| domain | range | |
| sin θ | (-∞,&infin) | [−1, 1] |
| cos θ | (-∞,&infin) | [−1, 1] |
| tan θ | (-∞,&infin), x ≠ π/2+nπ, n=0,±1,±2,.... | (−&infin, ∞) |
| csc θ | (-∞,&infin), x ≠ nπ, n=0,±1,±2,.... | (−∞, −1]∪[1,&infin) |
| sec θ | (-∞,&infin), x ≠ &pi/2+nπ, n=0,±1,±2,.... | (−∞, −1]∪[1,&infin) |
| cot θ | (-∞,&infin), x ≠ nπ, n=0,±1,±2,.... | (−&infin, ∞) |
Since the trigonometric functions are not one-to-one the domain must be restricted In order to define an inverse function.
| domain | range | domain | range | |||
| sin θ | [−&pi/2, π/2] | [−1, 1] | sin−1 θ | [−1, 1] | [−&pi/2, π/2] | |
| cos θ | [0, π] | [−1, 1] | cos−1 θ | [−1, 1] | [0, π] | |
| tan θ | (−&pi/2, π/2) | (−&infin, ∞) | tan−1 θ | (−&infin, ∞) | (−&pi/2, π/2) |
The range of the function becomes the domain of the inverse function, and the domain of the function is the range of the inverse function.
The inverse functions for csc θ, sec θ, and cot θ are a liitle trickier in order to preserve desired calculus properties and the relationship to the reciprocals of the other functions.
The result of evaluating, as an example, sin−1 x will be an angle in quadrant I (if positive) or quadrant IV (if negative).
Knowing the sign of cos−1 x permits the correct angle in the range [0, 2π] to be computed.
If θ = sin−1 x, then the correct angle is given by:
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