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!
 A: Hint: $$\sin \theta + \cos \theta \equiv \sqrt{2} \sin \left(\theta + \frac{\pi}{4}\right).$$
Or alternatively, in degrees $$\sqrt{2} \sin \left(\theta + 45^{\circ}\right).$$

Using your method, however, we get that $$\sin 2\theta < -
\frac{1}{2} \implies 210^{\circ} < 2\theta < 330^{\circ}.$$
A: From that point you can do
$$\sin2\theta<-\frac12$$
Since $\theta$ is between $0$ and $180^o$, $2\theta$ is between $0$ and $360^o$. Then,
$$210^o<2\theta<330^o$$
Now, dividing by $2$, you obtain your book's result.
A: You can use the "R-cos-alpha" method for this. (No idea of the "actual" name.)
Note, this method is applicable to any problem where you have linear combinations of sin and cosine. $A \sin(\theta)+B \cos(\theta) = C$
Take the extremities $c = \pm\frac{1}{\sqrt{2}}$, $c = \sin(\theta) + \cos(\theta)$
Then, $c = \sin(\theta) + \cos(\theta) = R \cos(\theta + \alpha)$
Expanding the cosine: $$c = R \left ( cos(\theta)\cos(\alpha) - sin(\theta)sin(\alpha) \right )$$
Compare coefficients of the variable $\theta$:
$$-1 = R \sin(\alpha); 1 = R\cos(\alpha)$$
Then, by division: $$\tan(\alpha) = -1; R^2 = 1^2 + 1^2 \rightarrow R = \frac{1}{\sqrt{2}}$$
Hence, $\cos(\theta + \alpha) = \frac{c}{R}$ and from this you can easily find all the solutions of $\theta$, by drawing a graph to help if required. Then shade the relevant range as required in your original problem.
I like this method - not enough people know about the "R-cos-alpha" method.
A: 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:
$$105^\circ < \theta < 165^\circ$$
