Have you ever obtained more than one possible angle while solving a trigonometric equation? This happens because trigonometric functions are periodic, meaning that different angles can produce the same value. Inverse trigonometric functions resolve this ambiguity by returning an angle from a specified principal range.
Physics Wallah’s Inverse Trigonometric Functions Notes help you revise the domain and range of all six inverse functions, understand their graphs and apply important identities to mathematical problems. Pay particular attention to principal values, sign changes and interval restrictions, as these determine whether an expression can be simplified directly or requires further evaluation.
An inverse exists only if the function is bijective:
One-to-one: different inputs give different outputs.
Onto: every element of the codomain is an output.
The full sine function is not one-to-one, but restricting it to x ∈ [−π/2, π/2] makes it one-to-one and onto [−1, 1]. The same idea applies to the other trigonometric functions: choose a suitable branch, and the domain and range are interchanged.
sin⁻¹x means the angle whose sine is x (inverse).
(sin x)⁻¹ = 1/sin x = cosec x (reciprocal).
sin⁻¹x: domain [−1, 1], range [−π/2, π/2]
cos⁻¹x: domain [−1, 1], range [0, π]
tan⁻¹x: domain R, range (−π/2, π/2)
cot⁻¹x: domain R, range (0, π)
sec⁻¹x: domain (−∞, −1] ∪ [1, ∞), range [0, π] except π/2
cosec⁻¹x: domain (−∞, −1] ∪ [1, ∞), range [−π/2, π/2] except 0
Examples: sin⁻¹(1/2) = π/6, but sin⁻¹(2) is undefined over R because 2 ∉ [−1, 1].
The excluded points exist because sec x = 1/cos x is undefined at cos x = 0 (x = π/2), and cosec x = 1/sin x is undefined at sin x = 0 (x = 0).
The graph of an inverse function is the reflection of the original (restricted) graph in the line y = x.
sin⁻¹x and tan⁻¹x are increasing and odd.
cos⁻¹x and cot⁻¹x are decreasing, and neither even nor odd.
All six inverse trigonometric functions are aperiodic.
Memory tip: Learn the ranges of sin⁻¹, cos⁻¹, tan⁻¹ and cot⁻¹ first. The ranges of sec⁻¹ and cosec⁻¹ follow from their restricted branches.
The result is x, provided x is in the domain of the inverse:
sin(sin⁻¹x) = x, for x ∈ [−1, 1]
cos(cos⁻¹x) = x, for x ∈ [−1, 1]
tan(tan⁻¹x) = x, for x ∈ R
cot(cot⁻¹x) = x, for x ∈ R
sec(sec⁻¹x) = x, for |x| ≥ 1
cosec(cosec⁻¹x) = x, for |x| ≥ 1
The result does not always equal x, because it must lie in the principal range:
sin⁻¹(sin x) lies in [−π/2, π/2], with period 2π.
cos⁻¹(cos x) lies in [0, π], with period 2π.
tan⁻¹(tan x) lies in (−π/2, π/2), with period π.
cot⁻¹(cot x) lies in (0, π), with period π.
Their graphs are repeating line segments.
Example 1: sin⁻¹(sin 3)
Since π/2 < 3 < π, use sin⁻¹(sin x) = π − x for π/2 < x ≤ π.
So sin⁻¹(sin 3) = π − 3.
Example 2: tan⁻¹(tan 5)
Since 3π/2 < 5 < 2π, subtract 2π to get an angle in the principal range.
So tan⁻¹(tan 5) = 5 − 2π.
Key point: Never simplify sin⁻¹(sin x) or tan⁻¹(tan x) to x without checking the principal range.
sin⁻¹(−x) = −sin⁻¹x
tan⁻¹(−x) = −tan⁻¹x
cosec⁻¹(−x) = −cosec⁻¹x
cos⁻¹(−x) = π − cos⁻¹x
cot⁻¹(−x) = π − cot⁻¹x
sec⁻¹(−x) = π − sec⁻¹x
Example: cos⁻¹(−1/2) = π − cos⁻¹(1/2) = π − π/3 = 2π/3
sin⁻¹x + cos⁻¹x = π/2, for −1 ≤ x ≤ 1
tan⁻¹x + cot⁻¹x = π/2, for x ∈ R
sec⁻¹x + cosec⁻¹x = π/2, for |x| ≥ 1
Both terms must have the same input, and it must satisfy the domain restrictions.
cosec⁻¹x = sin⁻¹(1/x), for |x| ≥ 1
sec⁻¹x = cos⁻¹(1/x), for |x| ≥ 1
cot⁻¹x = tan⁻¹(1/x), for x > 0
cot⁻¹x = π + tan⁻¹(1/x), for x < 0
The sign condition matters because cot⁻¹ has range (0, π), while tan⁻¹ has range (−π/2, π/2).
If xy < 1: tan⁻¹x + tan⁻¹y = tan⁻¹[(x + y)/(1 − xy)]
If xy > 1: tan⁻¹x + tan⁻¹y = π + tan⁻¹[(x + y)/(1 − xy)]
If xy = 1: tan⁻¹x + tan⁻¹y = π/2
tan⁻¹x − tan⁻¹y = tan⁻¹[(x − y)/(1 + xy)]
This holds when the difference lies in the principal range. Otherwise, add or subtract the appropriate multiple of π.
Exam tip: Check the signs of x and y, the value of xy, and the quadrant of the result before applying a formula.
Let θ = sin⁻¹x, so sin θ = x. Since θ is in the principal range, cos θ = √(1 − x²). Then:
sin⁻¹x = π/2 − cos⁻¹x
sin⁻¹x = tan⁻¹[x/√(1 − x²)], for 0 < x < 1
Steps:
Set the inverse expression equal to θ.
Write the corresponding trigonometric ratio.
Draw a right triangle with the known sides.
Find the missing side using the Pythagorean theorem.
Write the required ratio and convert it to the desired inverse function.
Check the principal range before finalising.
Consider Sₙ = Σ (k = 1 to n) tan⁻¹[1/(1 + k(k + 1))]
Since 1 + k(k + 1) = k² + k + 1, and
tan⁻¹(k + 1) − tan⁻¹k = tan⁻¹[1/(k² + k + 1)],
each term becomes tan⁻¹(k + 1) − tan⁻¹k.
Expanding:
Sₙ = (tan⁻¹2 − tan⁻¹1) + (tan⁻¹3 − tan⁻¹2) + … + (tan⁻¹(n + 1) − tan⁻¹n)
The middle terms cancel:
Sₙ = tan⁻¹(n + 1) − π/4 = tan⁻¹[n/(n + 2)]
Limit: As n → ∞, n/(n + 2) → 1, so Sₙ → tan⁻¹(1) = π/4.
Memory tip: Expand the first few terms, spot the cancellation pattern, and keep only the uncancelled terms.
Understanding principal ranges, composition properties and inverse tangent identities helps you simplify expressions accurately and solve more complex inverse trigonometric problems. Physics Wallah’s Inverse Trigonometric Functions Notes provide a revision base for practising these identities, interpreting graphs and applying the correct restrictions in mathematical problems.
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JEE Syllabus |
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JEE Main Formulas |
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JEE Advanced Formulas |
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JEE Advanced Sample Paper |
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