Upon multiplying $\cos(20^\circ)\cos(40^\circ)\cos(80^\circ)$ by the sine of a certain angle, it gets reduced. What is that angle?

So if $P = \cos(20^\circ)\cos(40^\circ)\cos(80^\circ)$, I can multiply $P$ by $\sin(X)$ so that the entire expression reduces to something manageable. I then take the simplified product and divide it by $\sin(X)$ and should get the numerical value of $\cos(20^\circ)\cos(40^\circ)\cos(80^\circ)$.

However, I can't figure out what angle to use. I know it probably has something to do with the double angle forumlae and proably something to do with the product sum formulae. I tried to reduce $\cos(20^\circ)\cos(40^\circ)\cos(80^\circ)$ to see if I got an answer but no luck. I've looked on the Internet and the answer is there but it's unclear how people figured out which sine angle to multiply it by so if you do provide an answer, it'd be really great if you could mention why you picked the angle you picked. Thanks.

If $P=\cos(20)\cos(40)\cos(80)$, assuming that you are working with degrees and not radians, we have $$ \begin{array}{rcl}\sin(20) P&=&(\sin (20)\cos(20))\cos(40)\cos(80)\\ &=&\frac12(\sin(40)\cos(40))\cos(80)\\ &=&\frac14\sin(80)\cos(80)\\&=&\frac18\sin(160)\\ &=&\frac18\sin(20),\end{array} $$ so $$P=\frac18.$$ I used that $\sin(180^\circ-\alpha)=\sin\alpha$, and that $2\sin\alpha\cos\alpha=\sin(2\alpha)$. The latter formula suggests to use $\sin(20)$, as we are given cosines of $20$, of $2$ times $20$, and of $4$ times $20$. But it is only the fact that $20=180-160$ that makes the idea work.

this is useful. $$\cos (60-\theta)\cos (60+\theta)\cos (\theta)=\frac{\cos 3\theta}{4}$$ $$ \sin (60-\theta)\sin (60+\theta)\sin (\theta)=\frac{\sin 3\theta}{4}$$ $$ \tan (60-\theta)\tan (60+\theta)\tan (\theta)=\tan 3\theta$$

We have $$\cos3\phi=4\cos^3\phi-3\cos\phi$$

Setting $\displaystyle \cos3\phi=\frac12, \cos3\phi=\cos60^\circ\implies3\phi=360^\circ n\pm60^\circ\iff \phi=120 n\pm20^\circ$

So, setting $\displaystyle n=0,1,2; \cos20^\circ,\cos140^\circ, \cos260^\circ$ are the roots of $\displaystyle4t^3-3t=\frac12\iff 8t^3-6t-1=0$

Using Vieta's formula, $\displaystyle \cos20^\circ\cdot\cos140^\circ\cdot\cos260^\circ=\frac18$

Now as $\displaystyle\cos(180^\circ\pm x)=-\cos x, $

$\displaystyle\cos140^\circ=\cos(180-40)^\circ=-\cos40^\circ$ and $\displaystyle\cos260^\circ=\cos(180+80)^\circ=-\cos80^\circ$

Another observation : With the identity $\displaystyle\sin8x=2\sin4x\cos4x=2(2\sin2x\cos2x)\cos4x=4(2\sin x\cos x)\cos2x\cos4x$

$\displaystyle\implies \cos x\cos2x\cos4x=\frac{\sin8x}{8\sin x} $

which will be $\displaystyle=\frac18$ if $\displaystyle\sin8x=\sin x,\sin x\ne0\iff x\ne180^\circ n$ where $n$ is any integer

Now, $\displaystyle\sin8x=\sin x\implies 8x=180^\circ n+(-1)^nx$

If $n$ is even, $=2m$(say), $\displaystyle8x=360^\circ m+x\iff x=\frac{360^\circ m}7$ where $7\not\mid m$ as $\sin x\ne0$

If $n$ is odd, $=2m+1$(say), $\displaystyle8x=(2m+1)180^\circ-x\iff x=(2m+1)20^\circ$ where $9\not\mid(2m+1)$ as $\sin x\ne0$