Solving $\sin \frac{\theta}{2} + \cos \frac{\theta}{2} = \sqrt{2}$ Does anyone have some tips for me how to go about the problem in the image? 
$$\sin \frac{\theta}{2} + \cos \frac{\theta}{2} = \sqrt{2}$$
I know it's supposed to be simple, but I can't figure out why the solution is: 
$90^{\circ}+ 720^{\circ}k$ where $k=1,2,3,...$  
Intuitively, I understand the answer, but that's not the point, I would like to understand the correct mathematical solution. Thank you very much!

Note that: $$\sin \frac{\theta}{2} + \cos \frac{\theta}{2} = \sqrt{2} \cos \left(\frac{\theta}{2} - \frac{\pi}{4} \right) \quad \quad \quad \quad (\star)$$
So your equation reduces down to $\cos \left(\frac{\theta}{2} - \frac{\pi}{4}\right) = 1 \iff \frac{\theta}{2} - \frac{\pi}{4} = 2\pi k$ where $k$ ranges over the integers. 
This can be re-written equivalently as $\theta = \frac{\pi}{2} + 4\pi k$ or in degrees: $\theta = 90^{\circ} + 720^{\circ} k$. 

$(\star):$ Assume that $\sin x + \cos x = R\cos (x - \alpha)$ then expanding the left hand, we have $\sin x + \cos x = R\cos x \cos \alpha + R\sin x \sin \alpha$. Now we have $R\cos \alpha = 1$ and $R\sin \alpha = 1$, squaring and adding gives us $R^2 (\cos^2 \alpha + \sin^2 \alpha) = 2 \Rightarrow R = \sqrt{2}$. Dividing the two equations gives $\frac{R\sin \alpha}{R\cos \alpha} = \tan \alpha = 1\Rightarrow \alpha = \frac{\pi}{4}$, whence $(\star)$. 
