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

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

Well done, Albert! Thankyou for your participation!
 
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