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Pure · Trigonometry

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Inverse trigonometric functions

The functions arcsin, arccos and arctan — why the domains of sin, cos and tan must be restricted to invert them, their domains, ranges and graphs, and evaluating exact values and composite expressions.

Pure · Trigonometry

Inverse trigonometric functions

The functions arcsin, arccos and arctan — why the domains of sin, cos and tan must be restricted to invert them, their domains, ranges and graphs, and evaluating exact values and composite expressions.

Why it works

sin⁡\sin, cos⁡\cos and tan⁡\tan each take the same value over and over, so they have no inverse as they stand — given sin⁡θ=12\sin\theta = \tfrac12 there are infinitely many θ\theta. To get a genuine inverse you restrict the domain to one stretch where the function is one-to-one, then reflect in y=xy = x.
FunctionDomainRange (principal values)
arcsin⁡x\arcsin x−1≤x≤1-1 \le x \le 1−π2≤y≤π2-\dfrac{\pi}{2} \le y \le \dfrac{\pi}{2}
arccos⁡x\arccos x−1≤x≤1-1 \le x \le 10≤y≤π0 \le y \le \pi
arctan⁡x\arctan xall real xx−π2<y<π2-\dfrac{\pi}{2} < y < \dfrac{\pi}{2}
So arcsin⁡\arcsin and arccos⁡\arccos only accept inputs in [−1,1][-1, 1] (since sin⁡\sin and cos⁡\cos never leave that range), while arctan⁡\arctan accepts any real number (since tan⁡\tan reaches every value). Each returns the one angle in its range — the principal value — not every possible angle.

Two notation traps. sin⁡−1x\sin^{-1}x means arcsin⁡x\arcsin x (the inverse function), not 1sin⁡x\dfrac{1}{\sin x} (that's cosec⁡x\operatorname{cosec} x). And the range of arccos⁡\arccos is [0,π][0, \pi], so arccos⁡\arccos of a negative number is an obtuse angle (second quadrant), e.g. arccos⁡(−12)=2π3\arccos(-\tfrac12) = \tfrac{2\pi}{3} — not a fourth-quadrant angle.

Composite values. For something like cos⁡(arcsin⁡35)\cos(\arcsin\tfrac35), set θ=arcsin⁡35\theta = \arcsin\tfrac35, so sin⁡θ=35\sin\theta = \tfrac35 with θ\theta in [−π2,π2][-\tfrac{\pi}{2}, \tfrac{\pi}{2}] (where cos⁡θ≥0\cos\theta \ge 0). Then cos⁡θ=1−sin⁡2θ=1−925=45\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \tfrac{9}{25}} = \tfrac45.