If $20x=\pi$, what is $\frac{\cos 4x - \cos 8x}{\cos 4x\cdot \cos 8x}$? If $20x=\pi$,
what is $$\frac{\cos 4x - \cos 8x}{\cos 4x\cdot \cos 8x}?$$
I've tried using the factor formula on the numerator but I haven't managed to get anywhere with it...
This is a multiple choice question with options $4$, $2$, $1$, $-1$, and $-2$.
Thanks in advance!
 A: $$\\ 20x=\pi \\ 10x=\frac { \pi  }{ 2 } \\ \frac { \cos { \left( 10x-6x \right) -\cos { \left( 10x-2x \right)  }  }  }{ \cos { \left( 10x-6x \right) \cos { \left( 10x-2x \right)  }  }  } =\frac { \sin { \left( 6x \right) -\sin { \left( 2x \right)  }  }  }{ \sin { \left( 6x \right) \sin { \left( 2x \right)  }  }  } =\frac { 2\sin { \left( 2x \right) \cos { \left( 4x \right)  }  }  }{ \sin { \left( 6x \right) \sin { \left( 2x \right)  }  }  } =2\frac { \cos { \left( 4x \right)  }  }{ \sin { \left( 6x \right)  }  } =2$$
A: You have $4x=\pi-16x$, so $\cos4x=\cos(\pi-16x)=-\cos16x$. By the sum-to-product formula,
$$
\cos16x+\cos8x=2\cos12x\cos4x
$$
and we have a first simplification:
$$
\frac{\cos 4x - \cos 8x}{\cos 4x\cdot \cos 8x}=
-\frac{2\cos12x}{\cos8x}
$$
Now $12x=\pi-8x$, so $\cos12x=-\cos8x$
A: If for some integer $n,5y=(2n+1)\pi,$
$$\cos3y=\cos\{(2n+1)\pi-2y\}=-\cos2y$$
$$-\cos2y=4\cos^3y-3\cos y=2\cos y(1+\cos2y)-3\cos y$$
$$\iff\cos y-\cos2y=2\cos y\cos2y$$
$$\iff\dfrac{\cos y-\cos2y}{\cos y\cos2y}=2$$
Here $y=4x,n=?$
