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Pre-Calculus Trigonometric functions

Inverse trig functions (arcsin, arccos, arctan)

20 practice questions 0 video lessons Theory + worked examples

Inverse Trigonometric Functions

Texas Precalculus (TEKS) • Standard P.2(H) • Trigonometric Functions

Inverse Trigonometric Functions is a topic in Trigonometric Functions in the Texas Essential Knowledge and Skills (Precalculus, §111.42). It is aligned to Standard P.2(H), which requires students to graph arcsine and arccosine with their domain limitations.

The inverse trigonometric functions \(\arcsin\), \(\arccos\), and \(\arctan\) undo the trigonometric functions on restricted domains, returning a single principal-value angle.

Texas Precalculus (TEKS) › Trigonometric Functions › Inverse Trigonometric Functions  —  Standard P.2(H)

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Theory

Trig functions repeat, so they are not one-to-one — to invert them we restrict the domain to a piece that is. The inverses return a single principal value:

  • \(\arcsin x\): range \(\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\).
  • \(\arccos x\): range \([0,\pi]\).
  • \(\arctan x\): range \(\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\).

Each answers “which angle (in the allowed range) has this sine / cosine / tangent?”

The range is the whole point. \(\arcsin\) and \(\arctan\) never leave QI/QIV; \(\arccos\) stays in QI/QII. Always pick the answer inside the range.
Restricted sine and its inverse arcsine Sine restricted to negative pi over 2 to pi over 2 is one-to-one; its reflection across y equals x is arcsine. x y sin x arcsin x
Sine restricted to \([-\dfrac{\pi}{2},\dfrac{\pi}{2}]\) reflects across \(y=x\) into \(\arcsin\).
Ranges of the inverse trig functions Arcsine outputs negative pi over 2 to pi over 2; arccosine 0 to pi; arctangent negative pi over 2 to pi over 2. arcsin[−π/2, π/2]arccos[0, π]arctan(−π/2, π/2)
The principal-value ranges of the three inverse functions.

The defining ranges:

\[\arcsin:[-\dfrac{\pi}{2},\dfrac{\pi}{2}],\quad \arccos:[0,\pi],\quad \arctan:(-\dfrac{\pi}{2},\dfrac{\pi}{2})\]
arcsine range negative pi over 2 to pi over 2; arccosine 0 to pi; arctangent open negative pi over 2 to pi over 2
For a composition like \(\sin(\arccos x)\), set the inner inverse as an angle, draw a right triangle, and read off the outer function.

How to evaluate an inverse trig expression

  1. Rephrase as “which angle has this value?”
  2. Restrict the answer to the function's range.
  3. Use reference angles and the correct quadrant for the range.
  4. For compositions, introduce an angle for the inner inverse and build a triangle.
Example 1 — Evaluate arcsine
Find \(\arcsin\dfrac{1}{2}\).
Solution

Ask: which angle in \(\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\) has sine \(\dfrac12\)?

\(\arcsin\dfrac{1}{2}\)\(=\)\(\dfrac{\pi}{6}\)
arcsine of one half is pi over 6
Example 2 — Arccosine of a negative
Find \(\arccos\!\left(-\dfrac{\sqrt2}{2}\right)\).
Solution

Arccosine outputs values in \([0,\pi]\); a negative input lands in Quadrant II.

\(\arccos\!\left(-\dfrac{\sqrt2}{2}\right)\)\(=\)\(\dfrac{3\pi}{4}\)
arccosine of negative root 2 over 2 is three pi over four
Example 3 — Evaluate arctangent
Find \(\arctan 1\).
Solution

Which angle in \(\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\) has tangent \(1\)?

\(\arctan 1\)\(=\)\(\dfrac{\pi}{4}\)
arctangent of 1 is pi over 4
Example 4 — A composition
Evaluate \(\sin\!\left(\arccos\dfrac{3}{5}\right)\).
Solution

Let \(\theta=\arccos\dfrac35\), so \(\cos\theta=\dfrac35\) with \(\theta\) in \([0,\pi]\). Build a right triangle: adjacent 3, hypotenuse 5, so opposite \(=4\).

\(\sin\theta\)\(=\)\(\dfrac{4}{5}\)
sine of arccosine of three fifths is four fifths

Common pitfalls

Respect the range. \(\arcsin\dfrac12=\dfrac{\pi}{6}\) only — not the infinitely many other angles with that sine.
\(\arccos\) of a negative is in QII (between \(\dfrac{\pi}{2}\) and \(\pi\)), never negative.
Inverse, not reciprocal. \(\sin^{-1}x=\arcsin x\), which is not \(\dfrac{1}{\sin x}=\csc x\).

Frequently asked questions

What is arcsin?

The inverse sine function: \(\arcsin x\) returns the angle in \(\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\) whose sine is \(x\).

Why do inverse trig functions have restricted ranges?

Because the trig functions repeat and are not one-to-one. Restricting the domain makes each invertible and gives one principal value.

Is sin to the minus 1 the same as cosecant?

No. \(\sin^{-1}x\) is arcsine, the inverse function; \(\csc x=\dfrac{1}{\sin x}\) is the reciprocal. They are different.

How do you evaluate something like sin(arccos x)?

Let the inner inverse be an angle, draw a right triangle with the given ratio, find the missing side, and read off the outer function.