trigonometry - Efficiently finding $theta$ such that $costheta = -frac12$ and $sintheta = frac{sqrt{3}}{2}$. I know the quadrant, but what are the angles?

I'm having trouble solving trigonometric equations. For example, let's say I'm solving a problem and I arrive at a trigonometric equation that says,
$$\cos\theta = -\frac12 \quad\text{and}\quad \sin\theta = \frac{\sqrt{3}}{2} $$
At this point, I get stuck and I don't have an efficient way to proceed apart from picking up a calculator.

I can figure that the quadrants (from the signs of the ratios) -- but I can't figure out the angles. What is a good way to figure out the angle? Specifically, how do I systematically solve $\sin$, $\cos$, and $\tan$ trigonometric equations? (I can reciprocate the other three into these ratios.)

I don't have trouble figuring out angles between $0^\circ$ to $90^\circ$ (since I have that memorized), but for angles in other quadrants, I get stuck.

Answer

If you have angle $\theta$ in quadrant $1$, you can find its "corresponding" angle in quadrant $2$ by $(\pi - \theta)$, in quadrant $3$ by $(\pi+\theta)$, and in quadrant $4$ by $(2\pi-\theta)$. For example, $\frac{\pi}{4}$ corresponds to $\frac{3\pi}{4}$, $\frac{5\pi}{4}$, and $\frac{7\pi}{8}$ in quadrants $2$, $3$, and $4$, respectively. (That's how I always think of them at least.)

Also, recall sine functions correspond to the height of the right triangle ($y$-axis), so they are positive in quadrants $1$ and $2$. Cosine functions correspond to base of the right triangle ($x$-axis), so they are positive in quadrants $2$ and $4$. (Tangent functions can be found through sine and cosine functions.)

You can use the following identities (which are derived from the aforementioned facts).

$$\sin\bigg(\frac{\pi}{2}+\theta\bigg) = \cos\theta \quad \sin\bigg(\frac{\pi}{2}-\theta\bigg) = \cos\theta$$

$$\cos\bigg(\frac{\pi}{2}+\theta\bigg) = -\sin\theta \quad \cos\bigg(\frac{\pi}{2}-\theta\bigg) = \sin\theta$$

$$\tan\bigg(\frac{\pi}{2}+\theta\bigg) = -\cot\theta \quad \tan\bigg(\frac{\pi}{2}-\theta\bigg) = \cot\theta$$

$$\sin\bigg(\pi+\theta\bigg) = -\sin\theta \quad \sin\bigg(\pi-\theta\bigg) = \sin\theta$$

$$\cos\bigg(\pi+\theta\bigg) = -\cos\theta \quad \cos\bigg(\pi-\theta\bigg) = -\cos\theta$$

$$\tan\bigg(\pi+\theta\bigg) = \tan\theta \quad \tan\bigg(\pi-\theta\bigg) = -\tan\theta$$

$$\sin\bigg(\frac{3\pi}{2}+\theta\bigg) = -\cos\theta \quad \sin\bigg(\frac{3\pi}{2}-\theta\bigg) = -\cos\theta$$

$$\cos\bigg(\frac{3\pi}{2}+\theta\bigg) = \sin\theta \quad \cos\bigg(\frac{3\pi}{2}-\theta\bigg) = -\sin\theta$$

$$\tan\bigg(\frac{3\pi}{2}+\theta\bigg) = -\cot\theta \quad \tan\bigg(\frac{3\pi}{2}-\theta\bigg) = \cot\theta$$

$$\sin\bigg(2\pi+\theta\bigg) = \sin\theta \quad \sin\bigg(2\pi-\theta\bigg) = -\sin\theta$$

$$\cos\bigg(2\pi+\theta\bigg) = \cos\theta \quad \cos\bigg(2\pi-\theta\bigg) = \cos\theta$$

$$\tan\bigg(2\pi+\theta\bigg) = \tan\theta \quad \tan\bigg(2\pi-\theta\bigg) = -\tan\theta$$

I certainly wouldn't recommend memorizing these though since knowing how the unit circle works basically means you know them already.

For example, in an equation you reach $$\cos \theta = -\frac{\sqrt{3}}{2}$$

You already know that $\cos {\frac{\pi}{6}} = \frac{\sqrt{3}}{2}$ and you also know cosine is negative in quadrants $2$ and $3$, so all you need to do is find the corresponding angle for ${\frac{\pi}{6}}$ in those quadrants.

$$\text{Quadrant II} \implies \theta = \pi-{\frac{\pi}{6}} = \frac{5\pi}{6}$$

$$\text{Quadrant III} \implies \theta = \pi+{\frac{\pi}{6}} = \frac{7\pi}{6}$$

This might take a bit of practice, but once you get this whole "corresponding" angle concept, it all becomes simple. Perhaps you can start by trying to visualize this by solving equations with a unit circle. You'll eventually get the hang of it.