Arctan 0 — Value in Degrees and Radians

Arctan 0 in Degrees and Radians

The value of arctan 0 is $0°$, which is also $0$ radians. It is the unique angle inside the inverse-tangent range $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ whose tangent equals $0$.

Quick Answer:

• Result: arctan 0 = 0° = 0 radians

• Notation: arctan 0 = tan⁻¹ 0 (the inverse function, NOT 1/tan)

• Method shown: read the angle whose tangent is 0 from the unit circle / special-angle table

• Exact form: 0 (the same value in both degrees and radians)

• Approximate value: 0 (exact, no rounding)

Tangent is $\dfrac{\sin}{\cos}$, so it is zero exactly when sine is zero and cosine is not. The angle in the inverse-tangent range where that happens is $0$ itself — the positive $x$-axis direction on the unit circle.

Tan-Inverse Reference Table

These are the tan-inverse values readers look up around $\arctan 0$. Each is shown in radians and degrees, with the tangent value that produces it.

Where Arctan 0 Appears

The angle $0$ marks a flat, level direction — zero slope. Is arctan 0 equal to 0 or pi? It is $0$, and that distinction is what makes inverse tangent the standard tool for recovering a heading from a slope. In computer graphics and navigation, atan2(y, x) returns the angle of a vector; when $y = 0$ and $x > 0$, that angle is $\arctan 0 = 0$ — pointing due east. In calculus, $\arctan x$ is the antiderivative of $\dfrac{1}{1 + x^2}$, and evaluating it at $0$ gives the lower limit of the integral that defines $\pi/4$. A road with zero rise over run has an incline of $\arctan 0 = 0°$ — perfectly level.

What Arctan Means

Arctangent answers "which angle has this tangent?" Written $\arctan x$ or $\tan^{-1} x$, it is the inverse of the tangent function. Tangent repeats every $\pi$ and is many-to-one — $\tan 0 = 0$, but $\tan \pi = 0$ too — so to define a single-valued inverse, mathematicians restrict tangent to the open interval $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$, where it rises steadily from $-\infty$ to $+\infty$. The inverse of that restricted tangent is the arctangent, with outputs confined to that interval.

Feed it $0$ and it returns the centre of that range: $0$.

How to Find Arctan 0

Method 1: Read it from the special-angle table

The tangent table gives $\tan 0° = 0$. Arctangent reverses the lookup:

$$\arctan 0 = 0° = 0 \text{ rad}.$$

Although $\tan \pi = 0$ as well, only $0$ lies inside the arctangent range, so the answer is unique.

Final answer: $\arctan 0 = 0° = 0$ rad.

Method 2: Use the ratio on the unit circle

Tangent is the ratio $\dfrac{\sin\theta}{\cos\theta} = \dfrac{y}{x}$. Setting this to $0$ needs $y = 0$ with $x \neq 0$. Inside $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$, the only such point is $(1, 0)$, at angle $0$:

$$\arctan 0 = \frac{0}{1} \to \theta = 0.$$

Final answer: the same $0° = 0$ rad, confirmed by the unit-circle ratio.

Examples of Arctan 0

Example 1

State $\arctan 0$ in both degrees and radians.

From the special-angle table, $\tan 0 = 0$, so $\arctan 0 = 0° = 0$ rad. The value is identical in both units.

Final answer: $0° = 0$ rad.

Example 2

A student writes $\arctan 0 = \pi$ because $\tan \pi = 0$. Check it.

Wrong attempt. Reasoning that any angle with tangent $0$ qualifies, the student notes $\tan \pi = 0$ and reports $\arctan 0 = \pi$ (that is, $180°$).

The break. Arctangent returns only angles in $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$. The value $\pi$ sits well outside that interval, so it cannot be an arctangent output — even though its tangent is indeed $0$. Returning $\pi$ would break the function's own range, the way two angles can share a tangent but only one is the principal value.

Correct. The angle with tangent $0$ that lies inside the range is $0$:

$$\arctan 0 = 0.$$

Final answer: $\arctan 0 = 0$, not $\pi$.

Example 3

Evaluate $\arctan 0 + \arctan 1$.

$\arctan 0 = 0$ and $\arctan 1 = \dfrac{\pi}{4}$, so the sum is $\dfrac{\pi}{4} = 45°$.

Final answer: $\dfrac{\pi}{4} = 45°$.

Example 4

Find the limit $\displaystyle\lim_{x \to 0} \arctan x$.

Arctangent is continuous everywhere, so the limit equals the function value at $0$: $\arctan 0 = 0$.

Final answer: $0$.

Example 5

A vector points along $(5, 0)$. What angle does it make with the positive $x$-axis?

The angle is $\arctan\left(\dfrac{0}{5}\right) = \arctan 0 = 0°$. The vector lies flat along the axis — zero elevation.

Final answer: $0° = 0$ rad.

Where Tan-Inverse Trips Students Up

Mistake 1: Returning $\pi$ instead of $0$

Where it slips in: A reader notices $\tan \pi = 0$ and reports $\arctan 0 = \pi$.

Don't do this: Pick any angle whose tangent is $0$ without checking the range.

The correct way: Arctangent outputs only angles in $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$, so the principal value is $0$.

Mistake 2: Reading $\tan^{-1}$ as a reciprocal

Where it slips in: A reader treats $\tan^{-1} 0$ as $\dfrac{1}{\tan 0}$.

Don't do this: Read the $-1$ as an exponent — which here would force a division by $0$.

The correct way: $\tan^{-1}$ is the inverse function, $\arctan$. The reciprocal of tangent is $\cot x = \dfrac{1}{\tan x}$, governed by the reciprocal identities — and $\cot 0$ is undefined, a different statement entirely from $\arctan 0 = 0$.

Mistake 3: Confusing arctan 0 with the asymptote

Where it slips in: A reader mixes up the input $0$ (which gives angle $0$) with the large-input behaviour (which approaches $90°$).

Don't do this: Assume arctangent of a small input is near its maximum.

The correct way: $\arctan x \to \dfrac{\pi}{2}$ only as $x \to +\infty$; at $x = 0$ the output is exactly $0$, the centre of the range. The graph passes through the origin.

What to Remember About Arctan 0

• Arctan 0 equals $0°$, or $0$ radians — the angle whose tangent is exactly $0$.

• On the unit circle it is the direction $(1, 0)$ along the positive $x$-axis, where $\dfrac{y}{x} = 0$.

• Although $\tan \pi = 0$, arctan 0 is $0$ because the range is $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$.

• $\tan^{-1}$ is the inverse function, not the reciprocal $\dfrac{1}{\tan}$.

Practice These Three

1. Find $\arctan\left(\dfrac{1}{\sqrt{3}}\right)$ in degrees and radians.

2. Evaluate $\arctan 0 + \arctan(-1)$.

3. A vector points along $(0, 4)$. Why does $\arctan\left(\dfrac{4}{0}\right)$ fail, and what angle is the vector actually at?

If Problem 2 gives $-\dfrac{\pi}{4}$, the $\arctan 0$ term contributes nothing, leaving $\arctan(-1)$.