Trig Functions

Inverse Trig Functions

Evaluate arcsin(x) and identify the principal value in the correct range: arcsin ∈ [-π/2, π/2].

Example

Use the principal-value range to evaluate an inverse sine exactly.

highlighted = computed this step

Step 1 — Set up

Set up the expression.

arcsin⁡(12)\arcsin( \frac{1}{2} )

Step 2 — Principal range

Use the arcsin range from -pi over 2 to pi over 2.

arcsin⁡ range [−π2,π2]\arcsin\text{ range } \hlmath{\left[-\frac{\pi}{2},\frac{\pi}{2}\right]}

Step 3 — Match sine value

Check that sine of pi over 6 equals 1 over 2.

sin⁡(π6)=12\sin( \hlmath{\frac{\pi}{6}} )= \frac{1}{2}

Step 4 — Result

The principal value is pi over 6.

arcsin⁡(12)=π6\arcsin( \frac{1}{2} )= \hlmath{\frac{\pi}{6}}
inverse-trig arcsin(x) returns the angle θ ∈ [-π/2, π/2] with sin(θ) = x. Principal range for arcsin: [-π/2, π/2]. Non-principal angles with the same sin value are excluded.