Principal Value of Inverse Trigonometric Functions: Concept, Method & Solved Examples

When a Class 12 board question asks you to “find the principal value of sin⁻¹(1/2)”, it is not just asking for a number. It is testing whether you understand what the principal value branch is, how to work within it, and how to recognise when a raw answer falls outside that branch and needs adjustment. This guide explains the concept thoroughly and walks you through worked examples for all six inverse trigonometric functions.

What Is the Principal Value of an Inverse Trig Function?

Each inverse trigonometric function has multiple possible branches — intervals on which the original function is one-one, each yielding a different inverse. For sin⁻¹ x, for example, valid branches exist on [−π/2, π/2], [π/2, 3π/2], [3π/2, 5π/2], and infinitely many others.

The principal value branch is the one that is chosen as the standard reference — the one that covers the most natural and symmetric portion of the number line around zero. Each inverse trig function has exactly one principal value branch:

The principal value of an inverse trig expression is the unique output that lies within this principal value branch. According to the NCERT textbook: “The value of an inverse trigonometric function which lies in the range of its principal branch is called the principal value of that function.”

The Standard Method for Finding Principal Values

The method is consistent across all six functions and can be applied in three clear steps:

Step 1: Write the inverse trig expression as an equation. If the expression is sin⁻¹(k) = y, then this means sin y = k.

Step 2: Find the angle y whose sine (or cosine, tangent, etc.) equals k.

Step 3: Verify that y lies within the principal value branch range. If it does, you are done. If it does not, use the relevant identity to shift it into the correct range.

Worked Examples: All Six Functions

Example 1 — sin⁻¹(1/√2)

Let sin⁻¹(1/√2) = y, so sin y = 1/√2.

We know that sin(π/4) = 1/√2. Since π/4 ∈ [−π/2, π/2], this is already within the principal branch.

Principal value = π/4

Example 2 — sin⁻¹(−1/2)

Let sin⁻¹(−1/2) = y, so sin y = −1/2.

We know sin(π/6) = 1/2, so sin(−π/6) = −1/2. Since −π/6 ∈ [−π/2, π/2], it lies within the principal branch.

Principal value = −π/6

Example 3 — cos⁻¹(−1/2)

Let cos⁻¹(−1/2) = y, so cos y = −1/2.

We know cos(π/3) = 1/2. For a negative cosine, we use the identity cos(π − θ) = −cos θ: cos(π − π/3) = −1/2, so cos(2π/3) = −1/2. Since 2π/3 ∈ [0, π], it lies within the principal branch.

Principal value = 2π/3

Example 4 — tan⁻¹(−√3)

Let tan⁻¹(−√3) = y, so tan y = −√3.

We know tan(π/3) = √3. Using tan(−θ) = −tan θ: tan(−π/3) = −√3. Since −π/3 ∈ (−π/2, π/2), it lies within the principal branch.

Principal value = −π/3

Example 5 — cot⁻¹(−1/√3)

Let cot⁻¹(−1/√3) = y, so cot y = −1/√3.

We know cot(π/3) = 1/√3. Using the identity cot(π − θ) = −cot θ: cot(π − π/3) = −1/√3, so cot(2π/3) = −1/√3. Since 2π/3 ∈ (0, π), it lies within the principal branch.

Principal value = 2π/3

Example 6 — cosec⁻¹(−√2)

Let cosec⁻¹(−√2) = y, so cosec y = −√2.

We know cosec(π/4) = √2. Using cosec(−θ) = −cosec θ: cosec(−π/4) = −√2. Since −π/4 ∈ [−π/2, π/2] and −π/4 ≠ 0, it lies within the principal branch.

Principal value = −π/4

When the Argument Falls Outside the Principal Branch

A common trap in problems like Exercise 2.2 and the Miscellaneous Exercise is that you may be asked to evaluate sin⁻¹(sin x) or cos⁻¹(cos x) where x is not in the principal branch.

For example: sin⁻¹(sin 3π/5)

At first glance it looks like the answer should be 3π/5. However, 3π/5 ∉ [−π/2, π/2], so it does not lie in the principal branch. The correct approach is to reduce:

sin(3π/5) = sin(π − 3π/5) = sin(2π/5)

Now 2π/5 ∈ [−π/2, π/2], so sin⁻¹(sin 3π/5) = 2π/5.

This type of question is extremely common in CBSE board exams and entrance tests. Recognising when to apply a reduction identity is the key skill.

Common Angles You Must Know by Heart

To find principal values efficiently under exam conditions, you need instant recall of these:

Having these at your fingertips means you never need to derive them mid-calculation. For negative values, apply the relevant sign identity (sin is odd, cos is positive in [0, π], etc.).

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Test Your Knowledge

Q1.What is the principal value of sin⁻¹(−1/2)?

5π / 6−π / 67π / 6π / 6

Q2.What is the principal value of cos⁻¹(−1/2)?

−π / 34π / 32π / 3π / 3

Q3.The value of tan⁻¹(−√3) in its principal value branch is:

2π / 35π / 3π / 3−π / 3