Question #c7520

Jul 14, 2016

Use the double-angle identity for sine and the unit circle to find solutions of $\theta = - \frac{\pi}{2} , \frac{\pi}{6} , \frac{\pi}{2} , \frac{5 \pi}{6}$, and $\frac{3 \pi}{2}$.

Explanation:

First, we use the important identity $\sin 2 \theta = 2 \sin \theta \cos \theta$:
$\sin 2 \theta - \cos \theta = 0$
$\to 2 \sin \theta \cos \theta - \cos \theta = 0$

Now we can factor out $\cos \theta$:
$2 \sin \theta \cos \theta - \cos \theta = 0$
$\to \cos \theta \left(2 \sin \theta - 1\right) = 0$

And using the zero product property, we obtain solutions of:
$\cos \theta = 0 \text{ and } 2 \sin \theta - 1 = 0 \to \sin \theta = \frac{1}{2}$

So, when does $\cos \theta = 0$ on the interval $- \frac{\pi}{2} \le \theta \le \frac{3 \pi}{2}$? The solutions can be found by using the unit circle and a property of the cosine function:
$\cos \left(- \theta\right) = \cos \theta$

If $\theta = \frac{\pi}{2}$, then:
$\cos \left(- \frac{\pi}{2}\right) = \cos \left(\frac{\pi}{2}\right)$

From the unit circle, we know that $\cos \left(\frac{\pi}{2}\right) = 0$, which also means $\cos \left(- \frac{\pi}{2}\right) = 0$; so two solutions are $- \frac{\pi}{2}$ and $\frac{\pi}{2}$. Also, the unit circle tells us that $\cos \left(\frac{3 \pi}{2}\right) = 0$, so we have another solution there.

Now, onto $\sin \theta = \frac{1}{2}$. Again, we will need the unit circle to find our solutions.

We know from the unit circle that $\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$, and $\sin \left(\frac{5 \pi}{6}\right) = \frac{1}{2}$, so we add $\frac{\pi}{6}$ and $\frac{5 \pi}{6}$ to the list of solutions.

Finally, we put all of our solutions together: $\theta = - \frac{\pi}{2} , \frac{\pi}{6} , \frac{\pi}{2} , \frac{5 \pi}{6}$, and $\frac{3 \pi}{2}$.

The Unit Circle