Inverse Trigonometric Functions

Answer

sin⁻¹x = y means sin y = x Domain: [-1, 1] (x must be between -1 and 1 inclusive) Range: [-π/2, π/2] (principal value) This is also called arcsin x. The inverse sine function exists only when we res

Answer

Step 1: Let sin⁻¹(-1/2) = θ Step 2: This means sin θ = -1/2 Step 3: We need θ in the range [-π/2, π/2] Step 4: sin(-π/6) = -1/2 Step 5: Since -π/6 lies in [-π/2, π/2] Therefore, sin⁻¹(-1/2) = -π/6

Answer

cos⁻¹x = y means cos y = x Domain: [-1, 1] (x must be between -1 and 1 inclusive) Range: [0, π] (principal value) Note: The range is different from sin⁻¹x. For cos⁻¹x, we take values from 0 to π to

Answer

Step 1: Let cos⁻¹(-1/√2) = θ Step 2: This means cos θ = -1/√2 Step 3: We need θ in the range [0, π] Step 4: cos(3π/4) = -1/√2 Step 5: Since 3π/4 lies in [0, π] Therefore, cos⁻¹(-1/√2) = 3π/4

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tan⁻¹x = y means tan y = x Domain: R (all real numbers) Range: (-π/2, π/2) (open interval, excluding endpoints) Note: Unlike sin⁻¹ and cos⁻¹, tan⁻¹ has domain as all real numbers because tan can tak

Answer

sin⁻¹(-x) = -sin⁻¹x Proof: Step 1: Let sin⁻¹(-x) = θ Step 2: Then -x = sin θ Step 3: So x = -sin θ = sin(-θ) Step 4: Therefore, sin⁻¹x = -θ Step 5: Hence, θ = -sin⁻¹x This shows that sin⁻¹ is an odd

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sin⁻¹x + cos⁻¹x = π/2 Proof: Step 1: Let sin⁻¹x = α Step 2: Then x = sin α = cos(π/2 - α) Step 3: Therefore, cos⁻¹x = π/2 - α Step 4: Adding: sin⁻¹x + cos⁻¹x = α + (π/2 - α) = π/2 This holds for all

Answer

cos(sin⁻¹x) = √(1 - x²) Step-by-step solution: Step 1: Let sin⁻¹x = θ Step 2: Then sin θ = x Step 3: We need to find cos θ Step 4: Using sin²θ + cos²θ = 1 Step 5: cos²θ = 1 - sin²θ = 1 - x² Step 6: c