Trigonometry, Polar Coordinates, and Vector Analysis

Posted on Jun 12, 2026 in Mathematics

Trigonometric Functions and Properties

Even Functions: cos(-t) = cos(t), sec(-t) = sec(t)

Odd Functions: sin(-t) = -sin(t), tan(-t) = -tan(t), csc(-t) = -csc(t), cot(-t) = -cot(t)

Example: A point P(x, y) is shown on the unit circle corresponding to a real number t. Find the values of the trigonometric functions at t.

A) P(-15/17, 8/17):

sin(t) = 8/17
cos(t) = -15/17
tan(t) = -8/15
csc(t) = 17/8
sec(t) = -17/15
cot(t) = -15/8

Graphs of Trigonometric Functions

Amplitude, Period, and Phase Shift:

For the equation y = A sin(Bx):

|A|: Amplitude (height from the x-axis to the curve)
B: Period = 2π/B (one cycle)

For the equation y = A sin(Bx – C):

C/B: Phase shift (the cycle starts at this point; if C/B < 0, shift to the left)
5 Key Points: Intercepts, maximum, and minimum points.

Function Transformations

f(x + 2): Shift left 2 units along the x-axis
f(x) – 3: Shift down 3 units along the y-axis
-f(x): Reflection over the x-axis
f(-x): Reflection over the y-axis
cf(x): Multiply each y-coordinate by c
f(cx): Divide each x-coordinate by c

Inverse Sine and Trigonometric Functions

y = sin^-1(x) means sin(y) = x

y = csc(x) means y = (sin(x))^-1 = 1/sin(x)

Example: Find the exact value of sin^-1(√2/2).

θ = sin^-1(√2/2); sin(θ) = √2/2, where -π/2 ≤ θ ≤ π/2; sin^-1(√2/2) = π/4

Exact Value Examples

A) cos(cos^-1(0.6)): cos(cos^-1(0.6)) = 0.6 (Note: cosine values must be between -1 and 1).

B) sin^-1(sin(3π/2)): sin^-1(sin(3π/2)) = sin^-1(-1) = -π/2

C) Find the exact value of cos(tan^-1(5/12)):

θ = tan^-1(5/12); tan(θ) = 5/12, where -π/2 < θ < π/2. cos(tan^-1(5/12)) = cos(θ) = 12/13

D) Find the exact value of cot[sin^-1(-1/3)]:

θ = sin^-1(-1/3) and sin(θ) = -1/3, where -π/2 < θ < π/2. cot[sin^-1(-1/3)] = cot(θ) = -2√2

Law of Sines and Oblique Triangles

Formula: a/sin(A) = b/sin(B) = c/sin(C)

Case Examples

1. One Solution: Solve triangle ABC if A = 43°, a = 81, and b = 62.

81/sin(43°) = 62/sin(B); 81sin(B) = 62sin(43°); sin(B) = (62sin(43°))/81; sin(B) ≈ 0.5220; B₁ ≈ 31°; B₂ = 180° – 31° = 149°.

C = 180° – B₁ – A ≈ 180° – 31° – 43° = 106°; c/sin(106°) = 81/sin(43°); c = (81sin(106°))/sin(43°) ≈ 114.2

2. No Solution: Solve triangle ABC if A = 75°, a = 51, b = 71.

51/sin(75°) = 71/sin(B); 51sin(B) = 71sin(75°); sin(B) = (71sin(75°))/51 ≈ 1.34. Since sine cannot exceed 1, there is no solution.

3. Two Solutions: When solving for B, if B₁ ≈ 48° and B₂ ≈ 180° – 48° = 132°, and adding either angle to the given angle results in a sum less than 180°, two triangles exist.

Area Formulas and Law of Cosines

Area of an Oblique Triangle: Area = (1/2)bc sin(A) = (1/2)ab sin(C) = (1/2)ac sin(B)

Law of Cosines: a² = b² + c² – 2bc cos(A)

Heron’s Formula for Area: Area = √[s(s – a)(s – b)(s – c)], where s is the semi-perimeter: s = (1/2)(a + b + c).

Polar Coordinates and Conversions

The point P = (r, θ) is located |r| units from the pole.

Multiple Representations: (r, θ) = (r, θ + 2nπ) or (r, θ) = (-r, θ + π + 2nπ)

Example: Find other representations of the point (2, π/3):

a) (2, 7π/3)
b) (-2, 4π/3)
c) (2, -5π/3)

Rectangular and Polar Relationships

x = r cos(θ)
y = r sin(θ)
x² + y² = r²
tan(θ) = y/x

Example (Polar to Rectangular): For point (2, 3π/2): x = 2 cos(3π/2) = 0, y = 2 sin(3π/2) = -2. Answer: (0, -2)

Converting Rectangular to Polar

Plot the point to determine the quadrant.
Find r using: r = √(x² + y²)
Find θ using: tan(θ) = y/x

Example: For point (-1, √3): Point is in Quadrant II. r = 2. tan(θ) = -√3. θ = π – π/3 = 2π/3. Answer: (2, 2π/3)

Equation Conversions

A) x + y = 5: r cos(θ) + r sin(θ) = 5; r(cos(θ) + sin(θ)) = 5; Answer: r = 5/(cos(θ) + sin(θ))

B) (x – 2)² + y² = 1: r²cos²(θ) – 4r cos(θ) + 4 + r²sin²(θ) = 1; r² – 4r cos(θ) + 3 = 0.

C) r = 5: r² = 25; Answer: x² + y² = 25

D) θ = π/4: tan(θ) = tan(π/4); y/x = 1; Answer: y = x

Complex Numbers and DeMoivre’s Theorem

Absolute Value: |z| = |a + bi| = √(a² + b²)

Polar Form: z = r(cos(θ) + i sin(θ)), where r is the modulus and θ is the argument.

Operations in Polar Form

Product: z₁z₂ = r₁r₂[cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)]
Quotient: z₁/z₂ = (r₁/r₂)[cos(θ₁ – θ₂) + i sin(θ₁ – θ₂)]
DeMoivre’s Theorem (Powers): zⁿ = rⁿ[cos(nθ) + i sin(nθ)]
Complex Roots: z_k = ⁿ√r [cos((θ + 2πk)/n) + i sin((θ + 2πk)/n)] for k = 0, 1, …, n-1.

Vector Operations

Unit Vectors: i (positive x-axis) and j (positive y-axis).

Rectangular Form: v = ai + bj. Magnitude ||v|| = √(a² + b²).

Equality: Vectors are equal if they have the same magnitude and direction.
Scalar Multiplication: kv = (ka)i + (kb)j.
Direction: v = ||v||cos(θ)i + ||v||sin(θ)j.
Unit Vector in Direction of v: u = v/||v||.

Trigonometric Identities

Product-to-Sum and Sum-to-Product

sin(a)sin(b) = (1/2)[cos(a – b) – cos(a + b)]
cos(a)cos(b) = (1/2)[cos(a – b) + cos(a + b)]
sin(a) + sin(b) = 2sin[(a + b)/2]cos[(a – b)/2]
cos(a) – cos(b) = -2sin[(a + b)/2]sin[(a – b)/2]

Half-Angle and Double-Angle

sin(a/2) = ±√[(1 – cos(a))/2]
sin(2θ) = 2sin(θ)cos(θ)
cos(2θ) = cos²(θ) – sin²(θ) = 2cos²(θ) – 1 = 1 – 2sin²(θ)
tan(2θ) = (2tan(θ))/(1 – tan²(θ))

Quadratic Relations and Conic Sections

General form: Ax² + Bxy + Cy² + Dx + Ey + F = 0

Classification (B = 0)

Circle: A = C
Ellipse: A and C have the same sign, A ≠ C
Parabola: A or C is zero
Hyperbola: A and C have different signs

Classification (B ≠ 0) using Discriminant (B² – 4AC)

Ellipse: Discriminant < 0
Hyperbola: Discriminant > 0
Parabola: Discriminant = 0

Advanced Practice Problems

Solving Equations: sin(2x) + cos(x) = 0 for 0 ≤ x < 2π. Solutions: π/2, 3π/2, 7π/6, 11π/6.
Multiple Angles: Express cos(3θ) as cos³(θ) – 3sin²(θ)cos(θ).
Tower Height: Two people 1100 feet apart observe a tower at angles 34° and 43.39°. Using the Law of Sines, the height is approximately 433 feet.
Heron’s Formula Application: A triangular lot (320′ x 510′ x 410′) at $4.50/sq ft costs approximately $294,968.
Inverse Composition: sec(cos^-1(1/x)) = x.