Mathematics / Calculus I Single-Variable Differential Calculus 100% Free Open Access
Chapter 2 • Theory & Derivations

Transcendental Functions & Hyperbolic Trigonometry

Comprehensive theory of transcendental mathematics: exponential functions, Euler's number e, natural logarithms, unit circle trigonometry, principal branch inverse trigonometric functions, hyperbolic geometry on x^2 - y^2 = 1, and rigorous logarithmic derivations of inverse hyperbolic functions.

§2.1 Exponential Functions, Euler's Number e & Logarithmic Foundations

1. Exponential Functions

For a fixed positive base $a > 0$ with $a \ne 1$, the exponential function with base $a$ is defined for all $x \in \mathbb{R}$ by $f(x) = a^x$.

Fundamental Exponential Laws: For all $x, y \in \mathbb{R}$ and $a, b > 0$:

$$a^{x+y} = a^x a^y, \qquad a^{x-y} = \frac{a^x}{a^y}, \qquad (a^x)^y = a^{x y}, \qquad (a b)^x = a^x b^x$$

The function is strictly positive: $\operatorname{Range}(a^x) = (0, \infty)$. It is strictly increasing if $a > 1$, and strictly decreasing if $0 < a < 1$.

2. Euler's Number $e$ & The Natural Exponential

The base of the natural exponential function, Euler's number $e \approx 2.718281828459...$, is uniquely defined by the fundamental limit:

$$e = \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n = \sum_{k=0}^\infty \frac{1}{k!} = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \dots$$

Geometrically, $y = e^x$ is the unique exponential curve whose tangent line at the $y$-intercept $(0, 1)$ has a slope of exactly $1$.

3. Logarithmic Functions

Because $f(x) = a^x$ is strictly monotonic on $\mathbb{R}$, it is a bijection from $\mathbb{R}$ to $(0, \infty)$. Its inverse function is the logarithm to base $a$:

$$y = \log_a(x) \iff a^y = x \quad (x > 0)$$

The inverse of the natural exponential $e^x$ is the natural logarithm $\ln(x) = \log_e(x)$:

$$\ln(e^x) = x \quad \forall x \in \mathbb{R}, \qquad e^{\ln(x)} = x \quad \forall x > 0$$

Logarithmic Properties & Change of Base:

$$\ln(x y) = \ln(x) + \ln(y), \quad \ln(x/y) = \ln(x) - \ln(y), \quad \ln(x^r) = r \ln(x), \quad \log_a(x) = \frac{\ln(x)}{\ln(a)}$$

§2.2 Trigonometric Functions, Exact Identities & Unit Circle Geometry

1. The Unit Circle Definition of Trigonometric Functions

In analytical calculus, angles are measured strictly in radians. Let $(x, y)$ be the terminal coordinates of an angle $\theta \in \mathbb{R}$ on the unit circle $x^2 + y^2 = 1$ measured counterclockwise from $(1, 0)$:

$$\cos\theta = x, \qquad \sin\theta = y, \qquad \tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{y}{x} \quad (x \ne 0)$$

The reciprocal functions are $\sec\theta = 1/\cos\theta$, $\csc\theta = 1/\sin\theta$, and $\cot\theta = 1/\tan\theta = \cos\theta/\sin\theta$.

2. Fundamental Trigonometric Identities

From the Pythagorean theorem $x^2 + y^2 = 1$ on the unit circle:

$$\sin^2\theta + \cos^2\theta = 1, \qquad 1 + \tan^2\theta = \sec^2\theta, \qquad 1 + \cot^2\theta = \csc^2\theta$$

Angle Addition & Double-Angle Formulas:

$$\begin{aligned} \sin(\alpha \pm \beta) &= \sin\alpha \cos\beta \pm \cos\alpha \sin\beta \ \cos(\alpha \pm \beta) &= \cos\alpha \cos\beta \mp \sin\alpha \sin\beta \ \sin(2\theta) &= 2\sin\theta\cos\theta \ \cos(2\theta) &= \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta \ \tan(\alpha \pm \beta) &= \frac{\tan\alpha \pm \tan\beta}{1 \mp \tan\alpha\tan\beta} \end{aligned}$$

Power-Reduction (Half-Angle) Formulas: Crucial for integration in calculus:

$$\sin^2\theta = \frac{1 - \cos(2\theta)}{2}, \qquad \cos^2\theta = \frac{1 + \cos(2\theta)}{2}$$

§2.3 Inverse Trigonometric Functions & Principal Value Branches

1. Restriction of Domains & Principal Branches

Because trigonometric functions are periodic, they fail the horizontal line test across $\mathbb{R}$. To construct inverses, we restrict each function to a standard principal interval on which it is strictly monotonic and surjective onto its range:

$$\begin{array}{l|c|c|l} \text{Function} & \text{Restricted Domain} & \text{Range} & \text{Inverse Function } f^{-1} \\\ \hline \sin x & [-\pi/2, \pi/2] & [-1, 1] & y = \arcsin x \iff x = \sin y, \quad y \in [-\pi/2, \pi/2] \ \cos x & [0, \pi] & [-1, 1] & y = \arccos x \iff x = \cos y, \quad y \in [0, \pi] \ \tan x & (-\pi/2, \pi/2) & (-\infty, \infty) & y = \arctan x \iff x = \tan y, \quad y \in (-\pi/2, \pi/2) \ \sec x & [0, \pi/2) \cup (\pi/2, \pi] & (-\infty, -1] \cup [1, \infty) & y = \operatorname{arcsec} x \iff x = \sec y \end{array}$$

2. Fundamental Identities of Inverse Trigonometric Functions

$$\arcsin(x) + \arccos(x) = \frac{\pi}{2} \quad \forall x \in [-1, 1], \qquad \arctan(x) + \operatorname{arccot}(x) = \frac{\pi}{2} \quad \forall x \in \mathbb{R}$$

Cancellation Cautions:

$$\sin(\arcsin x) = x \quad \forall x \in [-1, 1], \quad \text{but} \quad \arcsin(\sin x) = x \iff x \in [-\pi/2, \pi/2]$$

For arguments outside the principal range, symmetry must be utilized: e.g., $\arcsin(\sin(5\pi/6)) = \arcsin(1/2) = \pi/6 \ne 5\pi/6$.

§2.4 Hyperbolic Functions: Symmetries, Identities & The Unit Hyperbola

1. Formal Definitions of Hyperbolic Functions

The hyperbolic functions are defined as the symmetric and antisymmetric linear combinations of the exponential functions $e^x$ and $e^{-x}$:

$$\begin{aligned} \text{Hyperbolic Sine: } & \sinh x = \frac{e^x - e^{-x}}{2} \quad (\text{Strictly Odd: } \sinh(-x) = -\sinh x) \ \text{Hyperbolic Cosine: } & \cosh x = \frac{e^x + e^{-x}}{2} \quad (\text{Strictly Even: } \cosh(-x) = \cosh x) \ \text{Hyperbolic Tangent: } & \tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{e^{2x} - 1}{e^{2x} + 1} \end{aligned}$$

Reciprocals: $\operatorname{sech} x = \frac{1}{\cosh x}$, $\operatorname{csch} x = \frac{1}{\sinh x}$, $\operatorname{coth} x = \frac{\cosh x}{\sinh x}$.

2. The Fundamental Hyperbolic Identity & The Unit Hyperbola

$$\cosh^2 x - \sinh^2 x = \left(\frac{e^x + e^{-x}}{2}\right)^2 - \left(\frac{e^x - e^{-x}}{2}\right)^2 = \frac{(e^{2x} + 2 + e^{-2x}) - (e^{2x} - 2 + e^{-2x})}{4} = \frac{4}{4} = 1$$

Geometric Analogy: Just as $(\cos t, \sin t)$ parametrizes the unit circle $x^2 + y^2 = 1$, the parametric coordinates $(x, y) = (\cosh t, \sinh t)$ trace the right branch of the unit equilateral hyperbola:

$$x^2 - y^2 = 1 \quad (x \ge 1)$$

Derived Identities:

$$1 - \tanh^2 x = \operatorname{sech}^2 x, \qquad \coth^2 x - 1 = \operatorname{csch}^2 x$$ $$\sinh(2x) = 2\sinh x\cosh x, \qquad \cosh(2x) = \cosh^2 x + \sinh^2 x = 2\cosh^2 x - 1 = 1 + 2\sinh^2 x$$

§2.5 Inverse Hyperbolic Functions & Derivation of Explicit Logarithmic Forms

1. Invertibility of Hyperbolic Functions

  • $\sinh x$ is strictly increasing on $\mathbb{R}$ with range $\mathbb{R}$. Its inverse $\operatorname{arsinh} x$ is defined for all $x \in \mathbb{R}$.
  • $\cosh x$ is even on $\mathbb{R}$ with range $[1, \infty)$. Restricting to $x \ge 0$ yields the principal branch of $\operatorname{arcosh} x$ for $x \ge 1$, with range $[0, \infty)$.
  • $\tanh x$ is strictly increasing with range $(-1, 1)$. Its inverse $\operatorname{artanh} x$ is defined on $(-1, 1)$.

2. Rigorous Derivation of Logarithmic Closed Forms

Theorem: For all $x \in \mathbb{R}$, $\operatorname{arsinh} x = \ln(x + \sqrt{x^2 + 1})$.

Proof: Let $y = \operatorname{arsinh} x$. Then $x = \sinh y = \frac{e^y - e^{-y}}{2}$. Multiplying by $2e^y$ yields:

$$2x e^y = e^{2y} - 1 \implies (e^y)^2 - 2x(e^y) - 1 = 0$$

This is a quadratic equation in $u = e^y$. By the quadratic formula:

$$e^y = \frac{2x \pm \sqrt{4x^2 - 4(1)(-1)}}{2} = x \pm \sqrt{x^2 + 1}$$

Since $e^y > 0$ for all real $y$, and $\sqrt{x^2+1} > \sqrt{x^2} = |x| \ge x$, the minus sign yields a strictly negative value $x - \sqrt{x^2+1} < 0$, which is inadmissible. Thus:

$$e^y = x + \sqrt{x^2 + 1} \implies y = \operatorname{arsinh} x = \ln(x + \sqrt{x^2 + 1}) \quad \blacksquare$$

Complete Logarithmic Catalog:

$$\begin{aligned} \operatorname{arcosh} x &= \ln\left(x + \sqrt{x^2 - 1}\right), \quad x \ge 1 \ \operatorname{artanh} x &= \frac{1}{2}\ln\left(\frac{1 + x}{1 - x}\right), \quad -1 < x < 1 \ \operatorname{arcoth} x &= \frac{1}{2}\ln\left(\frac{x + 1}{x - 1}\right), \quad |x| > 1 \end{aligned}$$
TIERED UNIVERSITY HONORS PROBLEMS

Step-by-Step Solved Examination Problems

Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.

Tier 1: Foundational Example 2.1: Exact Evaluation of Inverse Trigonometric Expressions

Find the exact real values of: (a) $\arcsin\left(\sin\left(\frac{7\pi}{6}\right)\right)$, (b) $\cos\left(2\arcsin\left(-\frac{3}{5}\right)\right)$, and (c) $\tan\left(\arccos\left(\frac{5}{13}\right)\right)$.

Step 1: Evaluate (a) via Principal Branch Interval
$$\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2} \implies \arcsin\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$$

Because $7\pi/6 otin [-\pi/2, \pi/2]$, we cannot simply cancel. We evaluate the inner sine first to get $-1/2$, whose principal arcsine value is $-\pi/6$.

Step 2: Evaluate (b) via Double-Angle Formula
$$\cos(2\theta) = 1 - 2\sin^2\theta \quad \text{where } \theta = \arcsin(-3/5) \implies \sin\theta = -\frac{3}{5}$$

Substitute $\sin heta$: $\cos(2 heta) = 1 - 2\left(-\frac{3}{5}\right)^2 = 1 - 2\left(\frac{9}{25}\right) = 1 - \frac{18}{25} = \frac{7}{25}$.

Step 3: Evaluate (c) via Right Triangle Geometry
$$\alpha = \arccos\left(\frac{5}{13}\right) \implies \cos\alpha = \frac{5}{13}, \quad \sin\alpha = \sqrt{1 - (5/13)^2} = \frac{12}{13}$$

Then $ an\alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{12/13}{5/13} = \frac{12}{5}$.

Final Answer & Physical Insight

\text{(a) } -\frac{\pi}{6}, \qquad \text{(b) } \frac{7}{25}, \qquad \text{(c) } \frac{12}{5}

Tier 2: Intermediate Exam Example 2.2: Derivation of the Explicit Logarithmic Formula for Inverse Hyperbolic Tangent

Let $y = \operatorname{artanh}(x)$ for $x \in (-1, 1)$. (a) Starting from the definition $\tanh(y) = \frac{e^{2y} - 1}{e^{2y} + 1} = x$, derive the closed logarithmic formula $\operatorname{artanh}(x) = \frac{1}{2}\ln\left(\frac{1 + x}{1 - x}\right)$. (b) Use this formula to compute the exact value of $\operatorname{artanh}(3/5)$ and $\operatorname{artanh}(0)$.

Step 1: Set up the Inversion Equation
$$x = \frac{e^{2y} - 1}{e^{2y} + 1} \implies x(e^{2y} + 1) = e^{2y} - 1$$

Expand and isolate the exponential term $e^{2y}$.

Step 2: Solve for $e^{2y}$
$$x e^{2y} + x = e^{2y} - 1 \implies 1 + x = e^{2y}(1 - x) \implies e^{2y} = \frac{1 + x}{1 - x}$$

Since $-1 < x < 1$, both $1+x > 0$ and $1-x > 0$, so the ratio is strictly positive.

Step 3: Take Natural Logarithms
$$2y = \ln\left(\frac{1 + x}{1 - x}\right) \implies y = \operatorname{artanh}(x) = \frac{1}{2}\ln\left(\frac{1 + x}{1 - x}\right)$$

Dividing by 2 yields the exact logarithmic formula.

Step 4: Compute Values
$$\operatorname{artanh}(3/5) = \frac{1}{2}\ln\left(\frac{1 + 3/5}{1 - 3/5}\right) = \frac{1}{2}\ln\left(\frac{8/5}{2/5}\right) = \frac{1}{2}\ln(4) = \ln(2)$$

For $x = 0$: $\operatorname{artanh}(0) = \frac{1}{2}\ln(1/1) = 0$.

Final Answer & Physical Insight

\operatorname{artanh}(x) = \frac{1}{2}\ln\left(\frac{1 + x}{1 - x}\right), \qquad \operatorname{artanh}(3/5) = \ln(2), \qquad \operatorname{artanh}(0) = 0

Tier 3: Honors / Proof Challenge Example 2.3: Hyperbolic Addition Theorem & Ultra-Relativistic Rapidity Invariance

In special relativity, the velocity $v$ of a particle is related to its rapidity $\theta \in \mathbb{R}$ by $v/c = \tanh\theta$. (a) Using the definitions of $\sinh$ and $\cosh$, prove the general hyperbolic addition formula: $\tanh(\theta_1 + \theta_2) = \frac{\tanh\theta_1 + \tanh\theta_2}{1 + \tanh\theta_1 \tanh\theta_2}$. (b) Deduce that the relativistic velocity addition law $v_{12} = \frac{v_1 + v_2}{1 + v_1 v_2 / c^2}$ corresponds to the simple linear addition of rapidities: $\theta_{12} = \theta_1 + \theta_2$. (c) Prove that if $v_1 < c$ and $v_2 < c$, then $v_{12} < c$ strictly.

Step 1: Expand $\sinh( heta_1 + heta_2)$ and $\cosh( heta_1 + heta_2)$
$$\begin{aligned} \sinh(\theta_1 + \theta_2) &= \sinh\theta_1 \cosh\theta_2 + \cosh\theta_1 \sinh\theta_2 \ \cosh(\theta_1 + \theta_2) &= \cosh\theta_1 \cosh\theta_2 + \sinh\theta_1 \sinh\theta_2 \end{aligned}$$

Expanding $e^{( heta_1+ heta_2)} \pm e^{-( heta_1+ heta_2)}$ directly from definition proves these two addition identities.

Step 2: Form the Ratio $ anh( heta_1 + heta_2)$
$$\tanh(\theta_1 + \theta_2) = \frac{\sinh\theta_1 \cosh\theta_2 + \cosh\theta_1 \sinh\theta_2}{\cosh\theta_1 \cosh\theta_2 + \sinh\theta_1 \sinh\theta_2} = \frac{\frac{\sinh\theta_1}{\cosh\theta_1} + \frac{\sinh\theta_2}{\cosh\theta_2}}{1 + \frac{\sinh\theta_1\sinh\theta_2}{\cosh\theta_1\cosh\theta_2}} = \frac{\tanh\theta_1 + \tanh\theta_2}{1 + \tanh\theta_1 \tanh\theta_2}$$

Dividing numerator and denominator by $\cosh heta_1 \cosh heta_2$ establishes the identity.

Step 3: Relate to Velocity Addition and Strict Subluminal Bound
$$\frac{v_{12}}{c} = \tanh(\theta_1 + \theta_2) = \frac{v_1/c + v_2/c}{1 + (v_1/c)(v_2/c)} \implies v_{12} = \frac{v_1 + v_2}{1 + v_1 v_2 / c^2}$$

Because $\operatorname{Range}( anh heta) = (-1, 1)$ for all finite real rapidities $ heta \in (-\infty, \infty)$, the combined velocity $v_{12}/c = anh( heta_1 + heta_2)$ lies strictly in $(-1, 1)$, proving $v_{12} < c$ for all subluminal speeds.

Final Answer & Physical Insight

\tanh(\theta_1 + \theta_2) = \frac{\tanh\theta_1 + \tanh\theta_2}{1 + \tanh\theta_1 \tanh\theta_2}, \quad \theta_{12} = \theta_1 + \theta_2, \quad |v_{12}| < c \text{ strictly}