I presume you quickly got to \(\displaystyle sin(\theta)+ cos(\theta)= \frac{1}{2}\). To go beyond that you will need a trig identity such as \(\displaystyle sin(a+ b)= sin(a)cos(b)+ cos(a)sin(b)\) taking \(\displaystyle b= \theta\). Of course, "sin(a)" and "cos(a)" can't both be one. Sin(a) and cos(a) are the same for \(\displaystyle a= \pi/4\) and then are both equal to \(\displaystyle \frac{\sqrt{2}}{2}\). Multiplying both sides by that, and taking \(\displaystyle a= \pi/4\), \(\displaystyle \frac{\sqrt{2}}{2}sin(b+ \pi/4)= \frac{\sqrt{2}}{2}sin(b)+ \frac{\sqrt{2}}{2}cos(b)\).
So, going back to the original equation, \(\displaystyle \frac{\sqrt{2}}{2}sin(\theta)+ \frac{\sqrt{2}}{2}cos(\theta)= sin(\theta+ \pi/4)= \frac{\sqrt{2}}{4}\)..edited. Can you solve that equation?
