Inverse Trigonometric Functions — Flashcards

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Step 1: Use the standard principal value. Step 2: sin(pi/6) = 1/2. Step 3: pi/6 lies in [-pi/2, pi/2]. Answer: sin^-1(1/2) = pi/6.

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Step 1: Use the odd property: sin^-1(-x) = -sin^-1(x). Step 2: sin^-1(sqrt(3)/2) = pi/3. Step 3: Apply the negative sign. Answer: -pi/3.

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Step 1: Use cos^-1(-x) = pi - cos^-1(x). Step 2: cos^-1(1/2) = pi/3. Step 3: Compute pi - pi/3 = 2pi/3. Answer: 2pi/3.

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Step 1: Use the odd property: tan^-1(-x) = -tan^-1(x). Step 2: tan^-1(1) = pi/4. Step 3: Apply the sign. Answer: -pi/4.

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Step 1: Use cot^-1(-x) = pi - cot^-1(x). Step 2: cot^-1(sqrt(3)) = pi/6. Step 3: Compute pi - pi/6 = 5pi/6. Answer: 5pi/6.

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Step 1: Find an angle whose secant is 2. Step 2: sec(pi/3) = 2. Step 3: pi/3 is in [0, pi] - {pi/2}. Answer: sec^-1(2) = pi/3.

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Step 1: Find an angle whose cosecant is 2. Step 2: cosec(pi/6) = 2. Step 3: pi/6 is in [-pi/2, pi/2] - {0}. Answer: cosec^-1(2) = pi/6.

Answer

Use it only when theta lies in [-pi/2, pi/2]. Example: sin^-1(sin(7pi/6)) is not 7pi/6. Since 7pi/6 = pi + pi/6, sin(7pi/6) = -sin(pi/6), so sin^-1(sin(7pi/6)) = -pi/6. The identity works only on the …