How can we demonstrate that $\sqrt{\phi}=5^{1/4}\cos\left({1\over 2}\tan^{-1}(2)\right)?$ 
$$\text{Let}\ \sqrt{\phi}=5^{1/4}\cos\left({1\over 2}\tan^{-1}(2)\right)\tag1$$
  Where $\phi={\sqrt5+1\over 2}$

An attempt:
$$\tan^{-1}(2)={\pi\over 2}-\tan^{-1}{1\over 2}=2\tan^{-1}{1\over \phi}\tag2$$
Then $(1)$ becomes
$$\sqrt{\phi}=5^{1/4}\cos\left({\pi\over 4}-{1\over 2}\tan^{-1}{1\over 2}\right)\tag3$$
Apply compound formula
$$\sqrt{\phi}=5^{1/4}\cdot{\sqrt{2}\over 2}\left[\cos\left({1\over 2}\tan^{-1}{1\over 2}\right)+ \sin\left({1\over 2}\tan^{-1}{1\over 2}\right)\right]\tag4$$
It seems to be getting more complicated than before, so how else can we prove $(1)?$
 A: $$\sqrt{\phi}=\frac{\sqrt{\sqrt5+1}}{ \sqrt2}=5^{1/4}\frac{\sqrt{1+\frac{1}{\sqrt5}}}{ \sqrt2}$$
On the other hand : $ \cos (  \arctan x)=  \frac{1}{\sqrt{1+x^2}}$ so : $$\cos (  \frac{1}{2}\arctan x)=\sqrt{\frac{\cos (  \arctan x)+1}{2}}=\sqrt{\frac{ \frac{1}{\sqrt{1+x^2}}+1}{2}}$$
So : $$\cos (  \frac{1}{2}\arctan (2))=\sqrt{\frac{ \frac{1}{\sqrt{1+2^2}}+1}{2}}=\frac{\sqrt{1+\frac{1}{\sqrt5}}}{ \sqrt2}$$
Finally : 
$$\sqrt{\phi}=5^{1/4}\cos\left({1\over 2}\tan^{-1}(2)\right)\tag1$$
A: Observe that
$$\cos x=2\cos^2\frac x2-1\implies\cos\frac x2=\pm\sqrt{\frac{\cos x+1}2}$$
and since $\;\cos\arctan x=\frac1{\sqrt{1+x^2}}\;$ , we then get:
$$\cos^2\left(\frac12\arctan2\right)=\frac{\cos\arctan2+1}2=\frac{\frac1{\sqrt5}+1}2$$
so finally
$$\sqrt5\cos^2\left(\frac12\arctan2\right)=\frac{1+\sqrt5}2=\phi$$
A: From (2) 
$$\alpha=\frac12\tan^{-1}(2)=\tan^{-1}{1\over \phi}\tag2$$
$$\frac{1}{\cos^2\alpha}=1+\tan^{2}\alpha=1+\frac{1}{\phi^2}=\frac{1+\phi^2}{\phi^2}$$
$$\cos^2\alpha=\frac{\phi^2}{1+\phi^2}=\frac{\phi^2}{2\phi^2-\phi}=\frac{\phi}{2\phi-1}=\frac{\phi}{\sqrt{5}}$$
