proof left as an exercise

[Albert Chan]

(06-08-2022 11:18 PM)Thomas Klemm Wrote:  We can use the triple angle formulae:

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

I noticed an easier way

sin(3θ)/sin(θ) = 4*cos(θ)^2 - 1
cos(3θ)/cos(θ) = 4*cos(θ)^2 - 3

sin(3θ)/sin(θ) = cos(3θ)/cos(θ) + 2

This is all is need for the proof:

2*cos(30°) / (1+4*sin(70°))
= 2*sin(60°) / (1+4*cos(20°))
= 2*sin(20°) * (cos(60°)/cos(20°) + 2) / (1+4*cos(20°))
= tan(20°) * (1+4*cos(20°)) / (1+4*cos(20°))
= tan(20°)