<s>[INST] Solve: $-\frac{1}{\sqrt{2}} < \sin \theta + \cos \theta < \frac{1}{\sqrt{2}}$ The question is:

Solve $$-\frac{1}{\sqrt{2}} \lt \sin\theta + \cos\theta < \frac{1}{\sqrt{2}}$$ for values of $\theta$ between $0^\circ$ and $180^\circ$.

I realized that:
$$\begin{align}
-\frac{1}{\sqrt{2}} < \sin\theta + \cosθ &< \frac{1}{\sqrt{2}} \\[4pt]
\left|\sin\theta + \cosθ\right| &< \frac{1}{\sqrt{2}} \\[4pt]
\left(\sin\theta + \cos\theta\right)^2 &< \frac12 \\[4pt]
\sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta &< \frac12 \\[4pt]
1 + 2\sin\theta\cos\theta &< \frac12 \\[4pt]
1 + \sin 2\theta &< \frac12 \\[4pt]
\sin 2\theta &< -\frac12
\end{align}$$ 
But I don't know where to go from here. Can someone help me figure out how to get to the answer in the book: $105^\circ < θ < 165^\circ$. Thanks!
 [/INST] To solve: $$\sin(2\theta) < -\frac{1}{2}$$
we treat $2\theta$ as a single variable. The domain in your problem is $ 0^\circ < \theta < 180^\circ$. So to get our new domain all we must do is multiply by $2$ to obtain: $0^\circ < 2\theta < 360^\circ$. 
Now we must ask ourselves what angles when put through the sine function give me $-\frac{1}{2}$? Thinking back to the unit circle we have that $$\sin(210^\circ) = -\frac{1}{2} = \sin(330^\circ)$$
So any values between these (again from the unit circle) give me a value less than $-\frac{1}{2}$.
$$210^\circ < 2\theta < 330^\circ$$
Dividing by $2$ we get the result:
$$105^\circ < \theta < 165^\circ$$
</s>