$(\tan^2(18^\circ))(\tan^2(54^\circ))$ is a rational number Assuming $$\cos(36^\circ)=\frac{1}{4}+\frac{1}{4}\sqrt{5}$$ How to prove that         $$\tan^2(18^\circ)\tan^2(54^\circ)$$ is a rational number? Thanks! 
 A: $$\tan18^\circ\tan54^\circ=\frac{2\sin18^\circ\sin54^\circ}{2\cos18^\circ\cos54^\circ}$$
$$=\frac{\cos36^\circ-\cos72^\circ}{\cos36^\circ+\cos72^\circ}$$ (applying $\cos(A\pm B)$ formulae)
$$=\frac{\frac{\sqrt5+1}4-\frac{\sqrt5+1}4}{\frac{\sqrt5+1}4+\frac{\sqrt5+1}4}$$ as $$\cos 72^\circ=2\cos^236^\circ-1=\frac{\sqrt5-1}4$$
$$\implies \tan18^\circ\tan54^\circ=\sqrt5$$
A: Use the fact that
$$ \tan^2{18^{\circ}} = \frac{1-\cos{36^{\circ}}}{1+\cos{36^{\circ}}} = 1-\frac{2}{5} \sqrt{5} $$
Then use the fact that
$$ \tan^2{54^{\circ}} = \frac{1}{\tan^2{36^{\circ}}} $$
so that
$$ \tan^2{18^{\circ}} \tan^2{54^{\circ}} = \frac{\tan^2{18^{\circ}}}{\tan^2{36^{\circ}}} =   \frac{1}{4} (1 -\tan^2{18^{\circ}})^2 = \frac{1}{5} $$
A: Hint:
Use:
$$\sin(3x) = 3 \sin (x) \cos ^2(x)-\sin ^3(x)$$
$$\cos(3x) = \cos ^3(x)-3 \sin ^2(x) \cos (x)$$
Then plug in $x=18^\circ$.
A: Use the definition of tan (sin/cos) along with half angle relation for cos (i.e. 18=36/2), sum of angles relation for cos (i.e. 54=36+18) and the fact that sin^2=1-cos^2 to get the value for tan^2.  Then massage the numbers based on the value for cos(36deg).  You should end up with a ration number.
