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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!

haqnatural
  • 21,578
choly
  • 43

3 Answers3

4

$$\\ 20x=\pi \\ 10x=\frac { \pi }{ 2 } \\ \frac { \cos { \left( 10x-6x \right) -\cos { \left( 10x-2x \right) } } }{ \cos { \left( 10x-6x \right) \cos { \left( 10x-2x \right) } } } =\frac { \sin { \left( 6x \right) -\sin { \left( 2x \right) } } }{ \sin { \left( 6x \right) \sin { \left( 2x \right) } } } =\frac { 2\sin { \left( 2x \right) \cos { \left( 4x \right) } } }{ \sin { \left( 6x \right) \sin { \left( 2x \right) } } } =2\frac { \cos { \left( 4x \right) } }{ \sin { \left( 6x \right) } } =2$$

haqnatural
  • 21,578
1

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 $\cos12x=-\cos8x$

egreg
  • 238,574
0

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 $y=4x,n=?$

lab bhattacharjee
  • 274,582