AP Precalculus Flashcards: Inverse Trigonometric Functions

Study Inverse Trigonometric Functions in AP Precalculus with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

AP Precalculus

Inverse Trigonometric Functions

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QUESTION
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What is the inverse of sin(θ)=x\text{sin}(\theta) = x?

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ANSWER

θ=arcsin(x)\theta = \text{arcsin}(x). Solving for θ\theta when sine equals xx.

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What this deck covers

This deck focuses on Inverse Trigonometric Functions, giving you a quick way to review the definitions, rules, and examples that matter most for AP Precalculus.

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Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

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Flashcard 1: What is the inverse of sin(θ)=x\text{sin}(\theta) = x?

Answer: θ=arcsin(x)\theta = \text{arcsin}(x). Solving for θ\theta when sine equals xx.

Flashcard 2: What is the range of the inverse sine function, arcsin(x)\text{arcsin}(x)?

Answer: The range is [π2,π2][-\frac{\text{π}}{2}, \frac{\text{π}}{2}]. Restricted to quadrants I and IV where sine is invertible.

Flashcard 3: Find arccot(√3)\text{arccot}(\text{√3}).

Answer: π6\frac{\text{π}}{6}. Since cot(π6)=3\cot(\frac{\pi}{6}) = \sqrt{3}.

Flashcard 4: What is arctan(√3)\text{arctan}(\text{√3})?

Answer: π3\frac{\text{π}}{3}. Since tan(π3)=3\tan(\frac{\pi}{3}) = \sqrt{3}.

Flashcard 5: Find the value of arctan(1)\text{arctan}(-1).

Answer: π4-\frac{\text{π}}{4}. Since tan(π4)=1\tan(-\frac{\pi}{4}) = -1.

Flashcard 6: Find arcsin(√22)\text{arcsin}(-\frac{\text{√2}}{2}).

Answer: π4-\frac{\text{π}}{4}. Since sin(π4)=22\sin(-\frac{\pi}{4}) = -\frac{\sqrt{2}}{2}.

Flashcard 7: What is the inverse of tan(θ)=x\text{tan}(\theta) = x?

Answer: θ=arctan(x)\theta = \text{arctan}(x). Solving for θ\theta when tangent equals xx.

Flashcard 8: What is the range of the inverse cosecant function, arccsc(x)\text{arccsc}(x)?

Answer: [π2,0) or (0,π2][-\frac{\text{π}}{2}, 0) \text{ or } (0, \frac{\text{π}}{2}]. Excludes 0 where cosecant is undefined.

Flashcard 9: Evaluate arccos(1)\text{arccos}(1).

Answer: 00. Since cos(0)=1\cos(0) = 1.

Flashcard 10: What is arccos(12)\text{arccos}(\frac{1}{2})?

Answer: π3\frac{\text{π}}{3}. Since cos(π3)=12\cos(\frac{\pi}{3}) = \frac{1}{2}.

Flashcard 11: Find arccsc(√2)\text{arccsc}(\text{√2}).

Answer: π4\frac{\text{π}}{4}. Since csc(π4)=2\csc(\frac{\pi}{4}) = \sqrt{2}.

Flashcard 12: Find arccos(√32)\text{arccos}(-\frac{\text{√3}}{2}).

Answer: 5π6\frac{5\text{π}}{6}. Since cos(5π6)=32\cos(\frac{5\pi}{6}) = -\frac{\sqrt{3}}{2} in quadrant II.

Flashcard 13: Evaluate arccot(1)\text{arccot}(-1).

Answer: 3π4\frac{3\text{π}}{4}. Since cot(3π4)=1\cot(\frac{3\pi}{4}) = -1.

Flashcard 14: What is the domain of the inverse cosine function, arccos(x)\text{arccos}(x)?

Answer: The domain is [1,1][-1, 1]. Cosine input values must be between -1 and 1.

Flashcard 15: Evaluate arcsec(2)\text{arcsec}(2).

Answer: π3\frac{\text{π}}{3}. Since sec(π3)=2\sec(\frac{\pi}{3}) = 2.

Flashcard 16: What is arcsec(1)\text{arcsec}(1)?

Answer: 00. Since sec(0)=1\sec(0) = 1.

Flashcard 17: Evaluate arcsec(2)\text{arcsec}(-2).

Answer: 2π3\frac{2\text{π}}{3}. Since sec(2π3)=2\sec(\frac{2\pi}{3}) = -2 in quadrant II.

Flashcard 18: What is arccot(0)\text{arccot}(0)?

Answer: π2\frac{\text{π}}{2}. Since cot(π2)=0\cot(\frac{\pi}{2}) = 0.

Flashcard 19: What is the value of arccos(0)\text{arccos}(0)?

Answer: π2\frac{\text{π}}{2}. Since cos(π2)=0\cos(\frac{\pi}{2}) = 0.

Flashcard 20: What is the domain of the inverse secant function, arcsec(x)\text{arcsec}(x)?

Answer: (,1] or [1,)(-\text{∞}, -1] \text{ or } [1, \text{∞}). Secant undefined where cosine equals zero.

Flashcard 21: Find the value of arcsin(0)\text{arcsin}(0).

Answer: 00. Since sin(0)=0\sin(0) = 0.

Flashcard 22: Evaluate arccsc(2)\text{arccsc}(-2).

Answer: π6-\frac{\text{π}}{6}. Since csc(π6)=2\csc(-\frac{\pi}{6}) = -2.

Flashcard 23: What is the range of the inverse tangent function, arctan(x)\text{arctan}(x)?

Answer: The range is (π2,π2)(-\frac{\text{π}}{2}, \frac{\text{π}}{2}). Asymptotic limits as x approaches ±±∞.

Flashcard 24: Evaluate arctan(1)\text{arctan}(1).

Answer: π4\frac{\text{π}}{4}. Since tan(π4)=1\tan(\frac{\pi}{4}) = 1.

Flashcard 25: Identify the principal value of arccos(1)\text{arccos}(-1).

Answer: π\text{π}. Since cos(π)=1\cos(\pi) = -1.

Flashcard 26: Find arcsin(√32)\text{arcsin}(\frac{\text{√3}}{2}).

Answer: π3\frac{\text{π}}{3}. Since sin(π3)=32\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}.

Flashcard 27: Evaluate arctan(0)\text{arctan}(0).

Answer: 00. Since tan(0)=0\tan(0) = 0.

Flashcard 28: What is the range of the inverse cotangent function, arccot(x)\text{arccot}(x)?

Answer: (0,π)(0, \text{π}). Full range from 0 to π\pi for all real inputs.

Flashcard 29: What is the domain of the inverse cotangent function, arccot(x)\text{arccot}(x)?

Answer: (,)(-\text{∞}, \text{∞}). Cotangent is defined for all real numbers.

Flashcard 30: Identify the principal value of arcsin(1)\text{arcsin}(-1).

Answer: π2-\frac{\text{π}}{2}. Since sin(π2)=1\sin(-\frac{\pi}{2}) = -1.

Flashcard 31: Find arccos(32)\text{arccos}(-\frac{\sqrt{3}}{2}).

Answer: 5π6\frac{5\pi}{6}. Since cos(5π6)=32\cos(\frac{5\pi}{6}) = -\frac{\sqrt{3}}{2} in quadrant II.

Flashcard 32: What is the principal value of arccsc(1)\text{arccsc}(-1)?

Answer: π2-\frac{\text{π}}{2}. Since csc(π2)=1\csc(-\frac{\pi}{2}) = -1.

Flashcard 33: What is the principal value of arcsec(1)\text{arcsec}(-1)?

Answer: π\text{π}. Since sec(π)=1\sec(\pi) = -1.

Flashcard 34: What is the principal value of arccsc(1)\text{arccsc}(1)?

Answer: π2\frac{\text{π}}{2}. Since csc(π2)=1\csc(\frac{\pi}{2}) = 1.

Flashcard 35: What is the inverse of cos(θ)=x\text{cos}(\theta) = x?

Answer: θ=arccos(x)\theta = \text{arccos}(x). Solving for θ\theta when cosine equals xx.

Flashcard 36: What is arcsin(12)\text{arcsin}(\frac{1}{2})?

Answer: π6\frac{\text{π}}{6}. Since sin(π6)=12\sin(\frac{\pi}{6}) = \frac{1}{2}.

Flashcard 37: Evaluate arccos(12)\text{arccos}(-\frac{1}{2}).

Answer: 2π3\frac{2\text{π}}{3}. Since cos(2π3)=12\cos(\frac{2\pi}{3}) = -\frac{1}{2} in quadrant II.

Flashcard 38: Find arcsin(12)\text{arcsin}(-\frac{1}{2}).

Answer: π6-\frac{\text{π}}{6}. Since sin(π6)=12\sin(-\frac{\pi}{6}) = -\frac{1}{2}.