🎯 Key Points
- Principal value ranges to memorize exactly: sin⁻¹x∈[-π/2,π/2], cos⁻¹x∈[0,π], tan⁻¹x∈(-π/2,π/2) — cos⁻¹ is the ONLY one starting at 0, not symmetric about origin like the others
- Complementary identities (sum to π/2): sin⁻¹x+cos⁻¹x=π/2, tan⁻¹x+cot⁻¹x=π/2, sec⁻¹x+cosec⁻¹x=π/2 — pairs of "co-functions" always sum to π/2
- Odd-function identities: sin⁻¹,tan⁻¹,cosec⁻¹ are ODD (f(-x)=-f(x)); cos⁻¹,cot⁻¹,sec⁻¹ are NOT odd (f(-x)=π-f(x)) — mixing these up is the most common error in this chapter
- tan⁻¹x+tan⁻¹y formula needs a +π or -π correction term when xy>1 — blindly applying the basic formula without checking this condition gives a wrong-quadrant answer
The graph of sin⁻¹x is confined to a narrow domain [-1,1] (since sine itself only takes values in that range) and range [-π/2,π/2] (the principal value branch chosen to make sine one-one and invertible there).
Inverse Trigonometric Functions
Trigonometric functions like sin x and cos x are periodic, so they are not one-one over their entire domain and cannot be inverted directly. Restricting the domain to an interval where the function is one-one and onto (called a branch) makes inversion possible. Each such restricted interval is called the principal value branch.
Why Restrict the Domain
A function has an inverse only if it is a bijection (one-one and onto). Since sin x repeats every 2π and takes each value in [-1, 1] infinitely many times, sin⁻¹x is defined as the inverse of sin x restricted to [-π/2, π/2], where sin x is one-one and onto [-1, 1]. The same idea, restrict to a suitable interval, applies to all six trig functions.
Principal Value Branches (Domain and Range)
- sin⁻¹x: domain [-1, 1], range [-π/2, π/2]
- cos⁻¹x: domain [-1, 1], range [0, π]
- tan⁻¹x: domain R (all reals), range (-π/2, π/2)
- cot⁻¹x: domain R, range (0, π)
- sec⁻¹x: domain R - (-1, 1), range [0, π] - {π/2}
- cosec⁻¹x: domain R - (-1, 1), range [-π/2, π/2] - {0}
Graphs (Brief Description)
- y = sin⁻¹x is an increasing curve through the origin, bounded between x = -1 and x = 1, with y ranging from -π/2 to π/2; it is the mirror image of y = sin x (restricted) about the line y = x.
- y = cos⁻¹x is a decreasing curve from (-1, π) to (1, 0).
- y = tan⁻¹x is an increasing S-shaped curve defined for all real x, with horizontal asymptotes y = π/2 and y = -π/2.
Basic Identities (Reciprocal Style)
- sin⁻¹(1/x) = cosec⁻¹x, for x ≥ 1 or x ≤ -1
- cos⁻¹(1/x) = sec⁻¹x, for x ≥ 1 or x ≤ -1
- tan⁻¹(1/x) = cot⁻¹x, for x > 0
Odd Function Identities
- sin⁻¹(-x) = -sin⁻¹x
- tan⁻¹(-x) = -tan⁻¹x
- cosec⁻¹(-x) = -cosec⁻¹x
- cos⁻¹(-x) = π - cos⁻¹x (cos⁻¹ is not odd)
- cot⁻¹(-x) = π - cot⁻¹x
- sec⁻¹(-x) = π - sec⁻¹x
Complementary (Sum to a Constant) Identities
- sin⁻¹x + cos⁻¹x = π/2, for x in [-1, 1]
- tan⁻¹x + cot⁻¹x = π/2, for all real x
- sec⁻¹x + cosec⁻¹x = π/2, for |x| ≥ 1
Addition and Subtraction Formulas
- tan⁻¹x + tan⁻¹y = tan⁻¹[(x+y)/(1-xy)], valid when xy < 1 (add π if xy > 1 and x, y > 0)
- tan⁻¹x - tan⁻¹y = tan⁻¹[(x-y)/(1+xy)], valid when xy > -1
- 2tan⁻¹x = tan⁻¹[2x/(1-x²)], for |x| < 1
- 2tan⁻¹x = sin⁻¹[2x/(1+x²)], for |x| ≤ 1
Quick Reference Values
- sin⁻¹(1/2) = π/6; sin⁻¹(1/√2) = π/4; sin⁻¹(1) = π/2
- cos⁻¹(1/2) = π/3; cos⁻¹(0) = π/2; cos⁻¹(1) = 0
- tan⁻¹(1) = π/4; tan⁻¹(√3) = π/3; tan⁻¹(0) = 0
Composition Properties (Inverse Applied with Itself)
Applying a trig function and its inverse cancels, but the order matters:
- Function of its inverse always cancels on the natural domain: sin(sin⁻¹x)=x for x∈[-1,1], cos(cos⁻¹x)=x for x∈[-1,1], tan(tan⁻¹x)=x for all real x.
- Inverse of the function returns x only when x lies in the principal value branch: sin⁻¹(sin x)=x only for x∈[-π/2,π/2]; cos⁻¹(cos x)=x only for x∈[0,π]; tan⁻¹(tan x)=x only for x∈(-π/2,π/2).
- Outside the principal branch you must reduce first, e.g. sin⁻¹(sin(3π/4)) = sin⁻¹(sin(π-3π/4)) = sin⁻¹(sin(π/4)) = π/4, NOT 3π/4.
Converting One Inverse Function into Another
For x∈[0,1] (a first-quadrant angle), one right triangle rewrites any inverse function in terms of another:
- sin⁻¹x = cos⁻¹√(1-x²) = tan⁻¹[x/√(1-x²)]
- cos⁻¹x = sin⁻¹√(1-x²) = tan⁻¹[√(1-x²)/x]
- tan⁻¹x = sin⁻¹[x/√(1+x²)] = cos⁻¹[1/√(1+x²)]
Idea: set θ=sin⁻¹x so sin θ=x=opposite/hypotenuse; the third side √(1-x²) then supplies every other ratio.
Sum and Difference Formulae for sin⁻¹ and cos⁻¹
- sin⁻¹x + sin⁻¹y = sin⁻¹[x√(1-y²) + y√(1-x²)], valid when x²+y²≤1
- sin⁻¹x - sin⁻¹y = sin⁻¹[x√(1-y²) - y√(1-x²)]
- cos⁻¹x + cos⁻¹y = cos⁻¹[xy - √(1-x²)√(1-y²)], valid when x+y≥0
- cos⁻¹x - cos⁻¹y = cos⁻¹[xy + √(1-x²)√(1-y²)]
Multiple-Angle Formulae
- 2sin⁻¹x = sin⁻¹[2x√(1-x²)], for |x|≤1/√2
- 2cos⁻¹x = cos⁻¹(2x²-1), for 0≤x≤1
- 2tan⁻¹x = tan⁻¹[2x/(1-x²)] = sin⁻¹[2x/(1+x²)] = cos⁻¹[(1-x²)/(1+x²)]
- 3sin⁻¹x = sin⁻¹(3x-4x³); 3cos⁻¹x = cos⁻¹(4x³-3x); 3tan⁻¹x = tan⁻¹[(3x-x³)/(1-3x²)]
🚀 JEE Advanced Edge
Substitution tricks for inverse-trig expressions with square roots: An expression like sin⁻¹(2x√(1-x²)) simplifies dramatically by substituting x=sinθ, turning it into sin⁻¹(2sinθcosθ)=sin⁻¹(sin2θ)=2θ=2sin⁻¹x (valid in the appropriate range) — recognizing the "double angle inside an inverse function" pattern converts an intractable-looking expression into a one-line simplification.
Why range restrictions matter when adding inverse-trig values: tan⁻¹x+tan⁻¹y=tan⁻¹[(x+y)/(1-xy)] is valid only when xy<1; if xy>1 and both x,y>0, the true sum exceeds π/2 (outside tan⁻¹'s range), so π must be ADDED to the formula's result to get the correct value — this range-awareness is exactly what separates correct JEE answers from formula-matching errors.
Worked problem: Evaluate tan⁻¹(1) + tan⁻¹(2) + tan⁻¹(3). Approach: First combine tan⁻¹(2)+tan⁻¹(3): since 2×3=6>1 and both positive, sum = π+tan⁻¹[(2+3)/(1-6)] = π+tan⁻¹(-1) = π-π/4 = 3π/4. Then add tan⁻¹(1)=π/4: total = 3π/4+π/4 = π.
Worked Example: Principal Value
Find the principal value of sin⁻¹(−√3/2).
The principal value of sin⁻¹ lies in [−π/2, π/2]. We need an angle θ in this range with sin θ = −√3/2. Since sin(π/3) = √3/2, the negative angle is θ = −π/3. Answer: −π/3. Principal values for sin⁻¹ and tan⁻¹ lie in [−π/2, π/2]; for cos⁻¹ in [0, π].
Worked Example: Identity Application
Simplify sin(2 sin⁻¹ x).
Let α = sin⁻¹ x, so sin α = x and cos α = √(1−x²) (since α ∈ [−π/2, π/2]). Then sin(2α) = 2 sin α cos α = 2x√(1−x²). This is a standard NEET question type: substitute the inverse trig angle, then use the double-angle or compound-angle formula.