Simplifying $\frac{1-4\cos80^\circ}{\tan20^\circ}$ I am working on simplifying the value
$$\frac{1-4\cos80^\circ}{\tan20^\circ}$$
I am looking for it to be expressed in terms of tangent, because the value seems to be $\tan40^\circ$. However, I can't find any obvious way to do so. How would  simplify it to express like such?
 A: Let $\theta = \pi/9 = 20^\circ$.  We wish to show that $$1 - 4 \cos 4\theta = \tan \theta \tan 2\theta. \tag{1}$$  This suggests the tangent angle addition identity $$\tan (\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \tag{2}$$ for the choice $$\alpha = \theta, \beta = 2\theta$$ which yields
$$(\tan 3\theta)(1 - \tan \theta \tan 2\theta) = \tan \theta + \tan 2\theta. \tag{3}$$  Since $3\theta = \pi/3 = 60^\circ$ hence $\tan 3\theta = \sqrt{3}$, we find
$$\begin{align}
1 - \tan \theta \tan 2\theta 
&= \frac{1}{\sqrt{3}}(\tan \theta + \tan 2\theta) \\
&= \frac{\sin \theta \cos 2\theta + \sin 2\theta \cos \theta}{\sqrt{3} \cos \theta \cos 2\theta} \\
&= \frac{\sin 3\theta}{\sqrt{3} \cos \theta \cos 2\theta} \\
&= \frac{1}{2\cos \theta \cos 2\theta}. \tag{4}
\end{align}$$
We now claim that $$8 \cos \theta \cos 2\theta \cos 4\theta = 1. \tag{5}$$  To this end, write
$$\begin{align}
8 \sin \theta \cos \theta \cos 2\theta \cos 4 \theta
&= 4 \sin 2\theta \cos 2\theta \cos 4\theta \\
&= 2 \sin 4\theta \cos 4\theta \\
&= \sin 8\theta \\
&= \sin(\pi - \theta) \\
&= \sin \theta,
\end{align}$$
and now dividing both sides by $\sin \theta$ yields the proof of $(5)$.  Therefore, $(4)$ is equivalent to
$$1 - \tan \theta \tan 2\theta = 4 \cos 4\theta,$$ which in turn is equivalent to $(1)$.

Something unsettled me about my solution above; I had the feeling it was unnecessarily complicated.  After some thought I noticed that it could be simplified as follows:  to show $(1)$, consider
$$\begin{align}
1 - \tan \theta \tan 2\theta 
&= \frac{\cos \theta \cos 2\theta - \sin \theta \sin 2\theta}{\cos \theta \cos 2\theta} \\
&= \frac{\cos 3\theta}{\cos \theta \cos 2\theta} \\
&= \frac{1}{2 \cos \theta \cos 2\theta},
\end{align}$$
which takes us directly to $(4)$ without the tangent addition identity.
A: $\frac{1-4\cos(80)}{\tan(20)}=\frac{(1-4\cos(80))\sin(40)}{1-\cos(40)}$. Use $\tan \frac{\theta}{2}=\frac{1-\cos\theta}{\sin\theta}$. (Try to turn 20 into 40.)
$=\tan(40)\frac{(1-4\cos(80))\cos(40)}{1-\cos(40)}$
$=\tan(40)\frac{\cos(40)-2\cos(120)-2\cos(40)}{1-\cos(40)}$. Use $\cos x \cdot \cos y=\frac{1}{2}(\cos(x+y)+\cos(x-y))$. (Try to turn 80 into 40)
$=\tan(40)\frac{1-\cos(40)}{1-\cos(40)}$. (Done, all 20, 80, 120 gone away. Only 40 left now.)
$=\tan(40)$
A: By letting $f=e^{\frac{\pi i}{18}}$,
$\begin{align}\frac{1-4\cos80}{\tan20}&=\frac{(f^2+f^{-2})-2(f^{10}+f^{-10})-2(f^6+f^{-6})}{f^7+f^{-7}}\\
&=\frac{\cos20-2\cos100-2\cos60}{\cos70}\\
&=\frac{\cos20+2\cos80-1}{\sin20}\\
&=\frac{1-2\sin^2 10+2\sin10-1}{2\sin10\cos10}\\
&=\frac{1-\sin 10}{\cos10}\\
&=\frac{1-\cos 80}{\sin80}\\
&=\tan40\\
\end{align}$
A: Let's call $t=\tan(x)$ with $x=20°=\frac{\pi}9$
From tan(3x) formula we get $\ \tan(3x)=\sqrt{3}=\dfrac{3t-t^3}{1-3t^2}$
Squaring this to get rid of the root, we get that $t$ is solution of the integer polynomial
$3(1-3t^2)^2=(3t-t^3)^2\iff$
$$t^6-33t^4+27t^2-3=0$$
Now we can use the $\tan(\theta/2)$ conversions from the initial equation

*

*$1-4\cos(4x)=5-8\cos(2x)^2=5-8\left(\frac{1-t^2}{1+t^2}\right)^2$

*$\tan(40°)=\tan(2x)=\frac{2t}{1-t^2}$
And we get
$\begin{align}(1-4\cos(4x))-\tan(x)\tan(2x)&=5-8\left(\frac{1-t^2}{1+t^2}\right)^2-\frac{2t^2}{1-t^2}\\\\
&=\frac{t^6-33t^4+27t^2-3}{(1-t^2)(1+t^2)^2}\end{align}$
Which is precisely $0$ according to the work above.
A: A way to prove that $\;\dfrac{1-4\cos80^o}{\tan20^o}=\tan40^o\;$ by using only double-angle formula for sine and angle sum identity for cosine.
$\;\dfrac{1-4\cos80^o}{\tan20^o}=\dfrac{\big(1-4\cos80^o\big)\sin20^o\cos20^o\cos40^o}{\tan20^o\sin20^o\cos20^o\cos40^o}=$
$=\dfrac{\sin20^o\!\cos20^o\!\cos40^o-4\sin20^o\!\cos20^o\!\cos40^o\!\cos80^o}{\sin^220^o\cos40^o}=$
$=\dfrac{\sin20^o\cos20^o\cos40^o-2\sin40^o\cos40^o\cos80^o}{\sin^220^o\cos40^o}=$
$=\dfrac{\sin20^o\cos20^o\cos40^o-\sin80^o\cos80^o}{\sin^220^o\cos40^o}=$
$=\dfrac{\sin20^o\cos20^o\cos40^o-\frac12\sin160^o}{\sin^220^o\cos40^o}=$
$=\dfrac{\sin20^o\cos20^o\cos40^o-\cos60^o\sin20^o}{\sin^220^o\cos40^o}=$
$=\dfrac{\sin20^o\big(\cos20^o\cos40^o-\cos(20^o+40^o)\big)}{\sin^220^o\cos40^o}=$
$=\dfrac{\sin20^o\big(\sin20^o\sin40^o\big)}{\sin^220^o\cos40^o}=\tan40^o\;.$
A: Generalization
Let's see how the problem can have come into being
Using $\dfrac{\cos3x}{\cos x}=4\cos^2x-3=2(1+\cos2x)-3=2\cos2x-1$   for $\cos x\ne0,$
$$F=\dfrac{1-4\cos4t}{\tan t} =\dfrac{1-2\left(\dfrac{\cos6t}{\cos2t}+1\right)}{\sin t}\cdot\cos t =-\dfrac{(\cos2t+2\cos6t)\cos t}{\cos2t\sin t}$$
Now if $2\cos6t=-1,$
$$F=-\dfrac{(\cos2t-1)\cos t}{\cos2t\sin t}=\dfrac{2\sin^2t\cos t}{\cos2t\sin t}=\tan2t\text{ if }\sin t\ne0$$
So, we need $6t=360^\circ n\pm120^\circ\implies t=60^\circ n\pm20^\circ$ where $n$ is any integer, which clearly keeps $\sin t\ne0$
