Inverse Trigonometric Functions — Practice

Question 1

Find the principal values of: (i) sin⁻¹(−√3/2)  (ii) cos⁻¹(cos(7π/6))  (iii) tan⁻¹(−√3).

Accepted Answer

(i) sin⁻¹(−√3/2): Since sin(−π/3) = −√3/2 and −π/3 ∈ [−π/2, π/2], the principal value is −π/3. (ii) cos⁻¹(cos(7π/6)): 7π/6 ∉ [0, π] (the principal range). cos(7π/6) = cos(2π − 7π/6) = cos(5π/6). Now 5π/6 ∈ [0, π], so cos⁻¹(cos(7π/6)) = cos⁻¹(cos(5π/6)) = 5π/6. (iii) tan⁻¹(−√3): Since tan(−π/3) = −√3 and −π/3 ∈ (−π/2, π/2), the principal value is −π/3.

Question 2

Prove that tan⁻¹(1/2) + tan⁻¹(2/11) = tan⁻¹(3/4).

Accepted Answer

LHS = tan⁻¹(1/2) + tan⁻¹(2/11). Check: xy = (1/2)(2/11) = 1/11 < 1. So we can use the addition formula: tan⁻¹x + tan⁻¹y = tan⁻¹((x+y)/(1−xy)). tan⁻¹(1/2) + tan⁻¹(2/11) = tan⁻¹((1/2 + 2/11)/(1 − 1/11)) = tan⁻¹((11/22 + 4/22)/(10/11)) = tan⁻¹((15/22)/(10/11)) = tan⁻¹(15/22 × 11/10) = tan⁻¹(165/220) = tan⁻¹(3/4) = RHS. ✓

Question 3

Prove that: sin⁻¹(12/13) + cos⁻¹(4/5) + tan⁻¹(63/16) = π.

Accepted Answer

Let α = sin⁻¹(12/13), β = cos⁻¹(4/5). From α = sin⁻¹(12/13): sin α = 12/13, so cos α = 5/13, tan α = 12/5. From β = cos⁻¹(4/5): cos β = 4/5, so sin β = 3/5, tan β = 3/4. Now find α + β using tan: tan(α+β) = (tan α + tan β)/(1 − tan α tan β) = (12/5 + 3/4)/(1 − (12/5)(3/4)) = (48/20 + 15/20)/(1 − 36/20) = (63/20)/((20−36)/20) = (63/20)/(−16/20) = −63/16. Since tan α = 12/5 > 0 and α ∈ [0, π/2], and tan β = 3/4 > 0 and β ∈ [0, π/2], we have α+β ∈ (0, π). Since tan(α+β) = −63/16 < 0, and α+β ∈ (0, π), we have α+β ∈ (π/2, π). Therefore α+β = π − tan⁻¹(63/16) (since tan(π−θ) = −tan θ). So: α + β + tan⁻¹(63/16) = π. That is: sin⁻¹(12/13) + cos⁻¹(4/5) + tan⁻¹(63/16) = π. ✓

Practice — Inverse Trigonometric Functions

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