$$\text{Let}\ \sqrt{\phi}=5^{1/4}\cos\left({1\over 2}\tan^{-1}(2)\right)\tag1$$ 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 $(1)?$