# 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!

• It's not the case for arbitrary $\theta$ that $\sin 2\theta + \cos 2\theta = 1$. – anomaly Jul 12 '15 at 23:25

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 $$\sin 2\theta < - \frac{1}{2} \implies 210^{\circ} < 2\theta < 330^{\circ}.$$

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.

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.

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: $$105^\circ < \theta < 165^\circ$$