How can I solve $\cos A = -\frac12$ without having to consult a table of angles and their cosines?

Question

I am learning Trigonometry for the first time and I am kinda hating it, anyways, I would like you to help me out understanding one thing:

I am well aware of the unit circle and how it works (because of it, we no longer need to rely on right triangles to know the $\sin$, $\cos$, and $\tan$ due to the trigonometric ratios).

So here is my question: I ran into this equation

$$\cos A = -\frac12$$

I know it is both in the third and fourth quadrant due to the negative sign, but how can I figure out the solution without looking up a table with the angles and their sines and cosines?

I mean, I look up the table, find the angles which has this cosine and I move on. I don't like it; it does not challenge my brain at all! Is there another way to tackle this? Because thinking algebraically is what makes me like Math the most, but trigonometry seems to be something that I need to always look up the information elsewhere but my brain.

Answer 1

When I learnt trigonometry we were made to learn the special angles (you can also learn the corresponding special triangles if that is easier). From which we have that $$\cos{(60^\circ)}=\frac12$$ So we can solve $\cos{(x)}=-\frac12$ by noticing that $$\cos{(180^\circ-x)}=-\cos{(x)}$$ and using $x=60^\circ$ we get $$\cos{(180^\circ-60^\circ)}=-\cos{(60^\circ)}$$ $$\cos{(120^\circ)}=-\frac12$$ Then we have that one solution is $x=120^\circ$. From which we can get other solutions by using $$\cos{(360^\circ-x)}=\cos{(x)}$$ $$\cos{(360^\circ+x)}=\cos{(x)}$$ Which gives $x=\pm120^\circ,\pm240^\circ,\dots$

Answer 2

For general values of $x$, there is not an easy way with pen-and-paper to find the result of $\cos(\theta)=x$. For very specific values however, memorizing the table is the easiest way to go.

The table is easily constructed. Make yourself three columns. Write the numbers $0,1,2,3,4$ going down one column and up the other. $$\begin{smallmatrix}0&4\\1&3\\2&2\\3&1\\4&0\end{smallmatrix}$$

Then square root each and divide the result by two.

Finally, simplify the results and label the table and remember that cosine of zero is $1$.

$$\begin{array}{c|c|c}&\sin(\theta)&\cos(\theta)\\\hline 0^\circ&0&1\\30^\circ&\frac{1}{2}&\frac{\sqrt{3}}{2}\\45^\circ&\frac{\sqrt{2}}{2}&\frac{\sqrt{2}}{2}\\60^\circ&\vdots&\vdots\\90^\circ\end{array}$$

From there, you can extrapolate to other quadrants by applying negative signs where appropriate.

If you are interested in how a calculator would do it for arbitrary values, see this related question. I will emphasize though that although it may be useful to understand how it works and that the algorithm exists, no one actually performs these sort of calculations by hand in practice.