How to Find the Inverse Cosine Without a Calculator

The inverse cosine, written $\arccos(x)$ or $\cos^{-1}(x)$, asks: which angle in the principal range has cosine $x$? You can often answer exactly from familiar unit-circle angles. If the input is not a familiar value, you can still estimate the angle by hand using identities and a series.

The notation $\cos^{-1}(x)$ means inverse cosine, not $1/\cos(x)$. The input must satisfy $-1\le x\le 1$, and the standard output is an angle from $0$ to $\pi$ radians (or from $0^\circ$ to $180^\circ$).

Start with exact unit-circle values

Memorize or derive a few common cosine values:

$x$	$\arccos(x)$ in radians	In degrees
$1$	$0$	$0^\circ$
$\sqrt{3}/2$	$\pi/6$	$30^\circ$
$\sqrt{2}/2$	$\pi/4$	$45^\circ$
$1/2$	$\pi/3$	$60^\circ$
$0$	$\pi/2$	$90^\circ$
$-1/2$	$2\pi/3$	$120^\circ$
$-1$	$\pi$	$180^\circ$

For example, $\cos(\pi/3)=1/2$, so $\arccos(1/2)=\pi/3$. This is an exact answer; there is no need to turn it into a decimal unless the problem asks for one.

Cosine is not one-to-one over all angles. For instance, many angles have cosine $1/2$. The inverse cosine returns just the principal value, here $\pi/3$, rather than every solution to $\cos\theta=1/2$.

Use symmetry for negative inputs

The identity $\cos(\pi-\theta)=-\cos\theta$ gives

$$\arccos(-x)=\pi-\arccos(x).$$

So you can first find the inverse cosine of the corresponding positive value. For example,

$$\arccos(-\sqrt{2}/2)=\pi-\arccos(\sqrt{2}/2)=\pi-\pi/4=3\pi/4.$$

This also means you only need to develop a hand-approximation method for nonnegative inputs.

Approximate an unfamiliar value with a series

For $0<x<1$, a useful half-angle identity is

$$\arccos(x)=2\arctan\!\left(\sqrt{\frac{1-x}{1+x}}\right).$$

It follows from $\tan^2(\theta/2)=(1-\cos\theta)/(1+\cos\theta)$, taking $\theta=\arccos(x)$. The square root is nonnegative, which selects the angle in the required range.

Then use the alternating arctangent series

$$\arctan(z)=z-\frac{z^3}{3}+\frac{z^5}{5}-\frac{z^7}{7}+\cdots.$$

For $0\le z<1$, the terms decrease in size, so stopping after a term gives an error smaller than the next omitted term. The series is especially convenient when the square root in the identity is a simple fraction.

Worked example: find $\arccos(0.8)$

Write $0.8=4/5$. The half-angle expression becomes

$$\arccos(0.8)=2\arctan\!\left(\sqrt{\frac{1-4/5}{1+4/5}}\right) =2\arctan\!\left(\sqrt{\frac{1}{9}}\right) =2\arctan(1/3).$$

Substitute $z=1/3$ into the series and keep five terms:

$$\begin{aligned} \arccos(0.8)&\approx 2\left(\frac13-\frac{1}{81}+\frac{1}{1215}-\frac{1}{15309}+\frac{1}{177147}\right)\\ &\approx 0.643502\text{ radians}. \end{aligned}$$

The next omitted term in the doubled series has size $2/(3^{11}\cdot 11)$, which is about $0.00000103$. Thus $0.64350$ radians is a reasonable five-decimal-place approximation. If you want degrees, multiply radians by $180/\pi$; this angle is about $36.87^\circ$.

Use bisection when the series is awkward

Another hand method is to narrow down the angle. For a positive input, start with an interval between $0$ and $\pi/2$. Since cosine decreases on this interval, compare the target with the cosine of the midpoint. If the midpoint’s cosine is too large, the desired angle is larger; otherwise it is smaller. Repeat with the appropriate half of the interval.

You can estimate each cosine using

$$\cos\theta\approx 1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\frac{\theta^6}{6!}+\cdots,$$

with $\theta$ in radians. For example, to bracket $\arccos(0.8)$, compare cosine estimates at $0.6$ and $0.7$ radians: they are about $0.8253$ and $0.7648$. The desired angle lies between them. Further bisection gives a tighter interval, though the arithmetic is longer than for a simple exact angle.

Common mistakes

• Treating $\cos^{-1}(x)$ as a reciprocal. In inverse-trigonometry notation it means $\arccos(x)$; the reciprocal is $1/\cos(x)$, also written $\sec(x)$.
• Ignoring the principal range. The answer from $\arccos$ is between $0$ and $\pi$, even though the equation $\cos\theta=x$ may have many solutions.
• Using an input outside the domain. A real angle cannot have a cosine greater than $1$ or less than $-1$, so real $\arccos(x)$ is undefined for such inputs.
• Mixing radians and degrees. The series and the half-angle formula use radians. Convert to degrees only at the end if needed.