Properties and Identities of Inverse Trigonometric Functions

Introduction

Inverse trigonometric functions have a rich set of properties and identities that are crucial for simplifying expressions and solving equations. These properties often mirror the identities of standard trigonometric functions but have important restrictions related to their principal value branches.

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1. Composition Properties 🔄

A. f(f⁻¹(x)) = x

This property shows the direct cancellation of a function and its inverse.

• $$\sin(\sin^{-1}x) = x, \quad \text{for } x \in [-1, 1]$$
• $$\cos(\cos^{-1}x) = x, \quad \text{for } x \in [-1, 1]$$
• $$\tan(\tan^{-1}x) = x, \quad \text{for } x \in \mathbb{R}$$
• Intuition: This is like asking, "What is the sine of the angle whose sine is x?" The answer is simply x. The operations undo each other, provided x is in the valid domain of the inverse function.

B. f⁻¹(f(x)) = x (with restrictions)

This composition is trickier because the output of an inverse trigonometric function is always restricted to its principal value branch.

• $$\sin^{-1}(\sin x) = x, \quad \text{only if } x \in [-\pi/2, \pi/2]$$
• $$\cos^{-1}(\cos x) = x, \quad \text{only if } x \in [0, \pi]$$
• $$\tan^{-1}(\tan x) = x, \quad \text{only if } x \in (-\pi/2, \pi/2)$$
• Explanation: If $x$ is outside the principal range, the function returns an equivalent angle that is inside the range.
• Example: Find the value of $\sin^{-1}(\sin(2\pi/3))$.
  1. The input angle is $x=2\pi/3$, which is not in the principal range of $\sin^{-1}$, $[-\pi/2, \pi/2]$. So the answer is not $2\pi/3$.
  2. First, evaluate the inner function: $\sin(2\pi/3) = \sqrt{3}/2$.
  3. The problem becomes: find $\sin^{-1}(\sqrt{3}/2)$.
  4. Now, find the angle in the principal range $[-\pi/2, \pi/2]$ whose sine is $\sqrt{3}/2$. That angle is $\pi/3$.
  5. Therefore, $\sin^{-1}(\sin(2\pi/3)) = \pi/3$.

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2. Reciprocal Identities 📏

These properties relate the inverse of a trigonometric function to the inverse of its reciprocal.

• $$\csc^{-1}x = \sin^{-1}(1/x), \quad \text{for } |x| \ge 1$$

• $$\sec^{-1}x = \cos^{-1}(1/x), \quad \text{for } |x| \ge 1$$

• $$\cot^{-1}x = \tan^{-1}(1/x), \quad \text{for } x > 0$$ (Note: For $x<0$, $\cot^{-1}x = \pi + \tan^{-1}(1/x)$)

• Proof (for csc⁻¹x):

  1. Let $\csc^{-1}x = \theta$. By definition, $\csc\theta = x$.
  2. Take the reciprocal of both sides: $\frac{1}{\csc\theta} = \frac{1}{x}$.
  3. Since $\sin\theta = \frac{1}{\csc\theta}$, we have $\sin\theta = 1/x$.
  4. Take the inverse sine of both sides: $\theta = \sin^{-1}(1/x)$.
  5. Therefore, $\csc^{-1}x = \sin^{-1}(1/x)$.

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3. Reflection Identities (Negative Arguments) зеркало

These rules explain how to handle negative inputs to inverse trigonometric functions.

Group 1 (Odd Functions Family):

• $$\sin^{-1}(-x) = -\sin^{-1}x, \quad x \in [-1,1]$$
• $$\tan^{-1}(-x) = -\tan^{-1}x, \quad x \in \mathbb{R}$$
• $$\csc^{-1}(-x) = -\csc^{-1}x, \quad |x| \ge 1$$
• Example: $\sin^{-1}(-1/2) = -\sin^{-1}(1/2) = -\pi/6$.

Group 2 (Cosine Family):

• $$\cos^{-1}(-x) = \pi - \cos^{-1}x, \quad x \in [-1,1]$$
• $$\sec^{-1}(-x) = \pi - \sec^{-1}x, \quad |x| \ge 1$$
• $$\cot^{-1}(-x) = \pi - \cot^{-1}x, \quad x \in \mathbb{R}$$
• Example: $\cos^{-1}(-1/2) = \pi - \cos^{-1}(1/2) = \pi - \pi/3 = 2\pi/3$.

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4. Cofunction Identities (Complementary Angles) ➕

These identities show a simple relationship between pairs of inverse cofunctions.

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

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

• $$\sec^{-1}x + \csc^{-1}x = \frac{\pi}{2}, \quad \text{for } |x| \ge 1$$

• Intuition: This is the inverse version of identities like $\cos(\theta) = \sin(\pi/2 - \theta)$. It means that if an angle's sine is $x$, then the cosine of its complementary angle is also $x$.

• Example: Verify for $x = \sqrt{3}/2$.

  • $\sin^{-1}(\sqrt{3}/2) + \cos^{-1}(\sqrt{3}/2) = \pi/3 + \pi/6 = \frac{2\pi+\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}$.

Example 1: Using Composition Property

Question: Find the value of $\sin^{-1}(\sin(2\pi/3))$.

Explanation:

1. Check the Principal Range: The property $\sin^{-1}(\sin x) = x$ is only valid when $x$ is in the principal value branch of $\sin^{-1}$, which is $[-\pi/2, \pi/2]$. The angle $2\pi/3$ is not in this interval, so the answer is not $2\pi/3$.

2. Evaluate the Inner Function: First, find the value of $\sin(2\pi/3)$. Since $2\pi/3$ is in the second quadrant, $\sin(2\pi/3) = \sin(\pi - \pi/3) = \sin(\pi/3) = \frac{\sqrt{3}}{2}$.

3. Evaluate the Inverse Function: The problem now becomes $\sin^{-1}(\frac{\sqrt{3}}{2})$. We need to find the angle in the principal range $[-\pi/2, \pi/2]$ whose sine is $\frac{\sqrt{3}}{2}$. That angle is $\pi/3$.

Answer: $\pi/3$.

Example 2: Using Reflection Property

Question: Find the value of $\cos^{-1}(-1/2)$.

Explanation:

1. State the Property: For negative arguments in inverse cosine, we use the reflection identity: $$\cos^{-1}(-x) = \pi - \cos^{-1}(x)$$
2. Apply the Property: In this case, $x=1/2$. So, $$\cos^{-1}(-1/2) = \pi - \cos^{-1}(1/2)$$
3. Evaluate and Solve: We know that the principal value of $\cos^{-1}(1/2)$ is $\pi/3$. $$\pi - \frac{\pi}{3} = \frac{3\pi - \pi}{3} = \frac{2\pi}{3}$$

Answer: $2\pi/3$.

Example 3: Using Cofunction Identity

Question: Find the value of $\sin(\sin^{-1}(1/5) + \cos^{-1}(1/5))$.

Explanation:

1. State the Property: We use the cofunction identity $$\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$$ which is valid for any $x \in [-1, 1]$.

2. Apply the Property: The expression inside the parenthesis, $\sin^{-1}(1/5) + \cos^{-1}(1/5)$, simplifies to $\pi/2$.

3. Evaluate: The problem is now to find $\sin(\pi/2)$, which is 1.

Answer: 1.

Example 4: Proving a $2\sin^{-1}x$ Identity

Question: Prove that $\sin^{-1}(2x\sqrt{1-x^2}) = 2\sin^{-1}x$ for $|x| \le 1/\sqrt{2}$.

Explanation:

1. Strategy: We use a trigonometric substitution. Let $x = \sin\theta$, which implies $\theta = \sin^{-1}x$.

2. Substitute into the Expression: Replace $x$ with $\sin\theta$ in the left-hand side (LHS).

   LHS = $\sin^{-1}(2\sin\theta\sqrt{1-\sin^2\theta})$.

3. Simplify: Using the identity $$\sqrt{1-\sin^2\theta} = \cos\theta$$ (for $\theta$ in the principal range of $\sin^{-1}$), we get:

   LHS = $\sin^{-1}(2\sin\theta\cos\theta) = \sin^{-1}(\sin(2\theta))$.

4. Check the Range: The given condition $|x| \le 1/\sqrt{2}$ means $-\frac{1}{\sqrt{2}} \le \sin\theta \le \frac{1}{\sqrt{2}}$.

   This implies $-\pi/4 \le \theta \le \pi/4$.

   Therefore, $- \pi/2 \le 2\theta \le \pi/2$.

   This interval is the principal value branch for $\sin^{-1}$.

5. Finalize: Since $2\theta$ is in the principal range, $\sin^{-1}(\sin(2\theta)) = 2\theta$. Substituting back $\theta = \sin^{-1}x$, we get $2\sin^{-1}x$, which is the right-hand side (RHS).

Answer: Proven.

Example 5: Proving a $2\tan^{-1}x$ Identity

Question: Prove that $2\tan^{-1}x = \cos^{-1}(\frac{1-x^2}{1+x^2})$ for $x \ge 0$.

Explanation:

1. Strategy: We start with the more complex RHS and use the substitution $x = \tan\theta$.

   This means $\theta = \tan^{-1}x$.

2. Substitute into the RHS:

   RHS = $\cos^{-1}\left(\frac{1-\tan^2\theta}{1+\tan^2\theta}\right)$.

3. Simplify: The expression inside is the double angle identity for cosine: $$\cos(2\theta) = \frac{1-\tan^2\theta}{1+\tan^2\theta}$$

   RHS = $\cos^{-1}(\cos(2\theta))$.

4. Check the Range: The condition $x \ge 0$ means $\tan\theta \ge 0$. For the principal value of $\tan^{-1}$, this means $0 \le \theta < \pi/2$.

   Therefore, $0 \le 2\theta < \pi$. This is the principal value branch for $\cos^{-1}$.

5. Finalize: Since $2\theta$ is in the principal range, $\cos^{-1}(\cos(2\theta)) = 2\theta$.

   Substituting back $\theta = \tan^{-1}x$, we get $2\tan^{-1}x$, which is the left-hand side (LHS).

Answer: Proven.

Example 6: Simplifying Expressions

Question: Simplify $\tan^{-1}\left(\frac{\cos x}{1-\sin x}\right)$ for $x \in (-\pi/2, \pi/2)$.

Explanation:

1. Strategy: Our goal is to transform the inner expression into the form $\tan(\theta)$ so it cancels with $\tan^{-1}$. We use cofunction and half-angle identities.

2. Transform using Cofunctions: $\cos x = \sin(\pi/2-x)$ and $\sin x = \cos(\pi/2-x)$. The expression becomes $$\tan^{-1}\left(\frac{\sin(\pi/2-x)}{1-\cos(\pi/2-x)}\right)$$

3. Apply Half-Angle Identities: Use the identities $$\sin(2A)=2\sin A\cos A$$ and $$1-\cos(2A)=2\sin^2A$$

   Let $A = (\pi/4-x/2)$.

   The expression becomes $$\tan^{-1}\left(\frac{2\sin(\pi/4-x/2)\cos(\pi/4-x/2)}{2\sin^2(\pi/4-x/2)}\right)$$ $$= \tan^{-1}\left(\frac{\cos(\pi/4-x/2)}{\sin(\pi/4-x/2)}\right)$$ $$= \tan^{-1}(\cot(\pi/4-x/2))$$.

4. Convert cot to tan:

Use the identity $\cot(\theta) = \tan(\pi/2 - \theta)$. $$\tan^{-1}\left(\tan\left(\frac{\pi}{2} - (\frac{\pi}{4}-\frac{x}{2})\right)\right) = \tan^{-1}\left(\tan(\frac{\pi}{4}+\frac{x}{2})\right)$$

5. Finalize: Since $x \in (-\pi/2, \pi/2)$, the argument $\pi/4+x/2$ is in $(0, \pi/2)$, which is within the principal range of $\tan^{-1}$. Therefore, the functions cancel.

Answer: $\pi/4+x/2$.

Example 7: Solving an Equation

Question: Solve the equation $2\tan^{-1}(\cos x) = \tan^{-1}(2\csc x)$.

Explanation:

1. Apply the $2\tan^{-1}u$ Identity: Use the identity $$2\tan^{-1}u = \tan^{-1}\left(\frac{2u}{1-u^2}\right)$$ on the left side, with $u=\cos x$. $$\tan^{-1}\left(\frac{2\cos x}{1-\cos^2x}\right) = \tan^{-1}(2\csc x)$$

2. Simplify and Equate Arguments: The term $1-\cos^2x$ is $\sin^2x$.

   Since the $\tan^{-1}$ functions are equal, their arguments must be equal. $$\frac{2\cos x}{\sin^2x} = 2\csc x = \frac{2}{\sin x}$$

3. Solve for x: Assuming $\sin x \ne 0$, we can multiply both sides by $\sin^2x/2$. $\cos x = \sin x$. Dividing by $\cos x$ (assuming $\cos x \ne 0$), we get $\frac{\sin x}{\cos x} = 1 \implies \tan x = 1$.

4. Find the Solution: The principal solution for $\tan x=1$ is $x=\pi/4$. At this value, neither $\sin x$ nor $\cos x$ are zero, so our assumptions were valid.

Answer: The principal solution is $x=\pi/4$.

Example 8: Using Reciprocal Identity

Question: Find the value of $\sin(\csc^{-1}(5/3))$.

Explanation: Method 1: Using Identities

1. Use the reciprocal identity $\csc^{-1}x = \sin^{-1}(1/x)$. The expression becomes $\sin(\sin^{-1}(1/(5/3))) = \sin(\sin^{-1}(3/5))$.
2. Use the composition property $\sin(\sin^{-1}y) = y$ for $y \in [-1,1]$. Since $3/5$ is in the valid range, the answer is $3/5$.

Method 2: Using a Right Triangle

1. Let $\theta = \csc^{-1}(5/3)$. This means $\csc\theta = 5/3$.

2. Since $\csc\theta = \frac{\text{Hypotenuse}}{\text{Opposite}}$, we can draw a right triangle with hypotenuse 5 and opposite side 3.

3. By Pythagoras' theorem, the adjacent side is $\sqrt{5^2-3^2}=\sqrt{16}=4$.

4. The problem asks for $\sin(\theta)$. From the triangle, $\sin\theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = 3/5$.

Answer: 3/5.

Example 9: Finding Value of a Sum

Question: Find the value of $\sin^{-1}(4/5) + \sin^{-1}(5/13) + \sin^{-1}(16/65)$.

Explanation:

1. Strategy: The addition formula for $\sin^{-1}$ is complex. It's easier to convert everything to $\tan^{-1}$ and use its simpler addition formula: $$\tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right)$$

2. Convert: Using a right triangle (or the identity $\sin^{-1}x = \tan^{-1}(x/\sqrt{1-x^2})$), we find:

   • $\sin^{-1}(4/5) = \tan^{-1}(4/3)$. (3-4-5 triangle)

   • $\sin^{-1}(5/13) = \tan^{-1}(5/12)$. (5-12-13 triangle)

   • $\sin^{-1}(16/65) = \tan^{-1}(16/63)$. (16-63-65 triangle)

3. Add the First Two Terms: $$\tan^{-1}(4/3)+\tan^{-1}(5/12)$$ $$= \tan^{-1}\left(\frac{4/3+5/12}{1-(4/3)(5/12)}\right)$$ $$= \tan^{-1}\left(\frac{(16+5)/12}{1-20/36}\right)$$ $$= \tan^{-1}\left(\frac{21/12}{16/36}\right) = \tan^{-1}\left(\frac{7/4}{4/9}\right) = \tan^{-1}(63/16)$$

4. Add the Third Term: The expression is now $$\tan^{-1}(63/16) + \tan^{-1}(16/63)$$

5. Use Cofunction Identity: Using the property $$\tan^{-1}(x) = \cot^{-1}(1/x)$$ we can rewrite the second term: $$\tan^{-1}(16/63) = \cot^{-1}(63/16)$$

   The sum becomes $$\tan^{-1}(63/16) + \cot^{-1}(63/16)$$

   Using the identity $$\tan^{-1}z + \cot^{-1}z = \pi/2$$ the final result is $\pi/2$.

Answer: $\pi/2$.

Example 10: Simplifying a Complex Expression

Question: Find the value of $\tan^{-1}\sqrt{3} - \sec^{-1}(-2)$.

Explanation: We evaluate each term separately based on their principal value branches.

1. Evaluate $\tan^{-1}\sqrt{3}$:

   We are looking for the angle $\theta$ in the interval $(-\pi/2, \pi/2)$ such that $\tan\theta = \sqrt{3}$. This angle is $\pi/3$.

2. Evaluate $\sec^{-1}(-2)$:

   We use the reflection identity $\sec^{-1}(-x) = \pi - \sec^{-1}(x)$.

   $\sec^{-1}(-2) = \pi - \sec^{-1}(2)$.

   We need the angle $\theta$ in $[0, \pi]$ such that $\sec\theta=2$ (or $\cos\theta=1/2$). This angle is $\pi/3$.

   So, $\sec^{-1}(-2) = \pi - \pi/3 = 2\pi/3$.

3. Calculate the Final Value: The expression is $\pi/3 - 2\pi/3 = -\pi/3$.

Answer: $-\pi/3$.