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Chapter 10 — Further Applications of Trigonometry

Trigonometry reaches far beyond right triangles. This chapter covers the Law of Sines and Law of Cosines for any triangle, polar coordinates as an alternative to Cartesian, the polar form of complex numbers (and De Moivre’s Theorem), parametric equations, and vectors — the language of physics and engineering.


Table of Contents


Glossary

Term Definition
Law of Sines $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$
Law of Cosines $c^2 = a^2 + b^2 - 2ab\cos C$
Polar coordinates Point described by $(r, \theta)$ — distance and angle from origin
Modulus $|z| = \sqrt{a^2 + b^2}$ for complex number $z = a + bi$
De Moivre’s Theorem $[r(\cos\theta + i\sin\theta)]^n = r^n(\cos n\theta + i\sin n\theta)$
Parametric equations $x = f(t)$, $y = g(t)$ where $t$ is a parameter
Vector A quantity with both magnitude and direction
Dot product $\mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 = |\mathbf{u}||\mathbf{v}|\cos\theta$

1 — Non-Right Triangles: Law of Sines

1.1 The Law of Sines

For any triangle with sides $a, b, c$ opposite angles $A, B, C$:

\[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\]

Use when you know:

  • AAS (two angles + non-included side)
  • ASA (two angles + included side)
  • SSA (two sides + non-included angle) — ambiguous case

Example (AAS): In triangle $ABC$, $A = 42°$, $B = 73°$, $a = 20$. Find $b$.

$C = 180° - 42° - 73° = 65°$

$\dfrac{20}{\sin 42°} = \dfrac{b}{\sin 73°}$

$b = \dfrac{20\sin 73°}{\sin 42°} \approx \dfrac{20(0.9563)}{0.6691} \approx 28.59$

1.2 The Ambiguous Case (SSA)

Given two sides and an angle opposite one of them (SSA), there may be:

  • No solution — the side is too short to form a triangle
  • One solution — exactly one triangle exists
  • Two solutions — two different triangles satisfy the given information

Procedure: Use the Law of Sines to find $\sin B$. If $\sin B > 1$, no solution. If $\sin B = 1$, one right triangle. If $0 < \sin B < 1$, check if the supplement $B’ = 180° - B$ also yields a valid triangle (i.e., $A + B’ < 180°$).

Example (Two solutions): $a = 10$, $b = 12$, $A = 40°$.

$\sin B = \dfrac{b\sin A}{a} = \dfrac{12\sin 40°}{10} = \dfrac{12(0.6428)}{10} \approx 0.7714$

$B_1 \approx 50.5°$ and $B_2 = 180° - 50.5° = 129.5°$

Check: $A + B_2 = 40° + 129.5° = 169.5° < 180°$ ✓ — both valid!

Solution 1: $B \approx 50.5°$, $C \approx 89.5°$
Solution 2: $B \approx 129.5°$, $C \approx 10.5°$

1.3 Area Using Sine

\[\text{Area} = \frac{1}{2}ab\sin C = \frac{1}{2}bc\sin A = \frac{1}{2}ac\sin B\]

(Half the product of two sides times the sine of the included angle.)


2 — Non-Right Triangles: Law of Cosines

2.1 The Law of Cosines

\(a^2 = b^2 + c^2 - 2bc\cos A\) \(b^2 = a^2 + c^2 - 2ac\cos B\) \(c^2 = a^2 + b^2 - 2ab\cos C\)

Use when you know:

  • SAS (two sides + included angle)
  • SSS (three sides)

Note: When $C = 90°$, this reduces to the Pythagorean Theorem.

Example (SAS): $a = 8$, $b = 11$, $C = 60°$. Find $c$.

$c^2 = 64 + 121 - 2(8)(11)\cos 60° = 185 - 176(0.5) = 185 - 88 = 97$

$c = \sqrt{97} \approx 9.85$

Example (SSS — find angles): $a = 5$, $b = 7$, $c = 10$.

$\cos C = \dfrac{a^2 + b^2 - c^2}{2ab} = \dfrac{25 + 49 - 100}{70} = \dfrac{-26}{70} \approx -0.3714$

$C \approx \cos^{-1}(-0.3714) \approx 111.8°$

2.2 Heron’s Formula

For a triangle with sides $a, b, c$, let the semi-perimeter be:

\[s = \frac{a + b + c}{2}\]

Then the area is:

\(\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}\)

Example: $a = 7$, $b = 9$, $c = 12$.

$s = \dfrac{7 + 9 + 12}{2} = 14$

Area $= \sqrt{14(14-7)(14-9)(14-12)} = \sqrt{14 \cdot 7 \cdot 5 \cdot 2} = \sqrt{980} \approx 31.3$

2.3 Which Law to Use?

Given Use
AAS or ASA Law of Sines
SSA Law of Sines (check ambiguous case)
SAS Law of Cosines
SSS Law of Cosines (then Law of Sines for remaining angles)

3 — Polar Coordinates

3.1 Plotting Polar Points

A point in polar coordinates is written $(r, \theta)$ where:

  • $r$ = directed distance from the origin (pole)
  • $\theta$ = angle from the positive $x$-axis (polar axis)

If $r < 0$, the point is plotted in the opposite direction.

Unlike Cartesian, polar representations are not unique:

$(r, \theta) = (r, \theta + 2n\pi) = (-r, \theta + \pi + 2n\pi)$

3.2 Converting Between Polar and Rectangular

Polar → Rectangular:

\[x = r\cos\theta, \qquad y = r\sin\theta\]

Rectangular → Polar:

\[r = \sqrt{x^2 + y^2}, \qquad \tan\theta = \frac{y}{x}\]

(Choose $\theta$ in the correct quadrant.)

Convert $(3, \pi/6)$ to rectangular.

$x = 3\cos\dfrac{\pi}{6} = 3 \cdot \dfrac{\sqrt{3}}{2} = \dfrac{3\sqrt{3}}{2}$

$y = 3\sin\dfrac{\pi}{6} = 3 \cdot \dfrac{1}{2} = \dfrac{3}{2}$

Rectangular: $\left(\dfrac{3\sqrt{3}}{2}, \dfrac{3}{2}\right)$

Convert $(-3, 3)$ to polar.

$r = \sqrt{9 + 9} = 3\sqrt{2}$

$\tan\theta = \dfrac{3}{-3} = -1$. Since the point is in Q II: $\theta = \dfrac{3\pi}{4}$.

Polar: $\left(3\sqrt{2}, \dfrac{3\pi}{4}\right)$

3.3 Polar Equations

Converting equations:

| Rectangular | Polar | |————-|——-| | $x^2 + y^2 = a^2$ | $r = a$ (circle) | | $y = x$ | $\theta = \pi/4$ (line) | | $x = a$ | $r\cos\theta = a$ | | $y = a$ | $r\sin\theta = a$ | | $x^2 + y^2 = ax$ | $r = a\cos\theta$ (circle through origin) | | $x^2 + y^2 = ay$ | $r = a\sin\theta$ (circle through origin) |


4 — Polar Graphs

4.1 Classic Polar Curves

| Equation | Shape | Notes | |———-|——-|——-| | $r = a$ | Circle centered at origin | Radius $a$ | | $\theta = \alpha$ | Line through origin | Angle $\alpha$ | | $r = a\cos\theta$ | Circle | Center $(a/2, 0)$, diameter $a$ | | $r = a\sin\theta$ | Circle | Center $(0, a/2)$, diameter $a$ | | $r = a + b\cos\theta$ | Limaçon | $a/b < 1$: inner loop; $= 1$: cardioid; $> 1$: dimpled or convex | | $r = a + a\cos\theta$ | Cardioid | Heart-shaped, passes through origin | | $r = a\cos(n\theta)$ | Rose | $n$ petals if $n$ odd; $2n$ petals if $n$ even | | $r^2 = a^2\cos(2\theta)$ | Lemniscate | Figure-eight, centered at origin | | $r = a\theta$ | Archimedean spiral | Distance increases linearly with angle |

Graphing strategy for polar curves:

  1. Make a $\theta/r$ table with values at $0, \pi/6, \pi/4, \pi/3, \pi/2, \ldots$
  2. Identify symmetry (symmetric about polar axis if $f(-\theta) = f(\theta)$; about $\theta = \pi/2$ if $f(\pi - \theta) = f(\theta)$; about origin if $f(\theta + \pi) = f(\theta)$ or $f(-\theta) = -f(\theta)$)
  3. Find where $r = 0$ (curve passes through origin)
  4. Find maximum $ r $

Classic Polar Curves — Visual Gallery


5 — Polar Form of Complex Numbers

5.1 Complex Plane and Modulus

A complex number $z = a + bi$ is plotted in the complex plane with $a$ on the horizontal (real) axis and $b$ on the vertical (imaginary) axis.

Modulus (absolute value):

\(|z| = \sqrt{a^2 + b^2}\)

5.2 Polar (Trigonometric) Form

\[z = r(\cos\theta + i\sin\theta) = r\,\text{cis}\,\theta\]

where $r = |z|$ and $\theta = \arg(z) = \tan^{-1}(b/a)$ (adjusted for quadrant).

Write $z = -1 + i\sqrt{3}$ in polar form.

$r = \sqrt{1 + 3} = 2$

$\tan\theta = \dfrac{\sqrt{3}}{-1}$. Since Q II: $\theta = \dfrac{2\pi}{3}$.

$z = 2!\left(\cos\dfrac{2\pi}{3} + i\sin\dfrac{2\pi}{3}\right)$

5.3 Multiplying and Dividing in Polar Form

If $z_1 = r_1\,\text{cis}\,\theta_1$ and $z_2 = r_2\,\text{cis}\,\theta_2$:

\[z_1 z_2 = r_1 r_2\,\text{cis}(\theta_1 + \theta_2)\] \[\frac{z_1}{z_2} = \frac{r_1}{r_2}\,\text{cis}(\theta_1 - \theta_2)\]

Multiply: multiply moduli, add arguments.
Divide: divide moduli, subtract arguments.

5.4 De Moivre’s Theorem

\[[r(\cos\theta + i\sin\theta)]^n = r^n(\cos n\theta + i\sin n\theta)\]

for any positive integer $n$.

Compute $(1 + i)^8$.

$r = \sqrt{2}$, $\theta = \pi/4$.

$(1 + i)^8 = (\sqrt{2})^8\left(\cos\dfrac{8\pi}{4} + i\sin\dfrac{8\pi}{4}\right) = 16(\cos 2\pi + i\sin 2\pi) = 16(1 + 0i) = 16$

5.5 Finding nth Roots

The $n$ distinct $n$th roots of $z = r\,\text{cis}\,\theta$ are:

\[z_k = r^{1/n}\,\text{cis}\!\left(\frac{\theta + 2k\pi}{n}\right), \qquad k = 0, 1, 2, \ldots, n-1\]

The roots are equally spaced around a circle of radius $r^{1/n}$, separated by $\dfrac{2\pi}{n}$ radians.

Find the cube roots of $8$.

$8 = 8\,\text{cis}\,0$, so $r = 8$, $\theta = 0$, $n = 3$.

$z_k = 8^{1/3}\,\text{cis}!\left(\dfrac{0 + 2k\pi}{3}\right) = 2\,\text{cis}!\left(\dfrac{2k\pi}{3}\right)$

$z_0 = 2\,\text{cis}\,0 = 2$

$z_1 = 2\,\text{cis}\,\dfrac{2\pi}{3} = 2!\left(-\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2}\right) = -1 + i\sqrt{3}$

$z_2 = 2\,\text{cis}\,\dfrac{4\pi}{3} = -1 - i\sqrt{3}$


6 — Parametric Equations

6.1 Parametric Curves

A parametric curve is defined by:

\[x = f(t), \qquad y = g(t)\]

where $t$ is a parameter (often representing time). As $t$ varies, the point $(x, y)$ traces out a curve in the plane.

6.2 Eliminating the Parameter

To obtain a rectangular equation, solve for $t$ in one equation and substitute into the other — or use identities.

Example: $x = 2t + 1$, $y = t^2$.

From $x = 2t + 1$: $t = \dfrac{x - 1}{2}$.

$y = \left(\dfrac{x - 1}{2}\right)^2 = \dfrac{(x - 1)^2}{4}$

This is a parabola opening upward with vertex at $(1, 0)$.

Example (using trig identity): $x = 3\cos t$, $y = 3\sin t$.

$\cos t = \dfrac{x}{3}$, $\sin t = \dfrac{y}{3}$.

$\cos^2 t + \sin^2 t = 1 \Rightarrow \dfrac{x^2}{9} + \dfrac{y^2}{9} = 1 \Rightarrow x^2 + y^2 = 9$

This is a circle of radius $3$ centered at the origin.

6.3 Parametric Equations for Conics and Projectile Motion

Circle: $x = h + r\cos t$, $y = k + r\sin t$

Ellipse: $x = h + a\cos t$, $y = k + b\sin t$

Projectile Motion (launched at angle $\alpha$ with speed $v_0$ from height $h$):

\(x = (v_0\cos\alpha)t\) \(y = -\frac{1}{2}gt^2 + (v_0\sin\alpha)t + h\)

where $g \approx 9.8\,\text{m/s}^2$ (or $32\,\text{ft/s}^2$).


7 — Vectors

7.1 Introduction to Vectors

A vector has both magnitude (length) and direction.

Notation: $\mathbf{v}$, $\vec{v}$, or $\langle a, b \rangle$.

A scalar has magnitude only (a real number).

Two vectors are equal if they have the same magnitude and direction, regardless of position.

7.2 Component Form

The vector from $P(x_1, y_1)$ to $Q(x_2, y_2)$:

\[\vec{PQ} = \langle x_2 - x_1, y_2 - y_1 \rangle\]

Magnitude:

\[\|\mathbf{v}\| = \sqrt{a^2 + b^2} \quad \text{for } \mathbf{v} = \langle a, b \rangle\]

Direction angle $\theta$:

\(\tan\theta = \frac{b}{a}, \quad \mathbf{v} = \|\mathbf{v}\|\langle\cos\theta, \sin\theta\rangle\)

7.3 Vector Operations

Let $\mathbf{u} = \langle u_1, u_2 \rangle$ and $\mathbf{v} = \langle v_1, v_2 \rangle$, and let $c$ be a scalar.

Operation Formula
Addition $\mathbf{u} + \mathbf{v} = \langle u_1 + v_1, u_2 + v_2 \rangle$
Subtraction $\mathbf{u} - \mathbf{v} = \langle u_1 - v_1, u_2 - v_2 \rangle$
Scalar multiplication $c\mathbf{u} = \langle cu_1, cu_2 \rangle$

Geometric interpretation:

  • Addition: Tip-to-tail method or parallelogram rule.
  • Scalar multiplication: Stretches (if $ c > 1$) or shrinks (if $ c < 1$) the vector; reverses direction if $c < 0$.

7.4 Unit Vectors

A unit vector has magnitude 1. The unit vector in the direction of $\mathbf{v}$ is:

\[\hat{\mathbf{v}} = \frac{\mathbf{v}}{\|\mathbf{v}\|}\]

Standard unit vectors:

\[\mathbf{i} = \langle 1, 0 \rangle, \qquad \mathbf{j} = \langle 0, 1 \rangle\]

Any vector: $\mathbf{v} = \langle a, b \rangle = a\mathbf{i} + b\mathbf{j}$

From magnitude and direction:

\(\mathbf{v} = \|\mathbf{v}\|\cos\theta\,\mathbf{i} + \|\mathbf{v}\|\sin\theta\,\mathbf{j}\)

7.5 The Dot Product

Algebraic definition:

\[\mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2\]

Geometric definition:

\[\mathbf{u} \cdot \mathbf{v} = \|\mathbf{u}\|\|\mathbf{v}\|\cos\theta\]

where $\theta$ is the angle between the vectors.

Angle between vectors:

\(\cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\|\|\mathbf{v}\|}\)

Key properties:

  • Orthogonal (perpendicular): $\mathbf{u} \cdot \mathbf{v} = 0$
  • Commutative: $\mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u}$
  • Distributive: $\mathbf{u} \cdot (\mathbf{v} + \mathbf{w}) = \mathbf{u} \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{w}$
  • $\mathbf{v} \cdot \mathbf{v} = |\mathbf{v}|^2$

Projection of $\mathbf{u}$ onto $\mathbf{v}$:

\[\text{proj}_{\mathbf{v}}\,\mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|^2}\,\mathbf{v}\]

Scalar component:

\(\text{comp}_{\mathbf{v}}\,\mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|}\)

Example: Find the angle between $\mathbf{u} = \langle 3, 4 \rangle$ and $\mathbf{v} = \langle -2, 5 \rangle$.

$\mathbf{u} \cdot \mathbf{v} = (3)(-2) + (4)(5) = -6 + 20 = 14$

$|\mathbf{u}| = \sqrt{9 + 16} = 5$, $|\mathbf{v}| = \sqrt{4 + 25} = \sqrt{29}$

$\cos\theta = \dfrac{14}{5\sqrt{29}} \approx 0.5199$

$\theta \approx 58.7°$


Key Takeaways

  1. Law of Sines works for AAS, ASA, and SSA (ambiguous); Law of Cosines works for SAS and SSS.
  2. Heron’s Formula gives the area from three sides alone using the semi-perimeter.
  3. Polar coordinates $(r, \theta)$ are not unique; learn to convert freely with $x = r\cos\theta$, $y = r\sin\theta$.
  4. Classic polar curves — cardioids, roses, lemniscates, limaçons — have distinctive equation forms.
  5. De Moivre’s Theorem makes powers and roots of complex numbers straightforward in polar form.
  6. The $n$th roots of a complex number are $n$ equally spaced points on a circle.
  7. Parametric equations add a parameter $t$ to decouple $x$ and $y$; eliminate $t$ to get a rectangular equation.
  8. Vectors: add component-wise, find magnitude with Pythagorean theorem, dot product reveals angle and orthogonality.
  9. Projection decomposes a vector into components parallel and perpendicular to another.

Practice Questions

Q1. In triangle $ABC$, $A = 35°$, $B = 82°$, $a = 15$. Find $b$.

Answer:
$C = 180° - 35° - 82° = 63°$

$\dfrac{15}{\sin 35°} = \dfrac{b}{\sin 82°}$

$b = \dfrac{15\sin 82°}{\sin 35°} = \dfrac{15(0.9903)}{0.5736} \approx 25.89$

Q2. Find the area of a triangle with $a = 8$, $b = 13$, $C = 48°$.

Answer:
Area $= \dfrac{1}{2}(8)(13)\sin 48° = 52\sin 48° \approx 52(0.7431) \approx 38.6$

Q3. Convert the polar point $\left(4, \dfrac{5\pi}{3}\right)$ to rectangular coordinates.

Answer:
$x = 4\cos\dfrac{5\pi}{3} = 4 \cdot \dfrac{1}{2} = 2$

$y = 4\sin\dfrac{5\pi}{3} = 4!\left(-\dfrac{\sqrt{3}}{2}\right) = -2\sqrt{3}$

Rectangular: $(2, -2\sqrt{3})$

Q4. Write $z = -2 - 2i$ in polar form.

Answer:
$r = \sqrt{4 + 4} = 2\sqrt{2}$

$\tan\theta = \dfrac{-2}{-2} = 1$. Q III: $\theta = \pi + \dfrac{\pi}{4} = \dfrac{5\pi}{4}$.

$z = 2\sqrt{2}!\left(\cos\dfrac{5\pi}{4} + i\sin\dfrac{5\pi}{4}\right)$

Q5. Find the angle between $\mathbf{u} = \langle 1, -3 \rangle$ and $\mathbf{v} = \langle 4, 2 \rangle$.

Answer:
$\mathbf{u} \cdot \mathbf{v} = 4 - 6 = -2$

$|\mathbf{u}| = \sqrt{10}$, $|\mathbf{v}| = \sqrt{20} = 2\sqrt{5}$

$\cos\theta = \dfrac{-2}{\sqrt{10} \cdot 2\sqrt{5}} = \dfrac{-2}{2\sqrt{50}} = \dfrac{-1}{\sqrt{50}} = \dfrac{-1}{5\sqrt{2}} \approx -0.1414$

$\theta \approx 98.1°$

Q6. Use De Moivre’s Theorem to compute $\left(\cos\dfrac{\pi}{6} + i\sin\dfrac{\pi}{6}\right)^{12}$.

Answer:
$r = 1$, $\theta = \dfrac{\pi}{6}$, $n = 12$.

$= \cos\dfrac{12\pi}{6} + i\sin\dfrac{12\pi}{6} = \cos 2\pi + i\sin 2\pi = 1 + 0i = 1$


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