Inverse Trigonometric Functions — Properties

Take a moment with this expression:

Three inverse tangents of three simple integers. What does the sum equal? It is not obvious — \tan^{-1} is not linear, so you cannot just add the arguments. But there is a formula that collapses pairs of \tan^{-1} terms into a single \tan^{-1}, and if you apply it twice, the entire expression simplifies to \pi. Three numbers in, one clean answer out. The formula that does this is the subject of this article.

Before the formulas, though, there is a layer of properties that sits underneath them — relations between the six inverse trig functions themselves. These are simpler, but they are the foundation on which the addition formulas rest.

Relations between inverse functions

Complementary relations

You already know from basic trigonometry that \sin\!\left(\frac{\pi}{2} - \theta\right) = \cos\theta. This means sine and cosine are complementary: one at angle \theta equals the other at \frac{\pi}{2} - \theta.

The same relationship holds for their inverses.

Proof of the first identity. Let \sin^{-1} x = \alpha. Then \sin\alpha = x, with \alpha \in [-\pi/2, \pi/2]. Since \cos(\pi/2 - \alpha) = \sin\alpha = x, you need to check that \pi/2 - \alpha falls in the principal range of \cos^{-1}, which is [0, \pi]. Since \alpha \in [-\pi/2, \pi/2], the value \pi/2 - \alpha ranges over [0, \pi]. So \cos^{-1} x = \pi/2 - \alpha, giving \sin^{-1} x + \cos^{-1} x = \alpha + (\pi/2 - \alpha) = \pi/2.

The proofs for the other two pairs follow the same pattern — the co-function identity from trigonometry translates directly into a complementary-sum identity for the inverses, as long as the principal-value ranges line up.

Negative-argument identities

What happens when you feed a negative number into an inverse trig function? The answer depends on which function you are using, and it depends on the symmetry (odd or even) of the original function.

Why the difference? Sine is an odd function (\sin(-\theta) = -\sin\theta), so its inverse inherits oddness. Cosine is an even function (\cos(-\theta) = \cos\theta), but \cos^{-1} is not even — it compensates through the shift \pi - \cos^{-1} x instead. The geometric reason: the principal branch of \cos^{-1} lives in [0, \pi], which is not symmetric about the origin. When you negate the input, the output reflects about \pi/2, the midpoint of that interval.

Reciprocal-argument identities

For positive x:

Proof. Let \tan^{-1} x = \alpha, so \tan\alpha = x with \alpha \in (0, \pi/2) since x > 0. Then \tan(\pi/2 - \alpha) = \cot\alpha = 1/x. Since \pi/2 - \alpha \in (0, \pi/2), this is the principal value. So \tan^{-1}(1/x) = \pi/2 - \alpha, and the sum is \pi/2.

For negative x, the sum is -\pi/2 instead. Combining both cases:

Similarly, \sin^{-1}(1/x) = \csc^{-1} x and \cos^{-1}(1/x) = \sec^{-1} x for |x| \geq 1.

Sum and difference formulas

These are the workhorses. They let you combine two inverse trig values into one.

The \tan^{-1} addition formula

Full derivation. Let \tan^{-1} a = \alpha and \tan^{-1} b = \beta. Then \tan\alpha = a and \tan\beta = b. Apply the tangent addition formula from compound angles:

So far, so clean. The temptation is to write \alpha + \beta = \tan^{-1}\!\frac{a+b}{1-ab} and call it done. But there is a subtlety: \tan^{-1} returns values in (-\pi/2, \pi/2), and the sum \alpha + \beta might fall outside that interval.

Since \alpha \in (-\pi/2, \pi/2) and \beta \in (-\pi/2, \pi/2), the sum \alpha + \beta lies in (-\pi, \pi). Three cases arise:

• If ab < 1: the sum \alpha + \beta lands in (-\pi/2, \pi/2), so \tan^{-1} returns it directly. No correction needed.
• If ab > 1 with a, b > 0: both angles are positive, and their sum overshoots \pi/2. The raw \tan^{-1} gives a value in (-\pi/2, 0) — the wrong quadrant. Add \pi to correct.
• If ab > 1 with a, b < 0: both angles are negative, and their sum undershoots -\pi/2. Subtract \pi to correct.

The case ab = 1 makes the denominator zero: \frac{a+b}{1-ab} is undefined, and the sum equals \pm\pi/2 directly.

The difference formula follows immediately by replacing b with -b:

The \sin^{-1} addition formula

valid when a^2 + b^2 \leq 1 (or when a, b \geq 0 and a^2 + b^2 \leq 1). Outside this range, a correction of \pi or -\pi applies, similar to the \tan^{-1} case.

Derivation. Let \sin^{-1} a = \alpha and \sin^{-1} b = \beta. Then \sin\alpha = a, \cos\alpha = \sqrt{1 - a^2} (non-negative because \alpha \in [-\pi/2, \pi/2]), and similarly \cos\beta = \sqrt{1 - b^2}. Now use the sine addition formula:

When \alpha + \beta stays in [-\pi/2, \pi/2], you can apply \sin^{-1} to both sides directly.

The \cos^{-1} addition formula

Derivation. Let \cos^{-1} a = \alpha, \cos^{-1} b = \beta. Then \cos\alpha = a, \sin\alpha = \sqrt{1 - a^2} (non-negative because \alpha \in [0, \pi]). Apply the cosine addition formula:

When \alpha + \beta \in [0, \pi], apply \cos^{-1} to both sides. The condition a + b \geq 0 ensures this. When a + b < 0, the sum exceeds \pi, and you write \cos^{-1} a + \cos^{-1} b = 2\pi - \cos^{-1}(ab - \sqrt{1-a^2}\sqrt{1-b^2}).

Multiple angle formulas

These express 2\tan^{-1} x, 3\tan^{-1} x, and similar multiples as a single inverse trig value.

Double-angle formula for \tan^{-1}

Setting a = b = x in the addition formula:

This is the inverse-trig version of \tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}.

There are two alternative forms using \sin^{-1} and \cos^{-1}:

Proof of the \sin^{-1} form. Let \tan^{-1} x = \theta, so \tan\theta = x. Then:

Since |x| \leq 1 implies |\theta| \leq \pi/4, the angle 2\theta lies in [-\pi/2, \pi/2], the principal range of \sin^{-1}. So 2\theta = \sin^{-1}\!\frac{2x}{1+x^2}.

Proof of the \cos^{-1} form. Similarly:

For x \geq 0, we have \theta \in [0, \pi/2), so 2\theta \in [0, \pi), which sits inside the principal range of \cos^{-1}. Hence 2\theta = \cos^{-1}\!\frac{1 - x^2}{1 + x^2}.

Triple-angle formula

Applying the addition formula twice:

This mirrors the identity \tan 3\theta = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}.

Simplification techniques

The real test of these formulas is using them to simplify complicated expressions. There are a few recurring patterns.

Pattern 1: Recognise a substitution

When you see \tan^{-1}\!\frac{2x}{1 - x^2}, recognise it as 2\tan^{-1} x (for |x| < 1).

When you see \sin^{-1}(2x\sqrt{1 - x^2}), try x = \sin\theta. Then 2x\sqrt{1-x^2} = 2\sin\theta\cos\theta = \sin 2\theta, so the expression becomes \sin^{-1}(\sin 2\theta) = 2\theta = 2\sin^{-1} x (when |x| \leq 1/\sqrt{2}).

Pattern 2: Convert everything to \tan^{-1}

Many problems become easier if you convert \sin^{-1} and \cos^{-1} into \tan^{-1} using right-triangle reasoning.

If \sin^{-1} x = \theta, then \sin\theta = x and \cos\theta = \sqrt{1-x^2}, so \tan\theta = \frac{x}{\sqrt{1-x^2}}, giving:

Similarly:

Pattern 3: Telescoping sums

A powerful technique for sums like \sum_{r=1}^{n} \tan^{-1}\!\frac{1}{1 + r + r^2}.

The key observation: \frac{1}{1 + r + r^2} = \frac{(r+1) - r}{1 + r(r+1)}. This has the form \frac{a - b}{1 + ab} with a = r+1 and b = r. By the difference formula:

So each term is a difference of two consecutive \tan^{-1} values. The sum telescopes:

Worked examples

Common confusions

• Forgetting the correction term when ab > 1. The formula \tan^{-1} a + \tan^{-1} b = \tan^{-1}\!\frac{a+b}{1-ab} without the \pm\pi correction is only valid when ab < 1. When ab > 1, the uncorrected formula gives an answer in the wrong quadrant. This is the single most common mistake in inverse-trig problems.

• Confusing \sin(\sin^{-1} x) = x with \sin^{-1}(\sin x) = x. The first is true for all x \in [-1, 1], always. The second is true only when x is already in [-\pi/2, \pi/2]. Outside that range, \sin^{-1}(\sin x) snaps x back to the principal branch, which changes the value.

• Assuming \cos^{-1} x is an even function. Since \cos is even, you might expect \cos^{-1}(-x) = \cos^{-1} x. But it is \pi - \cos^{-1} x, not \cos^{-1} x. The principal branch of \cos^{-1} is [0, \pi], which is not symmetric about 0.

• Dropping domain restrictions when simplifying. The formula 2\tan^{-1} x = \sin^{-1}\!\frac{2x}{1+x^2} holds for |x| \leq 1. For |x| > 1, the right side no longer simplifies to 2\tan^{-1} x — it wraps around. Always check whether your input falls in the validity range.

Going deeper

Why the correction term exists: a topological view

The tangent function has period \pi. So \tan^{-1}, which inverts it, can only return values in a single period — the interval (-\pi/2, \pi/2). When the true sum \alpha + \beta leaves that interval, the raw \tan^{-1} wraps it back in, landing in the wrong period. The correction \pm\pi is a "winding number" — it counts how many times you have gone around the period.

This is the same phenomenon that makes complex logarithms multi-valued, and it is the same \pm\pi that appears in the argument of a complex number. When you study complex numbers and the polar form, the correction term will reappear as the branch-cut adjustment for \arg(z_1 z_2).

Madhava's arctangent series and a formula for \pi

The infinite series

was discovered by Madhava of Sangamagrama in the 14th century, centuries before it appeared in European mathematics. Setting x = 1:

This series converges very slowly. But using the addition formula, you can accelerate it. Observe that \frac{\pi}{4} = \tan^{-1}\!\frac{1}{2} + \tan^{-1}\!\frac{1}{3}, and both \tan^{-1} series converge much faster for arguments less than 1. A further refinement is the identity \frac{\pi}{4} = 4\tan^{-1}\!\frac{1}{5} - \tan^{-1}\!\frac{1}{239}, which converges rapidly because 1/5 and 1/239 are small. The addition formulas you learned above are the engine behind every fast \pi computation that uses arctangent series.

Interplay with trigonometric equations

These identities are the bridge between trigonometric identities (which manipulate angles) and trigonometric equations (which solve for angles). An equation like \tan^{-1}(x+1) + \tan^{-1}(x-1) = \tan^{-1}\!\frac{8}{31} becomes solvable by applying the addition formula to the left side and comparing.

Where this leads next

• Inverse Trigonometric Functions — the definitions, domains, principal values, and graphs that this article builds on.
• Trigonometric Identities — the forward identities (\sin, \cos, \tan) whose inverses appear throughout this article.
• Compound Angles — the addition formulas for \sin(\alpha+\beta), \cos(\alpha+\beta), \tan(\alpha+\beta) that are the foundation for everything here.
• Trigonometric Equations — using these identities to solve equations involving inverse trig functions.
• Complex Numbers — Introduction — where the branch-cut corrections from the \tan^{-1} addition formula reappear as the argument of a product of complex numbers.