How to find the inverse cosine without a calculator

How to find the inverse of:

$$\cos(c)=\frac{1}{3}$$

In other words, i'm trying to solve for c and without a calculator. If it's hard or not possible, then how would you go about solving inverses in general. For example, lets say:

$$\cos(c)=\frac{3+1}{2\sqrt{2}}$$

how would we solve for c, which in this case is 105

The first step is to stop and think about the problem itself. What is the cosine? If you recall, the formula for cosine (remember SOHCAHTOA?) is adjacent over hypotenuse. So, with your $\frac{1}{3}$ example, $1$ represents the length of the adjacent side, and $3$ represents the hypotenuse.

In other words, the hypotenuse is going to be 3 times longer than the adjacent side. Picture that on the unit circle. The hypotenuse, as always on the unit circle, would have a length of 1, and the measurement along the $x$ axis would only be $0.\overline{3}$.

That's a pretty steep angle. Just from this mental picture, you can already realize that it's going to be much greater than $45^{\circ}$, and probably even more than $60^{\circ}$.

Ronald Doerfler, in his book Dead Reckoning: Calculating Without Instruments, teaches an estimation method for finding $arccos(x)$, or $cos^{-1}(x)$ (both just mean finding the original angle given its cosine) in degrees.

His estimation formula for $cos^{-1}(x)$ is: $\sqrt{7(1000-1000x)}-\frac{1}{2}$

This looks bad, but can be done mentally with practice. Using your $\frac{1}{3}$ example, let's step through that one step at a time.

This works better with decimals, so we'll switch from $\frac{1}{3}$ to $0.\overline{3}$.

Step 1: $1000\times0.\overline{3}=333.\overline{3}$, which we'll round to $333$.

Step 2: $1000-333=667$. Subtracting from 1000 is easy. If you're not already familiar with the mental method for this, this video will give you a quick refresher.

Step 3: $667\times7$? Work this out mentally from left to right! $600\times7=4200$, $60\times7=420$, and $7\times7=49$, so we've got $4200+420+49=4620+49=4669$.

Step 4: $\sqrt{4669}$?!? How are you supposed to do that in your head?!? First, you should be familiar with mentally squaring 2-digit numbers. I know $65^{2}=4225$ and $70^{2}=4900$, so I can quickly figure that $\sqrt{4669}$ is somewhere between $65$ and $70$.

Mentally figure that $67^{2}=4489$. Hmmm...perhaps $68^{2}$ or $69^{2}$ would be closer. $68^{2}=4624$ and $69^{2}=4761$, so the answer is obviously $68$ point something.

We can even quickly refine that, using other mental square root estimation techniques. $4669-4624=45$, so we can use the linked technique to realize that $\sqrt{4669}\approx68\frac{45}{137}$, or about $68\frac{1}{3}$.

Step 5: $68\frac{1}{3}-\frac{1}{2}=67\frac{5}{6}$, so this method has given us an estimated angle of about $68^{\circ}$.

Through experience, I've found a way to improve on Ronald Doerfler's above estimate.

Before step 1, take note of the tenths digit of your original $x$. With $0.\overline{3}$, the tenths digit is obviously $3$.

Whenever this digit is less than 6 (for 6 through 9, no adjustment is needed), you're going to add $6$ minus this digit, to the number of degrees, as a last step.

Since the original tenths digit was $3$ in this case, we'll add $6-3=3$ more degrees. $68+3=71$, so we have an adjusted estimate of about $71^{\circ}$.

If you actually work out $cos^{-1}(\frac{1}{3})$ with a calculator, you'll find the answer is $70.53^{\circ}$, so our mental estimate of $71^{\circ}$ is quite close!

There is not a general method (although you can approximate the $\arccos$ function, with Taylor for example) to find an exact solution. In fact, $\arccos(\frac{1}{3})$ isn't even a "clean angle", since it isn't a rational multiple of $\pi$, see also the answer at How do i prove that $\frac{1}{\pi} \arccos(1/3)$ is irrational?.

$$\arccos x =\dfrac{\pi}{2}-\displaystyle\sum_{n=0}^{\infty} \dfrac{(2n)!}{2^{2n}(n!)^2(2n+1)}x^{2n+1}=\dfrac{\pi}{2}-\biggl(x+\dfrac{1}{2}\dfrac{x^3}{3}+\dfrac{1}{2\cdot4}\dfrac{x^5}{5}+\underbrace{\cdots}_{o(x)}\biggr), \mid x\mid \lt 1$$

$$\arccos\biggl(\dfrac{1}{3}\biggr)\approx\dfrac{\pi}{2}-\biggl(\dfrac{1}{3}+\dfrac{1}{162}+\dfrac{1}{9720}+o\biggl(\dfrac{1}{3}\biggr)\biggr)\approx 1.231290=70.54773^{\circ}$$