It ain't that hard for me

$\cos^2{10^\circ} + \cos^2{50^\circ} + \cos^2{70^\circ}$

$= \cos^2{10^\circ} + (\cos{60^\circ}\cos{10^\circ} + \sin{60^\circ}\sin{10^\circ} )^2 + (\cos{60^\circ}\cos{10^\circ} - \sin{60^\circ}\sin{10^\circ} )^2$

$=\cos^2{10^\circ} + (\frac {1}{2} \cos{10^\circ} + \frac {\sqrt{3}}{2}\sin{10^\circ} )^2 + (\frac {1}{2} \cos{10^\circ} - \frac {\sqrt{3}}{2}\sin{10^\circ} )^2$

$=\cos^2{10^\circ} + \frac {1}{4} \cos^2{10^\circ} + \frac {\sqrt{3}}{2} \sin {10^\circ} \cos {10^\circ} + \frac {3}{4}\sin^2{10^\circ} + \frac {1}{4} \cos^2{10^\circ} - \frac {\sqrt{3}}{2} \sin {10^\circ} \cos {10^\circ} + \frac {3}{4}\sin^2{10^\circ}$

$=\cos^2{10^\circ} + \frac {1}{2} \cos^2{10^\circ} + \frac {3}{2}\sin^2{10^\circ} = \frac {3}{2} (\cos^2{10^\circ} + \sin^2{10^\circ} ) = \frac {3}{2} = \boxed {1.5}$

$\cos^2{10^\circ}+\cos^2{50^\circ}+\cos^2{70^\circ}\\=\frac{3}{2}+\frac{1}{2}\left(\cos20^\circ+\cos100^\circ+\cos140^\circ\right)\\=\frac{3}{2}+\frac{1}{2}\left(2\cos60^\circ\cos40^\circ+\cos140^\circ\right)=\frac{3}{2}+\frac{1}{2}\left(\cos40^\circ+\cos140^\circ\right)\\=\frac{3}{2}+\frac{1}{2}\left(2\cos90^\circ\cos50^\circ\right)=~\frac{3}{2}$

i feel using complex numbers would be better...