Chapter 2 Inverse Trigonometric Functions

Notes

The answer is not 2π/3 because the principal range of sin⁻¹x is [-π/2, π/2]. Answer: π/3 Q12. Find cos⁻¹(cos 5π/3). Solution: cos 5π/3 = 1/2 Now: cos⁻¹(1/2) = π/3 Because the principal range of cos⁻¹x is [0, π]. Answer: π/3 Q13. Find tan⁻¹(tan 3π/4). Solution: tan 3π/4 = -1 Now: tan⁻¹(-1) = -π/4 Because the principal range of tan⁻¹x is (-π/2, π/2). Answer: -π/4 Q14. Find: 2sin⁻¹(1/2) + cos⁻¹(1/2) Solution: sin⁻¹(1/2) = π/6 cos⁻¹(1/2) = π/3 Therefore: 2(π/6) + π/3 = π/3 + π/3 = 2π/3 Answer: 2π/3 Q15. Prove that: cos⁻¹(-x) = π - cos⁻¹x Solution: Let: cos⁻¹x = θ Then: cos θ = x Now: cos(π - θ) = -cos θ = -x Therefore: cos⁻¹(-x) = π - θ Substituting θ = cos⁻¹x: cos⁻¹(-x) = π - cos⁻¹x Hence proved. Q16. Prove that: sin⁻¹(-x) = -sin⁻¹x Solution: Let: sin⁻¹x = θ Then: sin θ = x Now: sin(-θ) = -sin θ = -x Therefore: sin⁻¹(-x) = -θ Substituting θ = sin⁻¹x: sin⁻¹(-x) = -sin⁻¹x Hence proved. Q17. Write the formula of tan inverse addition. Solution: If xy < 1: tan⁻¹x + tan⁻¹y = tan⁻¹((x + y)/(1 - xy)) If x > 0, y > 0 and xy > 1: tan⁻¹x + tan⁻¹y = π + tan⁻¹((x + y)/(1 - xy)) Q18. Find: tan⁻¹(1/2) + tan⁻¹(1/3) Solution: Using formula: tan⁻¹x + tan⁻¹y = tan⁻¹((x + y)/(1 - xy)) Here: x = 1/2 y = 1/3 xy = 1/6 < 1 Now: (x + y)/(1 - xy) = (1/2 + 1/3)/(1 - 1/6) = (5/6)/(5/6) = 1 Therefore: tan⁻¹1 = π/4 Answer: π/4 Q19. Find: tan⁻¹2 + tan⁻¹3 Solution: Here: x = 2 y = 3 xy = 6 > 1 and both are positive. Formula: tan⁻¹x + tan⁻¹y = π + tan⁻¹((x + y)/(1 - xy)) Now: (x + y)/(1 - xy) = 5/(1 - 6) = 5/(-5) = -1 Therefore: tan⁻¹2 + tan⁻¹3 = π + tan⁻¹(-1)…