Arcsin — Cheat sheet

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Definition

y=arcsin⁡x  ⟺  sin⁡y=xy = \arcsin x \iff \sin y = x with y∈[−π2,π2]y \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]. D=[−1,1]D = [-1, 1], R=[−π2,π2]R = \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right].

Values worth knowing

xxarcsin⁡x\arcsin x
−1-1−π2-\dfrac{\pi}{2}
−12-\dfrac{1}{2}−π6-\dfrac{\pi}{6}
0000
12\dfrac{1}{2}π6\dfrac{\pi}{6}
22\dfrac{\sqrt{2}}{2}π4\dfrac{\pi}{4}
11π2\dfrac{\pi}{2}

Inverting

Trapped insideApply
sin⁡(…)\sin(\ldots)arcsin⁡\arcsin to both sides
arcsin⁡(…)\arcsin(\ldots)sin⁡\sin to both sides

Non-standard ratios

Let α\alpha be the inverse expression, draw the right triangle with that opposite side and hypotenuse, and use Pythagoras for the third side. Read any function of α\alpha off it.

The complementary sum

x2+y2=1⇒arcsin⁡x+arcsin⁡y=π2x^2 + y^2 = 1 \Rightarrow \arcsin x + \arcsin y = \dfrac{\pi}{2} - the two legs of a unit-hypotenuse triangle, so the angles are complementary.

Properties

PropertyNote
Oddarcsin⁡(−x)=−arcsin⁡x\arcsin(-x) = -\arcsin x
Increasingacross the whole domain
Closed range±π2\pm\dfrac{\pi}{2} are reached
Outside [−1,1][-1,1]undefined