MHB Rational trigonometric expression show tan^218°⋅tan^254°∈Q

[lfdahl]

Show, that $$\tan^2 18^{\circ} \cdot \tan^254^{\circ} \in \Bbb{Q}.$$

 

[Albert1]

Show, that $$\tan^2 18^{\circ} \cdot \tan^254^{\circ} \in \Bbb{Q}.$$

my solution:

Spoiler

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

 

[lfdahl]

my solution:

Spoiler

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

Well done, Albert! Thankyou for your participation!