Golden ratio?!?

Given:

$\sqrt{2}\sin^2{\frac{C}{2}} + \cos{\frac{C}{2}} = \sqrt{2}$

Knowing that: $1 - \sin^2{\frac{C}{2}} = \cos^2{\frac{C}{2}}$ reduces the above equation to:

$\sqrt{2}\cos^2{\frac{C}{2}} = \cos{\frac{C}{2}}$

Based on the given information, the angle $C$ cannot be equal to 180 degrees as that would make $\cos{\frac{C}{2}} =0$ . This leaves:

$\cos{\frac{C}{2}} = \frac{1}{\sqrt{2}}$

This means that the triangle ABC is a right-angled triangle and $\angle C = \pi/2$ . Given that the sides follow a geometric progression, let $a = a$ , $b = ar$ and $c = ar^2$ . Using Pythagoras theorem:

$a^2r^4 = a^2 + a^2 r^2$

The ratio $r$ must be a real number. Solving for r^2 gives:

$r^2 = \phi$

This implies:

$a = a$ $b = a\sqrt{\phi}$ $c = a\phi$

Now using the sine-rule leads to:

$\frac{\sin{C}}{c} = \frac{\sin{A}}{a}$

$\frac{1}{a\phi} = \frac{\sin{A}}{a} \implies \sin{A} = \frac{1}{\phi}$

Now,

$\phi = \frac{\sqrt{5}+1}{2} \implies \frac{1}{\phi} = \frac{2}{\sqrt{5}+1} = \frac{\sqrt{5}-1}{2} = \frac{\sqrt{5}+1}{2} - 1 = \phi -1$

Therefore:

$\boxed{\sin{A} = \phi -1}$