Inverse Trigonometric Functions

INVERSE TRIGONOMETRIC FUNCTIONS

Introduction and Basic Concepts

**Inverse trigonometric functions** are the inverses of trigonometric functions, denoted as sin⁻¹x, cos⁻¹x, tan⁻¹x, cot⁻¹x, sec⁻¹x, and cosec⁻¹x (also called arc functions: arcsin, arccos, etc.).

**Why inverses need restricted domains:**

Trigonometric functions are **not one-one and onto** over their natural domains ℝ or ℝ\{certain values}. Recall from Chapter 1: a function has an inverse **only if it is bijective** (both one-one and onto).

For example:

sin: ℝ → [−1, 1] is onto but **not one-one** (sin(π/6) = sin(5π/6) = 1/2)

To make sin inverse exist, we **restrict domain** to [−π/2, π/2]

**Key principle:** When we restrict the domain of a trigonometric function to an interval where it is bijective, we get an inverse function. Multiple domain restrictions are possible, giving different **branches**. The branch with the principal value range is called the **principal value branch**.

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sin⁻¹x (Arcsine Function)

**Definition:** sin⁻¹: [−1, 1] → [−π/2, π/2]

**Domain and Range (Principal Value Branch):**

Domain: [−1, 1]

Range: [−π/2, π/2]

**Understanding the restriction:**

If we restrict sin: [−π/2, π/2] → [−1, 1], this restricted sine becomes bijective. Its inverse is sin⁻¹.

**Fundamental properties:**

If y = sin⁻¹x, then sin y = x and −π/2 ≤ y ≤ π/2

sin(sin⁻¹x) = x for all x ∈ [−1, 1]

sin⁻¹(sin x) = x for all x ∈ [−π/2, π/2]

**Important:** sin⁻¹(−x) = −sin⁻¹x (odd function)

**Example:** Find sin⁻¹(1/2)

Let y = sin⁻¹(1/2). Then sin y = 1/2 with y ∈ [−π/2, π/2].

Since sin(π/6) = 1/2 and π/6 ∈ [−π/2, π/2], we have **sin⁻¹(1/2) = π/6**.

**Graphical representation:** The graph of y = sin⁻¹x is obtained by reflecting y = sin x (restricted to [−π/2, π/2]) across the line y = x.

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cos⁻¹x (Arccosine Function)

**Definition:** cos⁻¹: [−1, 1] → [0, π]

**Domain and Range (Principal Value Branch):**

Domain: [−1, 1]

Range: [0, π]

**Understanding the restriction:**

Restricting cos: [0, π] → [−1, 1] makes cosine bijective. Note: cos is **decreasing** on [0, π], making it one-one.

**Fundamental properties:**

If y = cos⁻¹x, then cos y = x and 0 ≤ y ≤ π

cos(cos⁻¹x) = x for all x ∈ [−1, 1]

cos⁻¹(cos x) = x for all x ∈ [0, π]

**Important:** cos⁻¹(−x) = π − cos⁻¹x (neither even nor odd; has this transformation property)

**Complementary property:**

$$\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}, \quad x \in [-1, 1]$$

**Proof:** Let α = sin⁻¹x. Then sin α = x where −π/2 ≤ α ≤ π/2.

Consider β = π/2 − α. Then:

cos β = cos(π/2 − α) = sin α = x

β ∈ [0, π] (since 0 ≤ π/2 − α ≤ π)

Therefore cos⁻¹x = β = π/2 − α = π/2 − sin⁻¹x

**Example:** Find cos⁻¹(−1/2)

Let y = cos⁻¹(−1/2). Then cos y = −1/2 with y ∈ [0, π].

Since cos(2π/3) = −1/2 and 2π/3 ∈ [0, π], we have **cos⁻¹(−1/2) = 2π/3**.

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tan⁻¹x (Arctangent Function)

**Definition:** tan⁻¹: ℝ → (−π/2, π/2)

**Domain and Range (Principal Value Branch):**

Domain: ℝ (all real numbers)

Range: (−π/2, π/2) [open interval]

**Understanding the restriction:**

Restricting tan: (−π/2, π/2) → ℝ makes tangent bijective. Tan is strictly increasing on this interval.

**Fundamental properties:**

If y = tan⁻¹x, then tan y = x and −π/2 < y < π/2

tan(tan⁻¹x) = x for all x ∈ ℝ

tan⁻¹(tan x) = x for all x ∈ (−π/2, π/2)

**Important:** tan⁻¹(−x) = −tan⁻¹x (odd function)

**Key angle addition property:**

For x, y ∈ ℝ with xy < 1:

$$\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x + y}{1 - xy}\right)$$

**Proof:** Let α = tan⁻¹x and β = tan⁻¹y. Then tan α = x and tan β = y where α, β ∈ (−π/2, π/2).

$$\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \frac{x + y}{1 - xy}$$

Since xy < 1 ensures α + β ∈ (−π/2, π/2), we have **tan⁻¹(tan(α + β)) = α + β**.

**Special case:** When xy > 1 and x > 0:

$$\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x + y}{1 - xy}\right) + \pi$$

**Complementary property:**

$$\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}, \quad x \in \mathbb{R}$$

**Example:** Find tan⁻¹(1)

Let y = tan⁻¹(1). Then tan y = 1 with y ∈ (−π/2, π/2).

Since tan(π/4) = 1, we have **tan⁻¹(1) = π/4**.

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cot⁻¹x (Arccotangent Function)

**Definition:** cot⁻¹: ℝ → (0, π)

**Domain and Range (Principal Value Branch):**

Domain: ℝ (all real numbers)

Range: (0, π) [open interval]

**Understanding the restriction:**

Restricting cot: (0, π) → ℝ makes cotangent bijective. Cot is strictly decreasing on this interval.

**Fundamental properties:**

If y = cot⁻¹x, then cot y = x and 0 < y < π

cot(cot⁻¹x) = x for all x ∈ ℝ

cot⁻¹(cot x) = x for all x ∈ (0, π)

**Important:** cot⁻¹(−x) = π − cot⁻¹x

**Complementary property:**

$$\cot^{-1}x + \tan^{-1}x = \frac{\pi}{2}, \quad x \in \mathbb{R}$$

This is the same as the tan⁻¹ and cot⁻¹ complementary relationship.

**Example:** Find cot⁻¹(√3)

Let y = cot⁻¹(√3). Then cot y = √3 with y ∈ (0, π).

Since cot(π/6) = √3 and π/6 ∈ (0, π), we have **cot⁻¹(√3) = π/6**.

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cosec⁻¹x (Arccosecant Function)

**Definition:** cosec⁻¹: ℝ − (−1, 1) → [−π/2, π/2] − {0}

**Domain and Range (Principal Value Branch):**

Domain: ℝ − (−1, 1) = (−∞, −1] ∪ [1, ∞)

Range: [−π/2, π/2] − {0}

**Understanding the restriction:**

Restricting cosec to [−π/2, π/2] − {0} (avoiding x = 0 where cosec is undefined) makes it bijective onto ℝ − (−1, 1).

**Fundamental properties:**

If y = cosec⁻¹x, then cosec y = x and y ∈ [−π/2, π/2] − {0}

cosec(cosec⁻¹x) = x for all x ∈ ℝ − (−1, 1)

cosec⁻¹(cosec x) = x for all x ∈ [−π/2, π/2] − {0}

**Important:** cosec⁻¹(−x) = −cosec⁻¹x (odd function)

**Relationship with sin⁻¹:**

$$\sin^{-1}x = \text{cosec}^{-1}\left(\frac{1}{x}\right), \quad x \in [-1, 1] \setminus \{0\}$$

**Example:** Find cosec⁻¹(2)

Let y = cosec⁻¹(2). Then cosec y = 2, so sin y = 1/2, with y ∈ [−π/2, π/2] − {0}.

Since sin(π/6) = 1/2, we have **cosec⁻¹(2) = π/6**.

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sec⁻¹x (Arcsecant Function)

**Definition:** sec⁻¹: ℝ − (−1, 1) → [0, π] − {π/2}

**Domain and Range (Principal Value Branch):**

Domain: ℝ − (−1, 1) = (−∞, −1] ∪ [1, ∞)

Range: [0, π] − {π/2}

**Understanding the restriction:**

Restricting sec to [0, π] − {π/2} (avoiding x = π/2 where sec is undefined) makes it bijective onto ℝ − (−1, 1).

**Fundamental properties:**

If y = sec⁻¹x, then sec y = x and y ∈ [0, π] − {π/2}

sec(sec⁻¹x) = x for all x ∈ ℝ − (−1, 1)

sec⁻¹(sec x) = x for all x ∈ [0, π] − {π/2}

**Important:** sec⁻¹(−x) = π − sec⁻¹x

**Relationship with cos⁻¹:**

$$\cos^{-1}x = \text{sec}^{-1}\left(\frac{1}{x}\right), \quad x \in [-1, 1] \setminus \{0\}$$

**Example:** Find sec⁻¹(2)

Let y = sec⁻¹(2). Then sec y = 2, so cos y = 1/2, with y ∈ [0, π] − {π/2}.

Since cos(π/3) = 1/2, we have **sec⁻¹(2) = π/3**.

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Summary Table of Domains and Ranges

| Function | Domain | Principal Range |

|----------|--------|-----------------|

| sin⁻¹x | [−1, 1] | [−π/2, π/2] |

| cos⁻¹x | [−1, 1] | [0, π] |

| tan⁻¹x | ℝ | (−π/2, π/2) |

| cot⁻¹x | ℝ | (0, π) |

| cosec⁻¹x | ℝ − (−1, 1) | [−π/2, π/2] − {0} |

| sec⁻¹x | ℝ − (−1, 1) | [0, π] − {π/2} |

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Important Properties and Relations

**Complementary pairs:**

sin⁻¹x + cos⁻¹x = π/2

tan⁻¹x + cot⁻¹x = π/2

sec⁻¹x + cosec⁻¹x = π/2 (for x ≠ 0)

**Double angle type properties:** For x ∈ [−1/√2, 1/√2]:

$$\sin^{-1}(2x\sqrt{1-x^2}) = 2\sin^{-1}x$$

Similarly for other functions with appropriate domain restrictions.

**Composition with original function:**

sin(sin⁻¹x) = x for x ∈ [−1, 1]

sin⁻¹(sin x) = x **only** for x ∈ [−π/2, π/2]

**Critical error to avoid:** sin⁻¹(sin x) ≠ x in general. For example:

sin⁻¹(sin(3π/5)) ≠ 3π/5 because 3π/5 ∉ [−π/2, π/2]

Instead, we use: sin(3π/5) = sin(2π − 3π/5) = sin(7π/5), but sin(7π/5) ∉ [−1, 1] range achievable directly. Rather, sin(3π/5) = sin(π − 3π/5) = sin(2π/5), so sin⁻¹(sin(3π/5)) = 2π/5.

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Important Addition and Subtraction Formulas

**For tan⁻¹ (most frequently used):**

$$\tan^{-1}x + \tan^{-1}y = \begin{cases}

\tan^{-1}\left(\frac{x+y}{1-xy}\right) & \text{if } xy < 1 \\

\tan^{-1}\left(\frac{x+y}{1-xy}\right) + \pi & \text{if } xy > 1, x > 0 \\

\tan^{-1}\left(\frac{x+y}{1-xy}\right) - \pi & \text{if } xy > 1, x < 0

\end{cases}$$

$$\tan^{-1}x - \tan^{-1}y = \tan^{-1}\left(\frac{x-y}{1+xy}\right) \text{ if } xy > -1$$

**For sin⁻¹:**

$$\sin^{-1}x + \sin^{-1}y = \begin{cases}

\sin^{-1}(x\sqrt{1-y^2} + y\sqrt{1-x^2}) & \text{if } x^2 + y^2 \leq 1 \\

\pi - \sin^{-1}(x\sqrt{1-y^2} + y\sqrt{1-x^2}) & \text{if } x^2 + y^2 > 1, x,y > 0

\end{cases}$$

**For cos⁻¹:**

$$\cos^{-1}x + \cos^{-1}y = \cos^{-1}(xy - \sqrt{(1-x^2)(1-y^2)})$$

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Simplification Techniques with Substitution

**When to use substitution:**

These methods convert complex inverse trigonometric expressions into simpler forms using trigonometric identities.

**Method 1 - Direct trigonometric substitution:**

If the expression contains √(1 − x²), try x = sin θ or x = cos θ.

If the expression contains √(1 + x²), try x = tan θ.

If the expression contains √(x² − 1), try x = sec θ or x = cosec θ.

**Example:** Simplify tan⁻¹(√(1 − x²)/x) for x > 0

Let x = sin α where α ∈ (0, π/2). Then √(1 − x²) = cos α.

$$\tan^{-1}\left(\frac{\cos \alpha}{\sin \alpha}\right) = \tan^{-1}(\cot \alpha) = \tan^{-1}\left(\tan\left(\frac{\pi}{2} - \alpha\right)\right) = \frac{\pi}{2} - \alpha = \frac{\pi}{2} - \sin^{-1}x$$

**Method 2 - Expression manipulation:**

Express complex fractions using algebraic manipulation before applying inverse function.

**Example:** Simplify tan⁻¹((1 − cos x)/(sin x)) for x ∈ (−π/2, π/2)

Using half-angle formulas:

1 − cos x = 2sin²(x/2)

sin x = 2sin(x/2)cos(x/2)

$$\frac{1 - \cos x}{\sin x} = \frac{2\sin^2(x/2)}{2\sin(x/2)\cos(x/2)} = \frac{\sin(x/2)}{\cos(x/2)} = \tan(x/2)$$

Therefore: tan⁻¹((1 − cos x)/(sin x)) = **x/2**

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Finding Principal Values

**Key steps for finding principal values:**

1. Let y = f⁻¹(a) where f is the inverse trigonometric function

2. This means f(y) = a

3. Find y such that f(y) = a **and y lies in the principal range of f⁻¹**

4. Use memorized values: sin(π/6) = 1/2, cos(π/3) = 1/2, tan(π/4) = 1, etc.

**Example:** Find the principal value of sin⁻¹(−√3/2)

Let y = sin⁻¹(−√3/2). We need sin y = −√3/2 with y ∈ [−π/2, π/2].

Since sin(−π/3) = −√3/2 and −π/3 ∈ [−π/2, π/2], the principal value is **−π/3**.

**Common mistake:** Forgetting to check if the angle is in the principal range. For instance, sin(4π/3) = −√3/2, but 4π/3 ∉ [−π/2, π/2], so sin⁻¹(sin(4π/3)) ≠ 4π/3.

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Solving Equations Involving Inverse Trigonometric Functions

**General strategy:**

1. Apply the inverse trigonometric function to both sides carefully

2. Use the fundamental relations: sin(sin⁻¹x) = x, etc., **only when valid**

3. Verify solutions satisfy original domain restrictions

**Example:** Solve sin⁻¹(2x) = π/6

Taking sin of both sides: 2x = sin(π/6) = 1/2

Therefore: x = 1/4

Check: sin⁻¹(1/2) = π/6 ✓

**Example:** Solve tan⁻¹x + tan⁻¹(2x) = π/4

Using addition formula with xy = 2x² < 1 (when |x| < 1/√2):

$$\tan^{-1}\left(\frac{x + 2x}{1 - 2x^2}\right) = \frac{\pi}{4}$$

$$\frac{3x}{1 - 2x^2} = \tan(\pi/4) = 1$$

$$3x = 1 - 2x^2$$

$$2x^2 + 3x - 1 = 0$$

Using quadratic formula: x = (−3 ± √(9 + 8))/4 = (−3 ± √17)/4

Check domain: Only x = (−3 + √17)/4 ≈ 0.28 satisfies |x| < 1/√2.

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Common Pitfalls and How to Avoid Them

**Pitfall 1: Confusing sin⁻¹x with (sin x)⁻¹**

sin⁻¹x is **the inverse function** with domain [−1, 1] and range [−π/2, π/2]

(sin x)⁻¹ = 1/sin x = cosec x is the **reciprocal function**

**Pitfall 2: Assuming sin⁻¹(sin x) = x for all x**

This is **only true** when x ∈ [−π/2, π/2]

For other values, first reduce x to the principal range using periodic/symmetry properties

**Pitfall 3: Forgetting domain restrictions in addition formulas**

tan⁻¹x + tan⁻¹y requires checking if xy < 1 or xy > 1

Different formulas apply; using the wrong one gives wrong answers

**Pitfall 4: Ignoring principal value requirement**

When the problem asks for "principal value," ensure the answer is in the principal range

Multiple angles give the same sine/cosine value; only one is in the principal range

**Pitfall 5: Algebraic errors in simplification**

When simplifying expressions like cot⁻¹((1 − x²)/(2x)), carefully apply identities

Verify by substituting a specific value like x = 1/2 into both original and simplified forms

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NCERT-Style Practice Problems with Solutions

**Problem 1:** Find sin⁻¹(cos(π/3))

Solution: cos(π/3) = 1/2

We need: sin⁻¹(1/2) = π/6 [since sin(π/6) = 1/2 and π/6 ∈ [−π/2, π/2]]

**Answer: π/6**

**Problem 2:** Prove that tan⁻¹(3/4) + tan⁻¹(1/7) = π/4

Solution: Using tan⁻¹x + tan⁻¹y = tan⁻¹((x+y)/(1−xy)) where xy = 3/28 < 1:

$$\tan^{-1}\left(\frac{3/4 + 1/7}{1 - 3/28}\right) = \tan^{-1}\left(\frac{25/28}{25/28}\right) = \tan^{-1}(1) = \frac{\pi}{4}$$ ✓

**Problem 3:** Simplify: cos⁻¹x − cos⁻¹(−x)

Solution: Using cos⁻¹(−x) = π − cos⁻¹x:

cos⁻¹x − (π − cos⁻¹x) = 2cos⁻¹x − π

**Problem 4:** Solve 2sin⁻¹x = π

Solution: sin⁻¹x = π/2, so x = sin(π/2) = 1

Check: 2sin⁻¹(1) = 2(π/2) = π ✓