y=arctan⁡xy = \arctan x

Inverse Tangent (Arctangent) y=arctan⁡xy = \arctan x

arctan⁡x\arctan x, the inverse tangent or arctangent, is the inverse of the tangent function tan⁡\tan1. Since tan⁡\tan has period π\pi, it is restricted to the interval (−π2,π2)\left( -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right), on which it increases, and the inverse of that restriction, the principal value, is arctan⁡\arctan.

Definition

y=arctan⁡xy = \arctan x is the value yy satisfying both of the following.

  • tan⁡y=x\tan y = x
  • −π2<y<π2-\dfrac{\pi}{2} < y < \dfrac{\pi}{2}

Domain and range

Because tan⁡\tan takes every real value on that interval, the domain of arctan⁡\arctan is all real numbers. The range is the open interval (−π2,π2)\left( -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right), and the endpoint values are never reached.

Symmetry and monotonicity

Since arctan⁡(−x)=−arctan⁡x\arctan(-x) = -\arctan x it is an odd function, and its graph is symmetric about the origin. The derivative is as follows.

ddxarctan⁡x=11+x2\frac{d}{dx}\arctan x = \frac{1}{1 + x^2}

It is always positive, so the function increases monotonically. The denominator never vanishes, so the function is smooth on the whole real line, with none of the vertical tangents that the inverse sine and cosine have at their endpoints.

Asymptotes

xxarctan⁡x\arctan x
→−∞\to -\infty→−π2\to -\dfrac{\pi}{2}
−3-\sqrt{3}−π3-\dfrac{\pi}{3}
−1-1−π4-\dfrac{\pi}{4}
0000
11π4\dfrac{\pi}{4}
3\sqrt{3}π3\dfrac{\pi}{3}
→+∞\to +\infty→π2\to \dfrac{\pi}{2}

The horizontal asymptotes y=π2y = \dfrac{\pi}{2} and y=−π2y = -\dfrac{\pi}{2} give a gentle S-shaped curve, and the values never cross those two lines.

Inflection point

The second derivative is y′′=−2x(1+x2)2y'' = -\dfrac{2x}{(1 + x^2)^2}, whose sign changes only at x=0x = 0. The origin is therefore the single inflection point, and it is also the center of symmetry. Near it arctan⁡x≈x\arctan x \approx x.

Series and the number π\pi

The expansion about the origin converges on the interval [−1,1][-1, 1].

arctan⁡x=x−x33+x55−⋯\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots

Substituting x=1x = 1 and using arctan⁡1=π4\arctan 1 = \dfrac{\pi}{4} gives the Leibniz series2.

π4=1−13+15−17+⋯\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots

Its convergence is very slow, so practical computations of π\pi use formulas built from smaller arguments such as arctan⁡15\arctan\dfrac{1}{5}.

Polar angle and atan2

To find the polar angle of a point (x,y)(x, y), using arctan⁡yx\arctan\dfrac{y}{x} on its own cannot tell the first quadrant from the third, or the second from the fourth. That ambiguity is why most programming languages provide a separate two-argument atan2, which returns the correct angle in every direction.

Applications

  • Recovering an angle from a slope
  • The integral ∫dx1+x2=arctan⁡x+C\int \frac{dx}{1 + x^2} = \arctan x + C
  • Phase calculations in signal processing and control engineering
  • Rotation and direction in computer graphics
  1. Inverse trigonometric functions, Wikipedia
  2. Leibniz formula for pi, Wikipedia