Inverse Trigonometric Functions — Flashcards

Answer

Step 1: Let y = sin⁻¹(1/√2). Then sin(y) = 1/√2 Step 2: The principal value branch of sin⁻¹ has range [-π/2, π/2] Step 3: Find angle in [-π/2, π/2] where sin equals 1/√2 Step 4: sin(π/4) = 1/√2 Step 5…

Answer

Step 1: Let y = cos⁻¹(-1/2). Then cos(y) = -1/2 Step 2: Principal value branch of cos⁻¹ has range [0, π] Step 3: Find angle in [0, π] where cosine equals -1/2 Step 4: cos(2π/3) = -1/2 (since cos(π/3) …

Answer

Step 1: Let y = tan⁻¹(-√3). Then tan(y) = -√3 Step 2: Principal value branch of tan⁻¹ has range (-π/2, π/2) Step 3: tan(π/3) = √3, so tan(-π/3) = -√3 Step 4: -π/3 lies in (-π/2, π/2) ✓ Step 5: Therefo…

Answer

Step 1: Let y = cot⁻¹(-1/√3). Then cot(y) = -1/√3 Step 2: Principal value branch of cot⁻¹ has range (0, π) Step 3: cot(π/3) = 1/√3, so cot(π - π/3) = cot(2π/3) = -1/√3 Step 4: 2π/3 lies in (0, π) ✓ St…

Answer

Step 1: Let y = cosec⁻¹(2). Then cosec(y) = 2 Step 2: cosec(y) = 2 means sin(y) = 1/2 Step 3: Principal value branch of cosec⁻¹ has range [-π/2, π/2] - {0} Step 4: sin(π/6) = 1/2 Step 5: π/6 lies in […

Answer

Step 1: Let y = sec⁻¹(-2). Then sec(y) = -2 Step 2: sec(y) = -2 means cos(y) = -1/2 Step 3: Principal value branch of sec⁻¹ has range [0, π] - {π/2} Step 4: cos(2π/3) = -1/2 Step 5: 2π/3 lies in [0, π…

Answer

Step 1: Find tan⁻¹(1) tan(π/4) = 1, so tan⁻¹(1) = π/4 Step 2: Find cos⁻¹(-1/2) cos(2π/3) = -1/2, so cos⁻¹(-1/2) = 2π/3 Step 3: Find sin⁻¹(-1/2) sin(-π/6) = -1/2, so sin⁻¹(-1/2) = -π/6 Step 4:…

Answer

Step 1: Let x = sin(θ), then sin⁻¹(x) = θ Step 2: Substitute into the expression sin⁻¹(2sin(θ)√(1-sin²(θ))) Step 3: Simplify √(1-sin²(θ)) = |cos(θ)| = cos(θ) (for given range) = sin⁻¹(2sin(θ)cos…