FAC1004 L14 — Properties of Inverse Trig Functions

This lecture covers the algebraic properties of inverse trigonometric functions, including composition rules, complementary relationships, negative argument properties, sum/difference formulas, double-angle identities, and interconversion between inverse trig functions.

Learning Objective

To learn the properties of inverse trigonometric functions and perform calculations involving these properties.

Domain and Range (Principal Values)

Function Domain Range (Principal Value)
$y = \sin^{-1} x$ $[-1, 1]$ $\displaystyle\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
$y = \cos^{-1} x$ $[-1, 1]$ $[0, \pi]$
$y = \tan^{-1} x$ $\mathbb{R}$ $\displaystyle\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
$y = \cot^{-1} x$ $\mathbb{R}$ $[0, \pi]$
$y = \sec^{-1} x$ $(-\infty, -1] \cup [1, \infty)$ $\displaystyle\left[0, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \pi\right]$
$y = \csc^{-1} x$ $(-\infty, -1] \cup [1, \infty)$ $\displaystyle\left[-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right]$

[!note] Notation This lecture uses $\csc^{-1}$ and $\csc$ notation. The slides also use $\text{cosec}^{-1}$ and $\text{cosec}$ interchangeably.

Property 1: Composition (Cancellation) Equations

These describe when an inverse trig function "cancels" its corresponding trig function, and vice versa.

Inverse followed by function: $$\sin^{-1}(\sin \theta) = \theta, \quad -\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$$ $$\cos^{-1}(\cos \theta) = \theta, \quad 0 \le \theta \le \pi$$ $$\tan^{-1}(\tan \theta) = \theta, \quad -\frac{\pi}{2} < \theta < \frac{\pi}{2}$$

Function followed by inverse: $$\sin(\sin^{-1} x) = x$$ $$\cos(\cos^{-1} x) = x$$ $$\tan(\tan^{-1} x) = x$$

[!important] Domain restriction The identities $\sin^{-1}(\sin \theta) = \theta$, $\cos^{-1}(\cos \theta) = \theta$, and $\tan^{-1}(\tan \theta) = \theta$ only hold when $\theta$ is within the principal range of the respective inverse function.

Property 2: Reciprocal Relationships

Inverse cosecant, cotangent, and secant can be expressed in terms of inverse sine, tangent, and cosine:

$$(\text{i}) \quad \csc^{-1} x = \sin^{-1}\left(\frac{1}{x}\right)$$

$$(\text{ii}) \quad \cot^{-1} x = \tan^{-1}\left(\frac{1}{x}\right)$$

$$(\text{iii}) \quad \sec^{-1} x = \cos^{-1}\left(\frac{1}{x}\right)$$

Proof of (i): Let $\csc^{-1} x = \theta \Rightarrow x = \csc \theta \Rightarrow \frac{1}{x} = \sin \theta \Rightarrow \theta = \sin^{-1}\left(\frac{1}{x}\right)$.

The other two follow by analogous methods.

Property 3: Negative Argument Properties

$$(\text{i}) \quad \sin^{-1}(-x) = -\sin^{-1} x$$

$$(\text{ii}) \quad \tan^{-1}(-x) = -\tan^{-1} x$$

Proof of (i): Let $-x = \sin \theta$, so $x = -\sin \theta = \sin(-\theta)$. Then $-\theta = \sin^{-1} x$, giving $\theta = -\sin^{-1} x$. Therefore $\sin^{-1}(-x) = -\sin^{-1} x$.

[!note] This lecture covers only $\sin^{-1}(-x)$ and $\tan^{-1}(-x)$. The property $\cos^{-1}(-x) = \pi - \cos^{-1} x$ is not covered in L14.

Property 4: Complementary Relationships

$$(\text{i}) \quad \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$$

$$(\text{ii}) \quad \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$$

$$(\text{iii}) \quad \csc^{-1} x + \sec^{-1} x = \frac{\pi}{2}$$

Proof of (i): Let $\sin^{-1} x = \theta \Rightarrow x = \sin \theta = \cos\left(\frac{\pi}{2} - \theta\right)$. Then $\cos^{-1} x = \frac{\pi}{2} - \theta$, so $\theta + \cos^{-1} x = \frac{\pi}{2}$, giving $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$.

The others follow similarly using cofunction identities.

Property 5: Sum and Difference Formulas

For inverse tangent:

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

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

Proof of (i): Let $\tan^{-1} x = \theta$ and $\tan^{-1} y = \phi$, so $x = \tan \theta$ and $y = \tan \phi$.

$$\text{L.H.S.} = \theta + \phi$$ $$\text{R.H.S.} = \tan^{-1}\left(\frac{\tan \theta + \tan \phi}{1 - \tan \theta \tan \phi}\right) = \tan^{-1}[\tan(\theta + \phi)] = \theta + \phi = \text{L.H.S.}$$

$\therefore$ The result holds.

Property 6: Double Angle Formula

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

Proof: Let $x = \tan \theta$.

$$2\tan^{-1} x = 2\tan^{-1}(\tan \theta) = 2\theta \quad \text{......(i)}$$

$$\sin^{-1}\left(\frac{2x}{1 + x^2}\right) = \sin^{-1}\left(\frac{2\tan \theta}{1 + \tan^2 \theta}\right) = \sin^{-1}(2\sin\theta\cos\theta) = \sin^{-1}(\sin 2\theta) = 2\theta \quad \text{......(ii)}$$

$$\cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) = \cos^{-1}\left(\frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}\right) = \cos^{-1}(\cos^2\theta - \sin^2\theta) = \cos^{-1}(\cos 2\theta) = 2\theta \quad \text{......(iii)}$$

$$\tan^{-1}\left(\frac{2x}{1 - x^2}\right) = \tan^{-1}\left(\frac{2\tan \theta}{1 - \tan^2 \theta}\right) = \tan^{-1}(\tan 2\theta) = 2\theta \quad \text{......(iv)}$$

From (i), (ii), (iii), and (iv): $$2\tan^{-1} x = \sin^{-1}\left(\frac{2x}{1 + x^2}\right) = \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) = \tan^{-1}\left(\frac{2x}{1 - x^2}\right)$$

Property 7: Interconversion Formulas

Part (i): Expressing $\sin^{-1} x$ in terms of other inverse functions

$$\sin^{-1} x = \cos^{-1}\left(\sqrt{1 - x^2}\right) = \tan^{-1}\left(\frac{x}{\sqrt{1 - x^2}}\right) = \sec^{-1}\left(\frac{1}{\sqrt{1 - x^2}}\right) = \cot^{-1}\left(\frac{\sqrt{1 - x^2}}{x}\right) = \csc^{-1}\left(\frac{1}{x}\right)$$

Proof: Let $\sin^{-1} x = \theta \Rightarrow \sin \theta = x$. By right-triangle or Pythagorean identity: $$\cos \theta = \sqrt{1 - x^2}, \quad \tan \theta = \frac{x}{\sqrt{1 - x^2}}, \quad \sec \theta = \frac{1}{\sqrt{1 - x^2}}, \quad \cot \theta = \frac{\sqrt{1 - x^2}}{x}, \quad \csc \theta = \frac{1}{x}$$

Therefore $\theta$ equals each of the inverse expressions above.

Part (ii): Expressing $\cos^{-1} x$ in terms of other inverse functions

$$\cos^{-1} x = \sin^{-1}\left(\sqrt{1 - x^2}\right) = \tan^{-1}\left(\frac{\sqrt{1 - x^2}}{x}\right) = \csc^{-1}\left(\frac{1}{\sqrt{1 - x^2}}\right) = \cot^{-1}\left(\frac{x}{\sqrt{1 - x^2}}\right) = \sec^{-1}\left(\frac{1}{x}\right)$$

Proof: Let $\cos^{-1} x = \theta \Rightarrow \cos \theta = x$. Then: $$\sin \theta = \sqrt{1 - x^2}, \quad \tan \theta = \frac{\sqrt{1 - x^2}}{x}, \quad \csc \theta = \frac{1}{\sqrt{1 - x^2}}, \quad \cot \theta = \frac{x}{\sqrt{1 - x^2}}, \quad \sec \theta = \frac{1}{x}$$

Worked Examples

Example 1: Evaluating a Composite Expression

Find the exact value of: $$\cos\left(\sin^{-1}\left(\frac{3}{5}\right) + \frac{\pi}{2}\right)$$

Solution: Using the cosine addition formula: $$\cos\left(\sin^{-1}\left(\frac{3}{5}\right) + \frac{\pi}{2}\right) = \cos\left(\sin^{-1}\left(\frac{3}{5}\right)\right)\cos\left(\frac{\pi}{2}\right) - \sin\left(\sin^{-1}\left(\frac{3}{5}\right)\right)\sin\left(\frac{\pi}{2}\right)$$

From Property 7(i), $\cos\left(\sin^{-1}\left(\frac{3}{5}\right)\right) = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.

Also, $\sin\left(\sin^{-1}\left(\frac{3}{5}\right)\right) = \frac{3}{5}$, $\cos\left(\frac{\pi}{2}\right) = 0$, and $\sin\left(\frac{\pi}{2}\right) = 1$.

$$= \left(\frac{4}{5}\right)(0) - \left(\frac{3}{5}\right)(1) = \boxed{-\frac{3}{5}}$$

Example 2: Proving an Inverse Tangent Identity

Prove that: $$\tan^{-1}\left(\frac{1}{7}\right) + \tan^{-1}\left(\frac{1}{13}\right) = \tan^{-1}\left(\frac{2}{9}\right)$$

Solution: Applying Property 5(i): $$\tan^{-1}\left(\frac{1}{7}\right) + \tan^{-1}\left(\frac{1}{13}\right) = \tan^{-1}\left(\frac{\frac{1}{7} + \frac{1}{13}}{1 - \frac{1}{7} \cdot \frac{1}{13}}\right)$$

$$= \tan^{-1}\left(\frac{\frac{13 + 7}{91}}{1 - \frac{1}{91}}\right) = \tan^{-1}\left(\frac{\frac{20}{91}}{\frac{90}{91}}\right) = \tan^{-1}\left(\frac{20}{90}\right) = \boxed{\tan^{-1}\left(\frac{2}{9}\right)}$$

Summary of All Properties

Property Identity
1(i) $\sin^{-1}(\sin \theta) = \theta$, $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$
1(ii) $\cos^{-1}(\cos \theta) = \theta$, $0 \le \theta \le \pi$
1(iii) $\tan^{-1}(\tan \theta) = \theta$, $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$
2(i) $\csc^{-1} x = \sin^{-1}\left(\frac{1}{x}\right)$
2(ii) $\cot^{-1} x = \tan^{-1}\left(\frac{1}{x}\right)$
2(iii) $\sec^{-1} x = \cos^{-1}\left(\frac{1}{x}\right)$
3(i) $\sin^{-1}(-x) = -\sin^{-1} x$
3(ii) $\tan^{-1}(-x) = -\tan^{-1} x$
4(i) $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$
4(ii) $\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$
4(iii) $\csc^{-1} x + \sec^{-1} x = \frac{\pi}{2}$
5(i) $\tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x + y}{1 - xy}\right)$
5(ii) $\tan^{-1} x - \tan^{-1} y = \tan^{-1}\left(\frac{x - y}{1 + xy}\right)$
6 $2\tan^{-1} x = \sin^{-1}\left(\frac{2x}{1 + x^2}\right) = \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) = \tan^{-1}\left(\frac{2x}{1 - x^2}\right)$
7(i) $\sin^{-1} x = \cos^{-1}(\sqrt{1-x^2}) = \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right) = \sec^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right) = \cot^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right) = \csc^{-1}\left(\frac{1}{x}\right)$
7(ii) $\cos^{-1} x = \sin^{-1}(\sqrt{1-x^2}) = \tan^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right) = \csc^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right) = \cot^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right) = \sec^{-1}\left(\frac{1}{x}\right)$

Related

Source File

LECTURE_NOTES_2526/L14 Properties of Inverse Trigonometric Functions Full Version.pdf