MISSION · Geometry
Inverse Trig Functions
Why the calculator picks one angle out of infinitely many
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The idea in one sentence
\(\sin\) sends infinitely many angles to the same value, so to have an inverse at all it must first be cut down to a stretch where it never repeats — and that cut is what decides the single answer your calculator gives.
The problem with inverting a wave
An inverse function has to send each output back to exactly one input. \(\sin\) cannot do that. \(\sin 30° = \sin 150° = \sin 390° = 0.5\), and there are infinitely many more. Asked to undo it, there is no single right answer.
So mathematics makes a choice. Restrict \(\sin\) to a stretch of angles where it climbs steadily through every possible value exactly once, and invert only that. The chosen stretch is
\[-90° \le \theta \le 90°\]On it, \(\sin\) runs from \(-1\) up to \(1\) without ever repeating. Its inverse on that stretch is \(\arcsin\), also written \(\sin^{-1}\), and the angle it returns is called the principal value.
The three ranges
Each function needs its own stretch, because each repeats differently.
| Function | Domain (what goes in) | Range (what comes out) |
|---|---|---|
| \(\arcsin x\) | \(-1 \le x \le 1\) | \(-90°\) to \(90°\) |
| \(\arccos x\) | \(-1 \le x \le 1\) | \(0°\) to \(180°\) |
| \(\arctan x\) | all real \(x\) | \(-90°\) to \(90°\), exclusive |
\(\arccos\) is the odd one out, and for a good reason. \(\cos\) is symmetric about \(0°\), so the stretch from \(-90°\) to \(90°\) would hit every value twice. Starting at \(0°\) and running to \(180°\) gives a stretch where \(\cos\) falls steadily from 1 to \(-1\), each value once.
\(\arctan\) accepts anything, because \(\tan\) produces every real number. The endpoints are excluded because \(\tan\) is undefined there — and that is why \(\arctan\) has horizontal asymptotes at \(\pm 90°\) instead of stopping.
The composition trap
\(\sin^{-1}(\sin 150°)\) is not \(150°\).
Work from the inside: \(\sin 150° = 0.5\), and \(\sin^{-1}(0.5) = 30°\), because \(30°\) is the answer inside the allowed range and \(150°\) is not. The function cannot return an angle it is not permitted to return.
The two orders behave differently, and it is worth stating precisely:
\(\sin(\sin^{-1}x) = x \quad \text{for every } x \text{ in } [-1, 1]\) \(\sin^{-1}(\sin\theta) = \theta \quad \text{only when } \theta \text{ is already in } [-90°, 90°]\)
Going out and back always works. Going back and out only works if you started inside the restricted stretch.
Getting the other solutions
Because the calculator returns one angle, solving a trig equation always needs a second step. The inverse hands you the principal value; symmetry and periodicity give the rest.
\(\sin\theta = 0.5\) for \(0° \le \theta \le 360°\): the calculator says \(30°\), and the sine curve’s symmetry about \(90°\) gives \(180° - 30° = 150°\) as well.
This is not the calculator being unhelpful. It is the price of \(\arcsin\) being a function at all.
Where they are actually used
- Finding an angle from two sides in a right triangle — the reverse of the trig ratios, and the reason elevation problems can be solved for the angle.
- Converting to polar coordinates, where \(\theta = \arctan\frac{y}{x}\) — with a correction, since \(\arctan\) only ever returns angles in two of the four quadrants.
- Integration. \(\int \frac{dx}{1 + x^2} = \arctan x + C\), which is why these functions appear in calculus at all.
Common mix-ups
- Reading \(\sin^{-1}x\) as \(\frac{1}{\sin x}\). The \(-1\) means inverse function, not reciprocal. The reciprocal is \(\csc x\).
- Expecting \(\sin^{-1}(\sin\theta) = \theta\) always. Only inside the range.
- Giving only the calculator’s answer when an equation asks for all solutions.
- Feeding in a value outside \([-1,1]\). \(\arcsin 2\) does not exist; no angle has a sine of 2.
- Degrees against radians. The ranges above are \([-\frac\pi2, \frac\pi2]\) and \([0, \pi]\) in radians. Check the mode.
Check yourself
Give \(\arcsin\frac12\) in degrees, the top of \(\arccos\)’s range, and \(\sin^{-1}(\sin 150°)\).
\(\arcsin\frac12 = 30°\). \(\arccos\) runs from \(0°\) to \(180°\). And \(\sin 150° = 0.5\), so the answer is \(30°\) — the principal value, not the \(150°\) you started with.
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