The discussion centers on proving that the expression $$\tan^2 18^{\circ} \cdot \tan^2 54^{\circ}$$ is a rational number, denoted as $$\Bbb{Q}$$. Participants confirm the validity of this expression through mathematical reasoning. Albert's contribution is acknowledged positively, emphasizing the collaborative nature of the discussion.
PREREQUISITESMathematicians, students studying trigonometry, and educators looking to deepen their understanding of rational trigonometric expressions.
my solution:lfdahl said:Show, that $$\tan^2 18^{\circ} \cdot \tan^254^{\circ} \in \Bbb{Q}.$$
Albert said:my solution:
let $y=\tan\, 18^{\circ} \cdot \cot\,36^{\circ}=\dfrac {2cos^218^o-1}{2cos^218^o}$
$=1-\dfrac{1}{2cos^218^o}$=$\dfrac{\sqrt 5}{5}$
so $y^2=\dfrac {1}{5}\in \Bbb{Q}$
(using $sin\,18^o=cos\,72^o=\dfrac {\sqrt 5-1}{4}---(1)$
(1) can be proved easily using geometry ,which I posted long time ago