proof left as an exercise

[Thomas Klemm]

We can use the triple angle formulae:

\(
\begin{align}
\sin(3 \theta) &= 3 \sin \theta - 4 \sin^{3} \theta \\
\cos(3 \theta) & = 4 \cos^{3}\theta - 3 \cos \theta \\
\end{align}
\)

The first formula leads to:

\(
\begin{align}
\sin(3 \theta)
&= 3 \sin \theta - 4 \sin^{3} \theta \\
&= \sin \theta \, [3 - 4 \sin^{2} \theta] \\
&= \sin \theta \, [4 \cos^{2} \theta - 1] \\
\end{align}
\)

The second formula leads to:

\(
2 \cos(3 \theta) - 1 = 8 \cos^{3}\theta - 6 \cos \theta - 1 \\
\)

For \( \theta = 20^\circ \) we get:

\(
\begin{align}
2 \cos(3 \theta) - 1 &= \\
2 \cos(3 \cdot 20^\circ) - 1 &= \\
2 \cos(60^\circ) - 1 &= \\
2 \cdot \tfrac{1}{2} - 1 &= 0 \\
\end{align}
\)

And thus:

\(
8 \cos^{3}(20^\circ) - 6 \cos(20^\circ) - 1 = 0
\)

From this we conclude:

\(
\begin{align}
8 \cos^{3}(20^\circ) - 2 \cos(20^\circ) &= 1 + 4 \cos(20^\circ) \\
2 \cos(20^\circ) [4 \cos^{2}(20^\circ) - 1] &= 1 + 4 \cos(20^\circ) \\
\end{align}
\)

Time to plug all that into the formula:

\(
\begin{align}
\frac{2\cos(30^{\circ})}{1+4\sin(70^{\circ})}
&= \frac{2\sin(60^{\circ})}{1+4\cos(20^{\circ})} \\
\\
&= \frac{2\sin(3 \cdot 20^{\circ})}{1+4\cos(20^{\circ})} \\
\\
&= \frac{2 \sin(20^\circ) [4 \cos^{2}(20^\circ) - 1]}{2 \cos(20^\circ) [4 \cos^{2}(20^\circ) - 1]} \\
\\
&= \frac{\sin(20^\circ)}{\cos(20^\circ)} \\
\\
&= \tan(20^\circ)
\end{align}
\)

---

Or then we use the product to sum identity:

\(
2 \cos \theta \cos \varphi = \cos(\theta + \varphi )+\cos(\theta - \varphi)
\)

We use \(\cos(60^\circ) = \frac{1}{2}\) and start with:

\(
\begin{align}
\cos(10^\circ)
&= 2 \cos(60^\circ) \cos(10^\circ) \\
&= \cos(70^\circ) + \cos(50^\circ) \\
\\
\cos(50^\circ) + \cos(10^\circ) &= \cos(70^\circ) + 2 \cos(50^\circ) \\
2 \cos(30^\circ) \cos(20^\circ) &= \sin(20^\circ) + 2 \sin(40^\circ) \\
&= \sin(20^\circ) + 4 \sin(20^\circ) \cos(20^\circ) \\
&= \sin(20^\circ)[1 + 4 \cos(20^\circ)] \\
\end{align}
\)

This leads to:

\(
\begin{align}
\frac{2 \cos(30^\circ)}{1 + 4 \cos(20^\circ)} = \frac{\sin(20^\circ)}{\cos(20^\circ)} = \tan(20^\circ)
\end{align}
\)

Or then:

\(
\begin{align}
\frac{2 \cos(30^\circ)}{1 + 4 \sin(70^\circ)} = \tan(20^\circ)
\end{align}
\)