Key takeaway: Inverse trigonometric functions such as arcsin, arccos and arctan return the angle whose sine, cosine or tangent equals a given number, within a restricted range.
Grade 12 – PrecalculusF-TF.B.6Trigonometry
Definition of Inverse Trigonometric Functions
The inverse trigonometric functions run the trig machine backward. Give the sine of an angle and arcsine returns the angle. Because sine, cosine and tangent repeat, many angles share the same output, so each inverse is defined only on a restricted piece of the original function where it is one-to-one.
That restriction fixes a single answer, called the principal value. Arcsine and arctangent produce angles from the right half of the circle, between negative and positive quarter turns. Arccosine produces angles from the upper half, between 0 and a half turn. The name arcsin comes from “the arc whose sine is”.
Formula and notation
The inverse functions and their output ranges:
\[\sin^{-1}x\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}],\quad \cos^{-1}x\in[0,\pi],\quad \tan^{-1}x\in(-\tfrac{\pi}{2},\tfrac{\pi}{2}).\]
The exponent \(-1\) marks the inverse, not a reciprocal. Inputs for arcsin and arccos must lie in \([-1,1]\).
Key facts
- Of the inverse trigonometric functions, arcsin and arccos accept only -1 to 1; arctan accepts every real number.
- sin⁻¹ x means arcsin x, not 1/sin x, which is the cosecant.
- Calculators return the principal value, and other solutions are found with symmetry.
- Their graphs are reflections of the restricted trig graphs over the line y = x.
Inverse Trigonometric Functions example
A 12-foot ladder leans on a wall and reaches a point 10.5 feet up. What angle does the ladder make with the ground?
-
The height is the side opposite the angle and the ladder is the hypotenuse, so \(\sin\theta=\frac{10.5}{12}=0.875\).
-
Undo the sine: \(\theta=\sin^{-1}(0.875)\).
-
A calculator in degree mode gives \(\theta\approx 61.04^{\circ}\).
The ladder makes an angle of about 61.0° with the ground.
Common mistake
Watch out
Nadia evaluates \(\sin^{-1}(\sin\frac{5\pi}{6})\) and answers \(\frac{5\pi}{6}\), assuming the functions cancel.
Why it is wrong
The output of arcsine must lie between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\), and \(\frac{5\pi}{6}\) is outside that range. Since \(\sin\frac{5\pi}{6}=\frac{1}{2}\), the answer is the principal angle with sine one half: \(\frac{\pi}{6}\).
Related terms and lessons
Browse the glossary: A–Z list · Precalculus vocabulary · Trigonometry