What this chapter is about
In earlier classes you learned trigonometric functions such as sin, cos and tan that take an angle and return a ratio. Now you meet the reverse idea: given a ratio, find the angle. These are the inverse trigonometric functions, written sin⁻¹, cos⁻¹, tan⁻¹ and so on. Because the original functions repeat their values, we must restrict their domains to make them one-to-one before an inverse can exist. Understanding these restrictions is the heart of the chapter.
Inverse trigonometric functions appear whenever you need to recover an angle from measurements—finding the angle of elevation from distances, calculating phase angles in alternating-current circuits, or determining angles in navigation. They also arise in integral calculus, where many antiderivatives are expressed in terms of tan⁻¹ or sin⁻¹.
After this chapter you should be able to state the principal-value branches of all six inverse trigonometric functions, evaluate them at common values, simplify expressions using their properties, and prove or apply the standard identities that relate them.
Key ideas
- A function must be one-to-one to possess an inverse; trigonometric functions are made one-to-one by restricting their domains to principal-value branches.
- The principal-value branch of sin⁻¹ has domain [−1, 1] and range [−π/2, π/2].
- The principal-value branch of cos⁻¹ has domain [−1, 1] and range [0, π].
- The principal-value branch of tan⁻¹ has domain (−∞, ∞) and range (−π/2, π/2).
- For cot⁻¹, cosec⁻¹ and sec⁻¹, the domains exclude values where the original functions are undefined, and ranges are chosen to make them one-to-one.
- Identities such as sin⁻¹ x + cos⁻¹ x = π/2 hold for every x in [−1, 1]; similar complementary relations exist for tan⁻¹ and cot⁻¹.
- Sum and difference formulas, for instance tan⁻¹ x + tan⁻¹ y = tan⁻¹[(x + y)/(1 − xy)] when xy < 1, help simplify composite expressions.
- Inverse trigonometric functions can be written in terms of each other, useful when converting an expression to a single function.
Formulas and facts to remember
- Function: sin⁻¹ x · Domain: [−1, 1] · Principal-value range: [−π/2, π/2]
- Function: cos⁻¹ x · Domain: [−1, 1] · Principal-value range: [0, π]
- Function: tan⁻¹ x · Domain: (−∞, ∞) · Principal-value range: (−π/2, π/2)
- Function: cot⁻¹ x · Domain: (−∞, ∞) · Principal-value range: (0, π)
- Function: sec⁻¹ x · Domain: (−∞, −1] ∪ [1, ∞) · Principal-value range: [0, π], excluding π/2
- Function: cosec⁻¹ x · Domain: (−∞, −1] ∪ [1, ∞) · Principal-value range: [−π/2, π/2], excluding 0
Complementary identities (for x in the respective domains):
- sin⁻¹ x + cos⁻¹ x = π/2
- tan⁻¹ x + cot⁻¹ x = π/2
- sec⁻¹ x + cosec⁻¹ x = π/2
Negative-argument relations:
- sin⁻¹(−x) = −sin⁻¹ x
- cos⁻¹(−x) = π − cos⁻¹ x
- tan⁻¹(−x) = −tan⁻¹ x
Sum formula for tan⁻¹ (when xy < 1):
tan⁻¹ x + tan⁻¹ y = tan⁻¹[(x + y)/(1 − xy)]
Double-angle form:
2 tan⁻¹ x = sin⁻¹[2x/(1 + x²)] = cos⁻¹[(1 − x²)/(1 + x²)] = tan⁻¹[2x/(1 − x²)], valid for |x| ≤ 1 in the first two.
Worked examples
Example 1. Find the principal value of sin⁻¹(−√3/2).
Step 1. We need θ in [−π/2, π/2] such that sin θ = −√3/2.
Step 2. We know sin(π/3) = √3/2. Because sine is an odd function, sin(−π/3) = −√3/2.
Step 3. Since −π/3 lies in [−π/2, π/2], the principal value is −π/3.
Example 2. Simplify tan⁻¹(1/2) + tan⁻¹(1/3).
Step 1. Use the sum formula: tan⁻¹ x + tan⁻¹ y = tan⁻¹[(x + y)/(1 − xy)] provided xy < 1.
Step 2. Here xy = (1/2)(1/3) = 1/6 < 1, so the formula applies.
Step 3. (x + y)/(1 − xy) = (1/2 + 1/3)/(1 − 1/6) = (5/6)/(5/6) = 1.
Step 4. tan⁻¹ 1 = π/4. Hence the sum equals π/4.
Example 3. Prove that cos⁻¹(4/5) + cos⁻¹(12/13) = cos⁻¹(33/65).
Step 1. Let α = cos⁻¹(4/5) and β = cos⁻¹(12/13). So cos α = 4/5 and cos β = 12/13.
Step 2. Find the sines. Since α and β lie in [0, π] and their cosines are positive, both angles are in the first quadrant. sin α = 3/5, sin β = 5/13 (using sin²θ + cos²θ = 1).
Step 3. Use cos(α + β) = cos α cos β − sin α sin β = (4/5)(12/13) − (3/5)(5/13) = 48/65 − 15/65 = 33/65.
Step 4. Because α + β lies in [0, π], we have α + β = cos⁻¹(33/65), which completes the proof.
Common mistakes
- Ignoring the restricted range and giving multiple angles → always pick the angle inside the principal-value branch.
- Using tan⁻¹ x + tan⁻¹ y formula when xy ≥ 1 without adjusting by ±π → check the condition xy < 1 first; add or subtract π when needed.
- Confusing sin⁻¹ x with 1/sin x → sin⁻¹ x is the inverse function (arc sine), not the reciprocal; write 1/sin x as cosec x.
- Applying sin⁻¹(−x) = π − sin⁻¹ x, which holds for cos⁻¹, to sine → for sine and tangent, use the odd-function rule: sin⁻¹(−x) = −sin⁻¹ x.
- Forgetting domain restrictions when simplifying expressions like sin(sin⁻¹ x) → such identities hold only when x is in the correct domain.
Quick revision
- sin⁻¹ x lives in [−π/2, π/2]; cos⁻¹ x lives in [0, π]; tan⁻¹ x lives in (−π/2, π/2).
- sin⁻¹ x + cos⁻¹ x = π/2 for all x in [−1, 1].
- tan⁻¹ a + tan⁻¹ b = tan⁻¹[(a + b)/(1 − ab)] when ab < 1.
- Odd-function rule: sin⁻¹(−x) = −sin⁻¹ x; tan⁻¹(−x) = −tan⁻¹ x.
- Even-style rule for cosine: cos⁻¹(−x) = π − cos⁻¹ x.
- Always verify the answer lies in the principal-value range.