Definition

A function takes an input and produces an output. Its inverse runs the process backward: given the output, it recovers the input. If $f(a) = b$, then $f^{-1}(b) = a$.

For an inverse to exist as a function, $f$ must be one-to-one: each output comes from exactly one input. If two different inputs gave the same output, the inverse would not know which one to return.

Trig functions fail this! Consider $\sin\theta = \tfrac{1}{2}$. This is true for $\theta = \tfrac{\pi}{6}$, but also $\tfrac{5\pi}{6}$, and $\tfrac{\pi}{6} + 2\pi$, and infinitely many more. Asking "what angle has sine $\tfrac{1}{2}$?" has no single answer. So sine, cosine, and tangent have no inverse on their full domains.

We will use both inverse (e.g., $\sin^{-1}$) and arc (e.g., $\arcsin$) notation throughout this module; both mean the same thing.

Principal Ranges

We repair the above caveat using principal ranges, i.e, an interval where the function is one-to-one. On that interval the function passes the horizontal-line test, and an inverse exists. The chosen interval is called the principal range, and its outputs are the principal values.

For $sin$, we restrict to $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$, where the function covers its entire range without repetition. This gives the inverse sine, written $\arcsin$ (or $\sin^{-1}$):

For cosine, $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$ would not work (cosine is symmetric there), so we use $[0, \pi]$.

For tangent, which covers every real value on $(-\tfrac{\pi}{2}, \tfrac{\pi}{2})$, we restrict to that open interval:

Don't Get Trapped!

It is tempting to think $\arcsin(\sin\theta) = \theta$ always. It does not. The output must land in the principal range, so if $\theta$ starts outside it, you must reduce. For example:

because $\sin\tfrac{3\pi}{4} = \tfrac{\sqrt{2}}{2}$, and the angle in $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$ with that sine is $\tfrac{\pi}{4}$, not $\tfrac{3\pi}{4}$. Going the other way, $\sin(\arcsin x) = x$ always holds for $x \in [-1,1]$, because that composition does not leave the range.

Evaluating Compositions

Mixed compositions like $\cos(\arcsin x)$ look intimidating but are actually easy to deal with. Set $\theta = \arcsin x$, so $\sin\theta = x$ with $\theta \in [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$. Recall the Pythagorean identity:

Substitute $x$ for $\sin\theta$:

We take the positive root because $\theta \in [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$, cosine is never negative. Therefore:

Similarly, for $\sin(\arccos x)$: Set $\theta = \arccos x$, so $\cos\theta = x$ with $\theta \in [0, \pi]$. Substitute into the identity:

Here we take the positive root because $\theta \in [0, \pi]$, where sine is never negative. Therefore:

For tangent compositions we use $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$. To find $\cos(\arctan x)$, set $\theta = \arctan x$, so $\tan\theta = x$ with $\theta \in (-\tfrac{\pi}{2}, \tfrac{\pi}{2})$. Divide the Pythagorean identity by $\cos^{2}\theta$:

Substitute $x$ for $\tan\theta$ and solve:

The positive root holds because cosine is positive on $(-\tfrac{\pi}{2}, \tfrac{\pi}{2})$. Then $\sin\theta = x\cos\theta$ gives the following results (you might want to memorize these!):

$\cos(\arctan x) = \frac{1}{\sqrt{1 + x^2}}, \qquad \sin(\arctan x) = \frac{x}{\sqrt{1 + x^2}}$

$\sin(\arctan x) = \frac{x}{\sqrt{1 + x^2}}, \qquad \cos(\arctan x) = \frac{1}{\sqrt{1 + x^2}}.$

Key Identities

Here are a couple nice identites that pop up from time to time in contests. This topic is pretty niche, but everything's worth knowing:

$\arcsin x + \arccos x = \frac{\pi}{2}, \qquad x \in [-1, 1].$

Proof: if $\theta = \arcsin x$ then $\cos\left(\tfrac{\pi}{2} - \theta\right) = \sin\theta = x$, and $\tfrac{\pi}{2} - \theta \in [0, \pi]$, so $\tfrac{\pi}{2} - \theta = \arccos x$.

It is also worth mentioning

$\arctan a + \arctan b = \arctan\left(\frac{a + b}{1 - ab}\right) + k\pi,$

where the correction term is

We bring in $k\pi$ as when $ab > 1$ the sum leaves the arctan range, so the output is off by $\pi$. When $ab < 1$ with both terms positive, you can drop $k$.

Telescoping with Arctangent

We use the above formula backwards to attain:

This holds because, with $a = n+1$ and $b = -n$, we get $\frac{a + b}{1 - ab} = \frac{1}{1 + n(n+1)} = \frac{1}{n^2 + n + 1}$, and $ab = -n(n+1) < 1$, as desired.

Telescoping Example

Evaluate

We use the identity we just showed:

As $N \to \infty$, $\arctan(N+1) \to \tfrac{\pi}{2}$ and $\arctan(1) = \tfrac{\pi}{4}$, so

Worked Examples

2008 AIME I · Problem 8: Find the positive integer $n$ such that $\arctan\frac{1}{3} + \arctan\frac{1}{4} + \arctan\frac{1}{5} + \arctan\frac{1}{n} = \frac{\pi}{4}.$

Every term here is a positive acute angle, so each $\arctan\frac{1}{k}$ is small. That means we can simply apply the addition formula!

Combine the first two terms with $a = \frac{1}{3}$, $b = \frac{1}{4}$:

Combine third term, $\arctan\frac{1}{5}$:

Since $\arctan\frac{1}{n} = \frac{\pi}{4} - \arctan\frac{23}{24}$, take the tangent of both sides. Using $\tan\frac{\pi}{4} = 1$ and the subtraction formula:

Therefore $n = \boxed{47}$.

2013 AIME I · Problem 8: The domain of $f(x) = \arcsin\!\left(\log_m(nx)\right)$ is a closed interval of length $\frac{1}{2013}$, where $m$ and $n$ are positive integers and $m > 1$. Find the remainder when the smallest possible $m + n$ is divided by $1000$.

The whole problem hinges on the domain of $\arcsin$. Its input must satisfy $-1 \le t \le 1$, so we need

Undo the logarithm by raising each side to the power of m:

Divide by $n$:

This is the domain. Its length is the right end minus the left:

Set that equal to $\frac{1}{2013}$:

For $n$ to be an integer, $m$ must divide $2013(m^2 - 1)$. But $m$ shares no factor with $m^2 - 1$, so $m$ must divide $2013 = 3 \cdot 11 \cdot 61$.

Now minimize the sum. Substitute the expression for $n$:

This grows as $m$ grows, so we want the smallest divisor of $2013$ with $m > 1$, namely $m = 3$. Then

giving $m + n = 5371$. The remainder when divided by $1000$ is $\boxed{371}$.

Practice Problems