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Inverse trigonometric functions

arcsin, arccos, arctan (which angle has this ratio) and their derivatives.

sin is not one-to-one, so its inverse is defined on a restricted range: arcsin returns an angle in [−π/2, π/2], arccos in [0, π], arctan in (−π/2, π/2). Picture it: reflect the sine curve across y = x and keep one branch. Think it: their derivatives are algebraic (1/√(1 − x²), 1/(1 + x²)), which is why they appear as antiderivatives of ordinary fractions.

Kushanda muenzaniso · derivative of arctan(x)

Differentiate atan(x)

\frac{d}{dx}\left[\operatorname{atan}{\left(x \right)}\right]

Step by step

  1. \frac{d}{dx}\left[\operatorname{atan}{\left(x \right)}\right]

    Start from the derivative to compute.

  2. \frac{1}{x^{2} + 1}

    (arctan u)′ = 1/(1+u²).

Kuratidza mhinduro
f'(x) = \frac{1}{x^{2} + 1}

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Maitiro: Inverse trigonometric functions

  1. Start from the derivative to compute.
  2. (arctan u)′ = 1/(1+u²).

Mabvunzo vanhu vanobvunza

Why radians instead of degrees?

A radian is the angle whose arc equals the radius, so it is a pure ratio rather than an arbitrary 1/360 of a turn. With radians the derivative of sin x is exactly cos x; with degrees a factor of π/180 appears everywhere.

Why does sin x = 1/2 have infinitely many solutions?

Sine repeats every full turn, and within one turn it reaches 1/2 twice (at π/6 and 5π/6). Add any whole number of turns to either and it is still true.

How do I remember the exact values?

Two triangles: the 45-45-90 with sides 1, 1, √2 and the 30-60-90 with sides 1, √3, 2. Every value for 30°, 45° and 60° is a ratio of those sides; symmetry gives the rest of the circle.

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