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.
Input $x$ | $\arctan x$ (rad) | $\arctan x$ (deg) | Note |
|---|---|---|---|
$-1$ | $-\dfrac{\pi}{4}$ | $-45°$ | tangent $-1$ |
$-\dfrac{1}{\sqrt{3}}$ | $-\dfrac{\pi}{6}$ | $-30°$ | tangent $-1/\sqrt{3}$ |
$0$ | $0$ | $0°$ | tangent $0$ |
$\dfrac{1}{\sqrt{3}}$ | $\dfrac{\pi}{6}$ | $30°$ | tangent $1/\sqrt{3}$ |
$1$ | $\dfrac{\pi}{4}$ | $45°$ | tangent $1$ |
$\sqrt{3}$ | $\dfrac{\pi}{3}$ | $60°$ | tangent $\sqrt{3}$ |
$\to +\infty$ | $\to \dfrac{\pi}{2}$ | $\to 90°$ | horizontal asymptote (never reached) |
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
Find $\arctan\left(\dfrac{1}{\sqrt{3}}\right)$ in degrees and radians.
Evaluate $\arctan 0 + \arctan(-1)$.
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 $\arctan(-1)$.
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