Comprehensive Study Guide for Differential Calculus: Trigonometry, Limits, Continuity, and Derivatives

Foundations and Scope of Differential Calculus

Calculus is fundamentally structured around breaking a mathematical or physical quantity into smaller pieces to analyze its behavior and reconstruct comprehensive properties:

• Calculus: The broad mathematical discipline of breaking up a continuous entity into infinitesimal pieces to perform analytical operations.
• Differential Calculus: The branch of calculus centered on breaking up an entity into smaller pieces and working directly with those individual, infinitesimal pieces (rates of change, slopes, and local behaviors).
• Integral Calculus: The branch of calculus centered on breaking up an entity into smaller pieces, analyzing those pieces, and integrating (accumulating or summing) them together to determine total quantities (areas, volumes, and accumulated change).

The primary foundational topics encompassed within Differential Calculus include:

• Trigonometry and Functions
• Limits and Continuity
• Derivatives and Differentiation Techniques
• Applications of Derivatives
• Introduction to Definite and Indefinite Integrals

Historical Context and Modern Perspective of Trigonometry

Derived from the Greek language, the word trigonometry literally translates to "measurement of triangles." In its classical formulation, trigonometry dealt strictly with the geometric relationships among the sides and angles of triangles. This geometric framework was essential in ancient and early modern developments in astronomy, navigation, and land surveying.

During the seventeenth century, concurrent with the invention of calculus and the formulation of modern physical sciences, a fundamental shift in perspective occurred. Classic geometric triangle ratios were reconceptualized as analytic functions whose domains consist of the entire set of real numbers. This analytical transition expanded the applications of trigonometry far beyond static triangles to dynamic physical phenomena involving vibrations, periodic cycles, and rotations, including:

• Sound waves
• Light rays and electromagnetic waves
• Planetary orbits
• Vibrating strings
• Pendulums
• Trajectories and orbits of subatomic particles

Fundamental Mathematical Constants

In mathematics, fundamental constants have values strictly fixed by unambiguous definitions rather than arbitrary convention:

• Archimedes' Constant ($\pi$): Defined strictly as the ratio of any circle's circumference ($C$) to its diameter ($d$):   $\pi = \frac{C}{d}$   Numerical approximation: $\pi \approx 3.14159\dots$(Note: Fractions such as $\frac{22}{7}$ are rational approximations, and $180^\circ$ represents its angular degree equivalence, but its true mathematical definition is the ratio $\frac{C}{d}$.)
• Euler's Number ($e$): The base of the natural logarithm, representing continuous growth.
• The Imaginary Unit ($i$): Defined by $i = \sqrt{-1}$, where $i^2 = -1$.
• Pythagoras' Constant ($\sqrt{2}$): The length of the hypotenuse of an isosceles right triangle with unit legs.

Angles, Rotational Geometry, and Angle Measurement

An angle is generated by rotating a ray (half-line) around its fixed endpoint:

• Vertex: The fixed endpoint of the ray.
• Initial Side: The starting position of the ray before rotation.
• Terminal Side: The final position of the ray after rotation.
• Standard Position: An angle positioned on a Cartesian coordinate plane such that its vertex coincides with the origin $(0, 0)$ and its initial side lies along the positive $x$-axis.
• Rotation Direction:
  • Counterclockwise rotation produces positive angles.
  • Clockwise rotation produces negative angles.
• Nomenclature: Angles are standardly labeled using Greek letters such as $\alpha$ (alpha), $\beta$ (beta), and $\theta$ (theta), or uppercase Latin letters such as $A$, $B$, and $C$.
• Coterminal Angles: Two angles that share the exact same initial and terminal sides. For any given angle $\theta$, there are infinitely many coterminal angles given by:   $\theta + n(360^\circ) \quad \text{or} \quad \theta + 2\pi n, \quad n \in \mathbb{Z}$

Degree Measure

A degree represents a rotational unit of measure denoted by the symbol $^\circ$. A single degree ($1^\circ$) is defined precisely as equivalent to a rotation of $\frac{1}{360}$ of a complete circular revolution about the vertex:

• Complete revolution: $360^\circ$
• Half revolution (straight angle): $180^\circ$
• Quarter revolution (right angle): $90^\circ$

Coordinate plane quadrants are numbered counterclockwise as I, II, III, and IV:

• Quadrant I: $0^\circ < \theta < 90^\circ$
• Quadrant II: $90^\circ < \theta < 180^\circ$
• Quadrant III: $180^\circ < \theta < 270^\circ$
• Quadrant IV: $270^\circ < \theta < 360^\circ$

Angle classifications by magnitude:

• Acute Angle: An angle strictly between $0^\circ$ and $90^\circ$.
• Obtuse Angle: An angle strictly between $90^\circ$ and $180^\circ$.
• Complementary Angles: Two positive angles $\alpha$ and $\beta$ whose sum is $90^\circ$ ($\frac{\pi}{2}\,\text{rad}$):   $\alpha + \beta = 90^\circ$
• Supplementary Angles: Two positive angles $\alpha$ and $\beta$ whose sum is $180^\circ$ ($\pi\,\text{rad}$):   $\alpha + \beta = 180^\circ$

Radian Measure and Arc Length

Radian measure relies on the geometry of a central angle whose vertex is at the center of a circle. One radian ($1\,\text{rad}$) is defined as the measure of a central angle $\theta$ that intercepts an arc $s$ equal in length to the radius $r$ of the circle.

Algebraic relationship: $\theta = \frac{s}{r}$ where $\theta$ is measured in radians, $s$ is arc length, and $r$ is radius. Consequently: $s = r\theta$

When an angle measure is given without explicit units, radian measure is universally implied (e.g., $\theta = 2$ denotes $\theta = 2\,\text{radians}$).

Because a complete revolution corresponds to an arc length equal to the full circumference $s = 2\pi r$: $\theta = \frac{2\pi r}{r} = 2\pi\,\text{rad} = 360^\circ$ Dividing both sides by 2 yields: $\pi\,\text{rad} = 180^\circ$ Because $2\pi \approx 6.28$, there are slightly more than 6 radius lengths along the circumference of a complete circle.

Conversion Formulas and Worked Examples

1. Degrees to Radians: Multiply by $\frac{\pi\,\text{rad}}{180^\circ}$
2. Radians to Degrees: Multiply by $\frac{180^\circ}{\pi\,\text{rad}}$

Conversion Examples

• Convert $135^\circ$ to radians:   $135^\circ \times \frac{\pi\,\text{rad}}{180^\circ} = \frac{3\pi}{4}\,\text{rad}$
• Convert $-270^\circ$ to radians:   $-270^\circ \times \frac{\pi\,\text{rad}}{180^\circ} = -\frac{3\pi}{2}\,\text{rad}$
• Convert $30^\circ$ to radians:   $30^\circ \times \frac{\pi}{180^\circ} = \frac{\pi}{6}\,\text{rad}$
• Convert $45^\circ$ to radians:   $45^\circ \times \frac{\pi}{180^\circ} = \frac{\pi}{4}\,\text{rad}$
• Convert $50^\circ$ to radians:   $50^\circ \times \frac{\pi}{180^\circ} = \frac{5\pi}{18}\,\text{rad}$
• Convert $\frac{\pi}{2}\,\text{rad}$ to degrees:   $\frac{\pi}{2}\,\text{rad} \times \frac{180^\circ}{\pi\,\text{rad}} = 90^\circ$
• Convert $2\,\text{rad}$ to degrees:   $2\,\text{rad} \times \frac{180^\circ}{\pi\,\text{rad}} = \frac{360^\circ}{\pi} \approx 114.59^\circ$
• Convert $\frac{\pi}{18}\,\text{rad}$ to degrees:   $\frac{\pi}{18} \times \frac{180^\circ}{\pi} = 10^\circ$
• Convert $4\,\text{rad}$ to degrees:   $4 \times \frac{180^\circ}{\pi} = \frac{720^\circ}{\pi}$

Arc Length and Subtended Angle Problems

• Finding Angle ($\theta$): Given $r = 5\,\text{cm}$ and arc length $s = 15\,\text{cm}$:   $\theta = \frac{s}{r} = \frac{15\,\text{cm}}{5\,\text{cm}} = 3\,\text{rad}$
• Finding Angle with $\pi$: Given $r = \pi\,\text{cm}$ and $s = 2\,\text{cm}$:   $\theta = \frac{s}{r} = \frac{2}{\pi}\,\text{rad}$
• Finding Arc Length ($s$): Given $r = 4\,\text{inches}$ and $\theta = 3.5\,\text{rad}$:   $s = r\theta = (4\,\text{inches})(3.5) = 14\,\text{inches}$
• Finding Arc Length with Degree Input: Given $r = 4\,\text{inches}$ and central angle $\theta = 240^\circ$:   First convert $\theta$ to radians:   $\theta = 240^\circ \times \frac{\pi}{180^\circ} = \frac{4\pi}{3}\,\text{rad}$   Then calculate $s$:   $s = r\theta = (4)\left(\frac{4\pi}{3}\right) = \frac{16\pi}{3}\,\text{inches}$

The Six Trigonometric Functions and Reference Angles

Right Triangle Definitions ($0 < \theta < \frac{\pi}{2}$)

For an acute angle $\theta$ within a right triangle, the three sides are designated relative to $\theta$ as Hypotenuse ($\text{hyp}$), Opposite ($\text{opp}$), and Adjacent ($\text{adj}$):

• Sine: $\sin(\theta) = \frac{\text{opp}}{\text{hyp}}$
• Cosine: $\cos(\theta) = \frac{\text{adj}}{\text{hyp}}$
• Tangent: $\tan(\theta) = \frac{\text{opp}}{\text{adj}}$
• Cosecant: $\csc(\theta) = \frac{\text{hyp}}{\text{opp}}$
• Secant: $\sec(\theta) = \frac{\text{hyp}}{\text{adj}}$
• Cotangent: $\cot(\theta) = \frac{\text{adj}}{\text{opp}}$

All ratios are strictly positive for acute angles ($0^\circ < \theta < 90^\circ$).

Cofunction Relationships for Complementary Angles

Cofunctions of complementary angles are identical:

• $\sin(90^\circ - \theta) = \cos(\theta)$
• $\cos(90^\circ - \theta) = \sin(\theta)$
• $\tan(90^\circ - \theta) = \cot(\theta)$
• $\cot(90^\circ - \theta) = \tan(\theta)$
• $\sec(90^\circ - \theta) = \csc(\theta)$
• $\csc(90^\circ - \theta) = \sec(\theta)$

In radians, replacing $90^\circ$ with $\frac{\pi}{2}$:

• $\sin\left(\frac{\pi}{2} - u\right) = \cos(u)$
• $\cos\left(\frac{\pi}{2} - u\right) = \sin(u)$
• $\tan\left(\frac{\pi}{2} - u\right) = \cot(u)$
• $\cot\left(\frac{\pi}{2} - u\right) = \tan(u)$
• $\sec\left(\frac{\pi}{2} - u\right) = \csc(u)$
• $\csc\left(\frac{\pi}{2} - u\right) = \sec(u)$

Special Right Triangles: $\theta = \frac{\pi}{6}$ ($30^\circ$)

In a $30^\circ\text{--}60^\circ\text{--}90^\circ$ right triangle, the side lengths satisfy: $\text{hyp} = 2 \times \text{opp}, \quad \text{adj} = \sqrt{3} \times \text{opp}$ Setting $\text{opp} = 1$, $\text{adj} = \sqrt{3}$, and $\text{hyp} = 2$:

• $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$
• $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$
• $\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$
• $\csc\left(\frac{\pi}{6}\right) = 2$
• $\sec\left(\frac{\pi}{6}\right) = \frac{2}{\sqrt{3}}$
• $\cot\left(\frac{\pi}{6}\right) = \sqrt{3}$

Notice that $\sin(30^\circ) = \frac{1}{2} = \cos(60^\circ)$ because $30^\circ$ and $60^\circ$ are complementary.

General Angle Definitions (Any Real Angle $\theta$)

Let $\theta$ be an angle in standard position. Let $(x, y)$ be any point on the terminal side of $\theta$ distinct from the origin, with distance $r = \sqrt{x^2 + y^2} > 0$:

• $\sin(\theta) = \frac{y}{r}$
• $\cos(\theta) = \frac{x}{r}$
• $\tan(\theta) = \frac{y}{x}, \quad x \ne 0$
• $\csc(\theta) = \frac{r}{y}, \quad y \ne 0$
• $\sec(\theta) = \frac{r}{x}, \quad x \ne 0$
• $\cot(\theta) = \frac{x}{y}, \quad y \ne 0$

Quadrantal Signs and the CAST Rule

The sign of each trigonometric function depends on the algebraic signs of the coordinates $x$ and $y$:

• Quadrant I (All positive - 'A'): $x > 0, y > 0 \implies$ All six functions are positive.
• Quadrant II (Sine positive - 'S'): $x < 0, y > 0 \implies \sin(\theta)$ and $\csc(\theta)$ are positive; others are negative.
• Quadrant III (Tangent positive - 'T'): $x < 0, y < 0 \implies \tan(\theta)$ and $\cot(\theta)$ are positive; others are negative.
• Quadrant IV (Cosine positive - 'C'): $x > 0, y < 0 \implies \cos(\theta)$ and $\sec(\theta)$ are positive; others are negative.

Reference Angles

For any angle $\theta$ in standard position, its reference angle $\theta'$ is the acute angle formed by the terminal side of $\theta$ and the horizontal $x$-axis ($0 < \theta' < \frac{\pi}{2}$):

• Quadrant I: $\theta' = \theta$
• Quadrant II: $\theta' = 180^\circ - \theta = \pi - \theta$
• Quadrant III: $\theta' = \theta - 180^\circ = \theta - \pi$
• Quadrant IV: $\theta' = 360^\circ - \theta = 2\pi - \theta$

To evaluate any trigonometric function $f(\theta)$: $f(\theta) = \pm f(\theta')$ where the sign ($+\text{ or } -$) is determined by the quadrant of $\theta$ using the CAST rule.

Reference Angle Examples

1. Find reference angle for $\theta = 300^\circ$:    Lies in Quadrant IV: $\theta' = 360^\circ - 300^\circ = 60^\circ$.    $\sin(300^\circ) = -\sin(60^\circ) = -\frac{\sqrt{3}}{2}$
2. Find reference angle for $\theta = 2.3\,\text{rad}$:    Since $\frac{\pi}{2} \approx 1.5708 < 2.3 < \pi \approx 3.1416$, $\theta$ is in Quadrant II:    $\theta' = \pi - 2.3 \approx 3.1416 - 2.3 = 0.8416$
3. Find reference angle and ratios for $\theta = -135^\circ$:    Coterminal with $-135^\circ + 360^\circ = 225^\circ$ (Quadrant III).    $\theta' = 225^\circ - 180^\circ = 45^\circ$
   • $\sin(-135^\circ) = -\sin(45^\circ) = -\frac{1}{\sqrt{2}}$
   • $\cos(-135^\circ) = -\cos(45^\circ) = -\frac{1}{\sqrt{2}}$
   • $\tan(-135^\circ) = +\tan(45^\circ) = 1$
4. Find ratios for $\theta = -210^\circ$:    $-210^\circ + 360^\circ = 150^\circ$ (Quadrant II).    $\theta' = 180^\circ - 150^\circ = 30^\circ$
   • $\sin(-210^\circ) = +\sin(30^\circ) = \frac{1}{2}$
   • $\cos(-210^\circ) = -\cos(30^\circ) = -\frac{\sqrt{3}}{2}$
   • $\tan(-210^\circ) = -\tan(30^\circ) = -\frac{1}{\sqrt{3}}$

Coordinate Evaluation Examples

• Point on Terminal Side: Let $(-3, 4)$ lie on the terminal side of $\theta$:   $x = -3, \quad y = 4, \quad r = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = 5$
  • $\sin(\theta) = \frac{y}{r} = \frac{4}{5}$
  • $\cos(\theta) = \frac{x}{r} = -\frac{3}{5}$
  • $\tan(\theta) = \frac{y}{x} = -\frac{4}{3}$
  • $\csc(\theta) = \frac{r}{y} = \frac{5}{4}$
  • $\sec(\theta) = \frac{r}{x} = -\frac{5}{3}$
  • $\cot(\theta) = \frac{x}{y} = -\frac{3}{4}$
• Quadrant and Value Deduction: Given $\tan(\theta) = -\frac{4}{3}$ and $0 \le \theta \le \pi$:   Since tangent is negative, $\theta$ must lie in Quadrant II. Thus, $x = -3$, $y = 4$, and $r = 5$, yielding the identical ratios as above.
• Quadrant III Ratios: Let $\sin(\theta) = -\frac{7}{15}$ with $\pi \le \theta \le \frac{3\pi}{2}$:   Here $y = -7$, $r = 15$. Because $\theta$ is in Quadrant III, $x < 0$:   $x = -\sqrt{r^2 - y^2} = -\sqrt{15^2 - (-7)^2} = -\sqrt{225 - 49} = -\sqrt{176}$
  • $\cos(\theta) = \frac{x}{r} = -\frac{\sqrt{176}}{15}$
  • $\tan(\theta) = \frac{y}{x} = \frac{-7}{-\sqrt{176}} = \frac{7}{\sqrt{176}}$
  • $\csc(\theta) = \frac{r}{y} = -\frac{15}{7}$
  • $\sec(\theta) = \frac{r}{x} = -\frac{15}{\sqrt{176}}$
  • $\cot(\theta) = \frac{x}{y} = \frac{\sqrt{176}}{7}$

Fundamental Trigonometric Identities and Analytic Simplification

An identity is an equation that remains true for all values of the variable for which the expressions are defined (e.g., $x + 3x + 41 = 4x + 41$ is an identity). In contrast, a standard conditional equation holds true only for specific values (e.g., $x + 3x = 2x + 6$ is true only when $x = 3$).

Comprehensive Catalog of Fundamental Identities

Reciprocal Identities

$\sin(\theta) = \frac{1}{\csc(\theta)}, \quad \csc(\theta) = \frac{1}{\sin(\theta)}$$\cos(\theta) = \frac{1}{\sec(\theta)}, \quad \sec(\theta) = \frac{1}{\cos(\theta)}$$\tan(\theta) = \frac{1}{\cot(\theta)}, \quad \cot(\theta) = \frac{1}{\tan(\theta)}$

Quotient Identities

$\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}, \quad \cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}$

Pythagorean Identities

$\sin^2(\theta) + \cos^2(\theta) = 1$$1 + \tan^2(\theta) = \sec^2(\theta)$$1 + \cot^2(\theta) = \csc^2(\theta)$

Even/Odd Identities

$\sin(-u) = -\sin(u), \quad \csc(-u) = -\csc(u)$$\cos(-u) = \cos(u), \quad \sec(-u) = \sec(u)$$\tan(-u) = -\tan(u), \quad \cot(-u) = -\cot(u)$

Angle Addition and Subtraction Identities

$\sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B)$$\sin(A - B) = \sin(A)\cos(B) - \cos(A)\sin(B)$$\cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B)$$\cos(A - B) = \cos(A)\cos(B) + \sin(A)\sin(B)$$\tan(A + B) = \frac{\tan(A) + \tan(B)}{1 - \tan(A)\tan(B)}$$\tan(A - B) = \frac{\tan(A) - \tan(B)}{1 + \tan(A)\tan(B)}$

Double-Angle Identities

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

Half-Angle Identities

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

Product-to-Sum Identities

$\sin(A)\cos(B) = \frac{1}{2}[\sin(A + B) + \sin(A - B)]$$\sin(A)\sin(B) = \frac{1}{2}[\cos(A - B) - \cos(A + B)]$$\cos(A)\cos(B) = \frac{1}{2}[\cos(A - B) + \cos(A + B)]$

Sum-to-Product Identities

$\sin(A) + \sin(B) = 2\sin\left(\frac{A + B}{2}\right)\cos\left(\frac{A - B}{2}\right)$$\sin(A) - \sin(B) = 2\sin\left(\frac{A - B}{2}\right)\cos\left(\frac{A + B}{2}\right)$$\cos(A) + \cos(B) = 2\cos\left(\frac{A + B}{2}\right)\cos\left(\frac{A - B}{2}\right)$$\cos(A) - \cos(B) = -2\sin\left(\frac{A + B}{2}\right)\sin\left(\frac{A - B}{2}\right)$

Analytical Examples and Identity Proofs

Exact Value of $\sin(15^\circ)$

$\sin(15^\circ) = \sin(45^\circ - 30^\circ) = \sin(45^\circ)\cos(30^\circ) - \cos(45^\circ)\sin(30^\circ)$$= \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right) = \frac{\sqrt{3} - 1}{2\sqrt{2}}$

Compound Evaluation Problem

If $\sin(A) = -\frac{1}{3}$ and $\tan(B) = \frac{5}{12}$, where $A, B \in \left[\frac{\pi}{2}, \frac{3\pi}{2}\right]$, evaluate $\cos(2A + B)$.

1. Determine Quadrants:
   • Interval is Quadrants II and III.
   • $\sin(A) < 0 \implies A$ is in Quadrant III: $y = -1, r = 3, x = -\sqrt{3^2 - (-1)^2} = -2\sqrt{2}$.      $\cos(A) = -\frac{2\sqrt{2}}{3}$
   • $\tan(B) > 0 \implies B$ is in Quadrant III: $x = -12, y = -5, r = \sqrt{(-12)^2 + (-5)^2} = 13$.      $\sin(B) = -\frac{5}{13}, \quad \cos(B) = -\frac{12}{13}$
2. Evaluate Components:    $\cos(2A) = 2\cos^2(A) - 1 = 2\left(-\frac{2\sqrt{2}}{3}\right)^2 - 1 = 2\left(\frac{8}{9}\right) - 1 = \frac{7}{9}$$\sin(2A) = 2\sin(A)\cos(A) = 2\left(-\frac{1}{3}\right)\left(-\frac{2\sqrt{2}}{3}\right) = \frac{4\sqrt{2}}{9}$
3. Apply Addition Identity:    $\cos(2A + B) = \cos(2A)\cos(B) - \sin(2A)\sin(B)$$= \left(\frac{7}{9}\right)\left(-\frac{12}{13}\right) - \left(\frac{4\sqrt{2}}{9}\right)\left(-\frac{5}{13}\right) = \frac{-84 + 20\sqrt{2}}{117} = \frac{20\sqrt{2} - 84}{117}$

Algebraic Simplification

Simplify $\sin(2A)\csc(A) - \cos(2A)\sec(A)$: $= \frac{\sin(2A)}{\sin(A)} - \frac{\cos(2A)}{\cos(A)} = \frac{\sin(2A)\cos(A) - \cos(2A)\sin(A)}{\sin(A)\cos(A)}$ Applying the subtraction identity to the numerator: $= \frac{\sin(2A - A)}{\sin(A)\cos(A)} = \frac{\sin(A)}{\sin(A)\cos(A)} = \frac{1}{\cos(A)} = \sec(A)$

Verification of Identity: $(\sec^2(\theta) - 1)\cos^2(\theta) = \sin^2(\theta)$

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

Identity Proof: $\sec(x) - \cos(x) = \tan(x)\sin(x)$

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

Identity Proof: $\sin(x + y)\sin(x - y) = \sin^2(x) - \sin^2(y)$

$\sin(x + y)\sin(x - y) = [\sin(x)\cos(y) + \cos(x)\sin(y)][\sin(x)\cos(y) - \cos(x)\sin(y)]$$= \sin^2(x)\cos^2(y) - \cos^2(x)\sin^2(y)$ Express completely in terms of sines using $\cos^2(\theta) = 1 - \sin^2(\theta)$: $= \sin^2(x)(1 - \sin^2(y)) - (1 - \sin^2(x))\sin^2(y)$$= \sin^2(x) - \sin^2(x)\sin^2(y) - \sin^2(y) + \sin^2(x)\sin^2(y) = \sin^2(x) - \sin^2(y)$

Exact Trigonometric Calculations Without a Calculator

• Problem: If $\cos(\theta) = \frac{3}{4}$ and $\pi \le \theta \le 2\pi$, find $\tan(\theta)$ and $\sin(2\theta)$.   Since $\cos(\theta) > 0$ and $\theta \in [\pi, 2\pi]$, $\theta$ lies strictly in Quadrant IV.   $\sin(\theta) = -\sqrt{1 - \cos^2(\theta)} = -\sqrt{1 - \frac{9}{16}} = -\frac{\sqrt{7}}{4}$$\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{-\frac{\sqrt{7}}{4}}{\frac{3}{4}} = -\frac{\sqrt{7}}{3}$$\sin(2\theta) = 2\sin(\theta)\cos(\theta) = 2\left(-\frac{\sqrt{7}}{4}\right)\left(\frac{3}{4}\right) = -\frac{3\sqrt{7}}{8}$

• Problem: If $\tan(x) = \frac{1}{3}$ and $\sec(y) = \frac{13}{12}$, where $x$ and $y$ lie between $\pi$ and $2\pi$, evaluate $\csc(x - y)$ and $\cos(2x)$.

  1. Quadrant of $x$: $\tan(x) = \frac{1}{3} > 0$ and $x \in (\pi, 2\pi) \implies x$ is in Quadrant III.      $x_{\text{coord}} = -3, \quad y_{\text{coord}} = -1, \quad r = \sqrt{(-3)^2 + (-1)^2} = \sqrt{10}$$\sin(x) = -\frac{1}{\sqrt{10}}, \quad \cos(x) = -\frac{3}{\sqrt{10}}$
  2. Quadrant of $y$: $\sec(y) = \frac{13}{12} > 0 \implies \cos(y) = \frac{12}{13} > 0$. For $y \in (\pi, 2\pi)$, $y$ is in Quadrant IV.      $\sin(y) = -\sqrt{1 - \left(\frac{12}{13}\right)^2} = -\frac{5}{13}$
  3. Evaluate $\sin(x - y)$:      $\sin(x - y) = \sin(x)\cos(y) - \cos(x)\sin(y)$$= \left(-\frac{1}{\sqrt{10}}\right)\left(\frac{12}{13}\right) - \left(-\frac{3}{\sqrt{10}}\right)\left(-\frac{5}{13}\right) = -\frac{12}{13\sqrt{10}} - \frac{15}{13\sqrt{10}} = -\frac{27}{13\sqrt{10}}$$\csc(x - y) = \frac{1}{\sin(x - y)} = -\frac{13\sqrt{10}}{27}$
  4. Evaluate $\cos(2x)$:      $\cos(2x) = 2\cos^2(x) - 1 = 2\left(-\frac{3}{\sqrt{10}}\right)^2 - 1 = 2\left(\frac{9}{10}\right) - 1 = \frac{18}{10} - 1 = \frac{4}{5}$

• Problem: Evaluate $\tan\left(\sin^{-1}\left(\frac{1}{3}\right)\right)$.   Let $\theta = \sin^{-1}\left(\frac{1}{3}\right)$, so $\sin(\theta) = \frac{1}{3}$ with $\theta \in \left(0, \frac{\pi}{2}\right)$.   Using a reference right triangle with $\text{opp} = 1$ and $\text{hyp} = 3$:   $\text{adj} = \sqrt{3^2 - 1^2} = \sqrt{8} = 2\sqrt{2}$$\tan(\theta) = \frac{\text{opp}}{\text{adj}} = \frac{1}{2\sqrt{2}}$

Solving Trigonometric Equations

Systematic Solution Strategy

1. Use trigonometric identities to write the equation in terms of a single trigonometric function.
2. Isolate that function using algebraic operations (factoring, collecting terms, quadratic formula). Suppose $\sin(x) = a$.
3. Ignore the algebraic sign of $a$ and solve $\sin(x) = |a|$ to establish the acute reference angle $x' \in \left[0, \frac{\pi}{2}\right]$.
4. Use the original sign of $a$ and the CAST rule to identify all quadrants where the equation is satisfied.
5. Compute all quadrant solutions within the specified interval.

Linear Equations

• Solve: $2\sin(x) - 1 = 0$ for $x \in [0, 2\pi]$:   $2\sin(x) = 1 \implies \sin(x) = \frac{1}{2}$   Reference angle: $x' = \frac{\pi}{6}$. Since $\sin(x)$ is positive, solutions lie in Quadrants I and II:   $x = \frac{\pi}{6}, \quad x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$

• Solve: $2\sin(x) + 1 = 0$ for $x \in [0, 2\pi]$:   $2\sin(x) = -1 \implies \sin(x) = -\frac{1}{2}$   Reference angle: $x' = \frac{\pi}{6}$. Since $\sin(x)$ is negative, solutions lie in Quadrants III and IV:   $x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}, \quad x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$

Quadratic-Type Trigonometric Equations

Trigonometric equations often appear in quadratic form $au^2 + bu + c = 0$, where $u$ is a trigonometric function.

• Solve: $2\cos^2(x) = 1 - \sin(x)$ on $[0, 2\pi]$:   Rewrite $\cos^2(x)$ as $1 - \sin^2(x)$:   $2(1 - \sin^2(x)) = 1 - \sin(x)$$2 - 2\sin^2(x) = 1 - \sin(x)$$2\sin^2(x) - \sin(x) - 1 = 0$   Factor the quadratic:   $(2\sin(x) + 1)(\sin(x) - 1) = 0$   Set each factor to zero:

  1. $2\sin(x) + 1 = 0 \implies \sin(x) = -\frac{1}{2} \implies x = \frac{7\pi}{6}, \frac{11\pi}{6}$
  2. $\sin(x) - 1 = 0 \implies \sin(x) = 1 \implies x = \frac{\pi}{2}$   Full solution set: $x = \frac{\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6}$

• Solve: $\sin^2(\theta) + 2\cos^2(\theta) = 1$ for $0 \le \theta \le 2\pi$:   Decompose the terms:   $\sin^2(\theta) + \cos^2(\theta) + \cos^2(\theta) = 1$$1 + \cos^2(\theta) = 1 \implies \cos^2(\theta) = 0 \implies \cos(\theta) = 0$   Within $[0, 2\pi]$:   $\theta = \frac{\pi}{2}, \frac{3\pi}{2}$

• Solve: $5\cos(x) - 2\sin^2(x) = 1$ on $[0, 2\pi]$:   Substitute $\sin^2(x) = 1 - \cos^2(x)$:   $5\cos(x) - 2(1 - \cos^2(x)) = 1$$5\cos(x) - 2 + 2\cos^2(x) = 1$$2\cos^2(x) + 5\cos(x) - 3 = 0$   Factor the quadratic expression:   $(2\cos(x) - 1)(\cos(x) + 3) = 0$   Setting each factor to zero:

  1. $\cos(x) + 3 = 0 \implies \cos(x) = -3$ (No solution, since $-1 \le \cos(x) \le 1$).
  2. $2\cos(x) - 1 = 0 \implies \cos(x) = \frac{1}{2}$.      Within $[0, 2\pi]$, cosine is positive in Quadrants I and IV:      $x = \frac{\pi}{3}, \quad x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$

Multiple-Angle Equations

• Solve: $2\cos(3t) - 1 = 0$ where $t \in [0, 2\pi]$:   $2\cos(3t) = 1 \implies \cos(3t) = \frac{1}{2}$   If $t \in [0, 2\pi]$, the argument $3t$ spans three full revolutions: $3t \in [0, 6\pi]$.   The reference angle is $\frac{\pi}{3}$. In Quadrants I and IV:
  • Cycle 1 ($[0, 2\pi]$): $3t = \frac{\pi}{3}, \frac{5\pi}{3}$
  • Cycle 2 ($[2\pi, 4\pi]$): $3t = 2\pi + \frac{\pi}{3} = \frac{7\pi}{3}, \quad 3t = 4\pi - \frac{\pi}{3} = \frac{11\pi}{3}$
  • Cycle 3 ($[4\pi, 6\pi]$): $3t = 4\pi + \frac{\pi}{3} = \frac{13\pi}{3}, \quad 3t = 6\pi - \frac{\pi}{3} = \frac{17\pi}{3}$   Dividing each value by 3:   $t = \frac{\pi}{9}, \frac{5\pi}{9}, \frac{7\pi}{9}, \frac{11\pi}{9}, \frac{13\pi}{9}, \frac{17\pi}{9}$