6x}{\cos 2x}+\dfrac{\sin 6x}{\sin 2x}&=\dfrac{\sin 2x\cos 6x+\cos 2x \sin 6x}{\cos 2x \sin 2x}\\&=\dfrac{\sin (2x+6x)}{\cos 2x \sin 2x}\\&=\dfrac{\sin 8x}{\left(\dfrac{\sin 4x}{2}\right)}\\&=2\left(\dfrac{2\sin 4x\cos 4x}{\sin 4x}\right)\\&=4\cos 4x\end{align*}

Aww..this doesn't seem like a promising step, what if we subtract them?

\begin{align*}\dfrac{\sin 6x}{\sin 2x}-\dfrac{\cos 6x}{\cos 2x}&=\dfrac{\sin 6x\cos 2x-\cos 6x \sin 2x}{\sin 2x \cos 2x}\\&=\dfrac{\sin (6x-2x)}{\sin 2x \cos 2x}\\&=\dfrac{\sin 4x}{\left(\dfrac{\sin 4x}{2}\right)}\\&=2\end{align*}

Hey, this is the righteous path that we are now one step away from the answer.

Since we know $\dfrac{\cos 6x}{\cos 2x}=\dfrac{1}{6}$, we get:

$\dfrac{\sin 6x}{\sin 2x}-\dfrac{\cos 6x}{\cos 2x}=2$

$\dfrac{\sin 6x}{\sin 2x}=\dfrac{\cos 6x}{\cos 2x}+2=\dfrac{1}{6}+2=\dfrac{13}{6}$

And we're done!