Trigonometry: Functions, Laws, and Applications

Posted on Mar 20, 2025 in Computing and Communications

Trigonometric Functions

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:

y = Asin(Bx):

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

y = Asin(Bx – C):

  • C/B: Phase shift (the cycle starts at this point; if C/B < 0, shift to the left)
  • x-values for the 5 key points: intercepts, maximum, and minimum points.

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 Function

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

Find the exact value, if possible:

A) cos(cos-1(0.6)); cos(cos-1(0.6)) = 0.6 (cos cannot be more than 1)

B) sin-1(sin(3π/2)); sin-1(sin(3π/2)) = sin-1(-1) = -π/2

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

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

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

Examples:

1 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

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.

Two Solutions: When solving for B: B1 ≈ 48° and B2 ≈ 180° – 48° = 132°. When adding either angle to the given angle, the sum does not exceed 180°, thus there are two triangles. Solve for both.

Area of an Oblique Triangle

Area = (1/2)bcsin(A) = (1/2)absin(C) = (1/2)acsin(B)

Law of Cosines

a² = b² + c² – 2bccos(A)

Heron’s Formula for the Area of a Triangle:

Area = √[s(s – a)(s – b)(s – c)], where a, b, and c are the sides and s = (1/2)(a + b + c) (semi-perimeter)

Polar Coordinates

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

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

Example: Find another representation of the point (2, π/3):

  • a) (2, π/3) = (2, π/3 + 2π) = (2, 7π/3)
  • b) (2, π/3) = (-2, π/3 + π) = (-2, 4π/3)
  • c) (2, π/3) = (2, π/3 – 2π) = (2, -5π/3)

Relationship between Polar and Rectangular Coordinates:

  • x = rcos(θ)
  • y = rsin(θ)
  • x² + y² = r²
  • tan(θ) = y/x

Example: Polar to Rectangular: Find the rectangular coordinate of the point (2, 3π/2):

x = 2cos(3π/2) = 2 * 0 = 0, y = 2sin(3π/2) = 2(-1) = -2. Answer: (0, -2)

Converting from Rectangular to Polar Coordinates:

  1. Plot the point.
  2. Find r using: r = √(x² + y²)
  3. Find θ using: tan(θ) = y/x

Example: Find the polar coordinate of the point (-1, √3):

1st) Plotting the point, we notice that it lies in Quadrant II. 2nd) Find r = 2. Find θ: tan(θ) = -√3. We know that π/3 = √3. Because θ lies in Quadrant II, θ = π – π/3 = 2π/3. Answer: (2, 2π/3)

Converting Equations from Rectangular to Polar Coordinates:

Examples: Convert each rectangular equation:

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

B) (x – 2)² + y² = 1 → (rcos(θ) – 1)² + (rsin(θ))² = 1 → r²cos²(θ) – 2rcos(θ) + 1 + r²sin²(θ) = 1 → r²cos²(θ) + r²sin²(θ) – 2rcos(θ) = 0 → r² – 2rcos(θ) = 0 → r(r – 2cos(θ)) = 0 → r = 0 or r = 2cos(θ)

Converting Equations from Polar to Rectangular Form:

Examples: Convert:

A) r = 5 → r² = 25 → Answer: x² + y² = 25

B) θ = π/4 → tan(θ) = tan(π/4) → tan(θ) = 1 → y/x = 1 → Answer: y = x

C) r = 3csc(θ) → r = 3/sin(θ) → rsin(θ) = 3 → Answer: y = 3

D) r = -6cos(θ) → r² = -6rcos(θ) → x² + y² = -6x → x² + 6x + y² = 0 → x² + 6x + 9 + y² = 9 → Answer: (x + 3)² + y² = 9

Complex Numbers in Polar Form; DeMoivre’s Theorem

Plotting each complex number: (z = a + bi) → a = x, b = y

Absolute Value of a Complex Number: (distance from origin to the point) (a + bi): |z| = |a + bi| = √(a² + b²)

Polar Form of a Complex Number (z = a + bi): z = r(cos(θ) + isin(θ)); a = rcos(θ), b = rsin(θ), r = √(a² + b²) (modulus), tan(θ) = b/a → θ = argument (in radians)

(Use exact values of θ when writing complex numbers in rectangular form)

Product of Two Complex Numbers in Polar Form: [z1 = r1(cos(θ1) + isin(θ1)) and z2 = r2(cos(θ2) + isin(θ2))]:

z1z2 = r1r2[cos(θ1 + θ2) + isin(θ1 + θ2)] (Multiply moduli (r) and add arguments (θ); leave θ in degree form) i² = -1

Quotient of Two Complex Numbers in Polar Form: [z1 = r1(cos(θ1) + isin(θ1)) and z2 = r2(cos(θ2) + isin(θ2))]:

z1/z2 = r1/r2[cos(θ1 – θ2) + isin(θ1 – θ2)] (Divide moduli (r) and subtract arguments (θ))

Power of Complex Numbers (DeMoivre’s Theorem): [z = r(cos(θ) + isin(θ))]: “n is a positive integer” zn = [r(cos(θ) + isin(θ))]n = rn(cos(nθ) + isin(nθ))

(If degrees are given, switch to absolute values) and (If given in rectangular form, switch to radians)

DeMoivre’s Theorem for Finding Complex Roots: [w = r(cos(θ) + isin(θ))]: “w ≠ 0, w has n distinct complex nth roots given by the formula”:

zk = n√r[cos((θ + 2πk)/n) + isin((θ + 2πk)/n)] (radians) or zk = n√r[cos((θ + 360°k)/n) + isin((θ + 360°k)/n)] (degrees) → k = 0, 1, 2, 3, …, n – 1

Examples: Find all complex roots; write in rectangular form: 5th root of 32; (0, 1, 2, 3, 4) → 32 + i0 = 32(cos(0) + isin(0)) → (0) 5√32(cos(0 + 360°(0)/5) + isin(0 + 360°(0)/5)) → 2(cos(0) + isin(0)) → 2… Continue with the rest: 1, 2, 3, and 4

Vectors

i and j Unit Vectors: i = its direction is along the positive x-axis, j = its direction is along the positive y-axis.

Representing Vectors in Rectangular Coordinates (v from (0, 0) to (a, b): v = ai + bj (a and b are real numbers called scalar components of v, a is horizontal, b is vertical) / The sum ai + bj is called a Linear Combination / The magnitude (length) of v = ai + bj: ||v|| = √(a² + b²).

Showing that 2 vectors are equal: Use the distance formula on both vectors = √[(x2 – x1)² + (y2 – y1)²].

Show that 2 vectors have the same direction: Use the slope formula on both vectors: m = (y2 – y1)/(x2 – x1).

Scalar Multiplication: Multiplying the vector by a real number changes the magnitude, not the direction; kv = (ka)i + (kb)j *k: real number*

Writing a vector in Terms of its Magnitude and Direction (v = ai + bj): cos(θ) = a/||v|| → a = ||v||cos(θ) and sin(θ) = b/||v|| → b = ||v||sin(θ) → v = ai + bj = ||v||cos(θ)i + ||v||sin(θ)j.

Finding the Unit vector that has the same Direction as a Given Nonzero Vector v: v/||v|| (v→i+and–j)(||v||→√(a²+b²)).

Representing Vectors in Rectangular Coordinates: v = ai + bj = (x2 – x1)i + (y2 – y1)j.

Adding and Subtracting Vectors in Terms of i and j (v = a1i + b2j and w = a2i + b2j): v + w = (a1 + a2)i + (b1 + b2)j and v – w = (a1 – a2)i + (b1 – b2)j.

Zero Vector = magnitude of 0 → 0 = 0i + 0j. cis = cos + isin (i = √-1)

Formulas

Product to Sum:

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

Sum to Product:

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

Half-Angle (the ± symbols indicate that I must determine the sign of the function based on the quadrant that it lies in):

  • sin(a/2) = ±√[(1 – cos(a))/2]
  • cos(a/2) = ±√[(1 + cos(a))/2]
  • tan(a/2) = ±√[(1 – cos(a))/(1 + cos(a))] → (1 – cos(a))/sin(a) → sin(a)/(1 + cos(a))

Double-Angle:

  • 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))

Power-Reducing:

  • sin²(θ) = (1 – cos(2θ))/2
  • cos²(θ) = (1 + cos(2θ))/2
  • tan²(θ) = (1 – cos(2θ))/(1 + cos(2θ))

Extra: sin²(x)cos²(x) = (sin(2x)/2)²

Sum and Differences:

  • cos(a+b)=cos(a)cos(b)-sin(a)sin(b)
  • cos(a-b)=cos(a)cos(b)+sin(a)sin(b)
  • sin(a+b)=sin(a)cos(b)+cos(a)sin(b)
  • sin(a-b)=sin(a)cos(b)-cos(a)sin(b).

Quadratic Relations

(Ax² + Bxy + Cy² + Dx + Ey + F = 0):

If B = 0 (no xy term):

  • Circle: A = C (have the same coefficient)
  • Ellipse: A and C have the same sign, but are not equal
  • Parabola: If one of A and C is zero
  • Hyperbola: A and C have different signs

If there is an xy term (define the discriminant of the relation B² – 4AC):

  • Ellipse: if the discriminant is negative
  • Hyperbola: if the discriminant is positive
  • Parabola: if the discriminant is equal to zero

Examples: x² + y² = 3 (circle); x² – 2y² = 3 → x² – 2y² – 3 (hyperbola); 9x² – 6xy + 8y + y² = 3 (parabola); x² + 2y² – 3x + 4y = 3 (ellipse)

Write an equation describing each of the following conic sections:

  • a) An ellipse where the foci are (2, -5) and (2, 3), and where the total distance to the two foci is 12 = {√[(x – 2)² + (y – (-5))²] + √[(x – 2)² + (y – 3)²] = 12}
  • b) A hyperbola where the foci are (2, -5) and (2, 3), and where any point on the hyperbola is 5 units closer to one focus than to the other: |both square roots| = 5
  • c) A parabola where the focus is the point (2, -5), and the directrix is the line x = 6: √[(x – 2)² + (y + 5)²] = |x – 6|

Center, Vertex, Focus:

  • Circle: Complete the square and have the equation in this form: (x – h)² + (y – k)² = r², then the center is (h, k).
  • Parabolas: If the equation has an x² term but no y², put the equation in the term y = ax² + bx + c; the vertex is the point [-b/2a, (4ac – b²)/4a]. If it has a y² but no x², then put the equation in the term x = ay² + by + c; the vertex is the point [(4ac – b²)/4a, -b/2a].
  • Ellipses: Complete the square on both x and y and put the equation in the form a(x – h)² + b(y – k)² = c (a, b, and c need to be positive). If a > b, the major axis is horizontal, and the foci are (h ± c, k), where c = √(a – b). If b > a, the major axis is vertical, and the foci are (h, k ± c), where c = √(b – a).
  • Hyperbolas: Complete the square on both x and y, and put the equation in the form a(x – h)² – b(y – k)² = c or a(y – k)² – b(x – h)² = c (a, b, and c, positive). In either case, let d = √(c/a + c/b). In the first case, the foci are (h ± d, k); in the second case, the foci are (h, k ± d).

Examples:

  • 0 ≤ x < 2π → sin(2x) + cos(x) = 0 → 2sin(x)cos(x) + cos(x) = 0 → cos(x)[2sin(x) + 1] = 0 → cos(x) = 0: (3π/2) and (π/2), sin(x) = -1/2: (11π/6) and (7π/6)
  • Express cos(3θ) in terms of trigonometric functions of θ → cos(2θ + θ) → cos(2θ)cos(θ) – sin(2θ)sin(θ) → [cos²(θ) – sin²(θ)]cos(θ) – 2sin(θ)cos(θ)sin(θ) → cos³(θ) – 3sin²(θ)cos(θ) or cos(θ) – 4sin²(θ)cos(θ)
  • Express tan(2π/3 – θ) in terms of trigonometric functions of θ: (tan(2π/3) – tan(θ))/(1 + tan(2π/3)tan(θ)) → Answer: [-√3 – tan(θ)]/[1 – √3tan(θ)]
  • Given B = 79°, a = 50.8192198409, c = 20.3002605154: Find A, C, and B. Use the Law of Cosines (A = 78°, C = 23°, B = 51).
  • Two people stand 1100 feet apart, with a transmission tower between them. The angle of elevation from one person to the top of the tower is 34 degrees. The other person’s angle of elevation is 43.3896048657 degrees (A). How tall is the tower? Use the Law of Sines: B = 180 – (34 + A) = 102.6103951 → sin(B)/1100 = sin(A)/y → y = (1100sin(A))/sin(B) → sin(34°) = x/y → ysin(34°) = x → Answer: 433 = x
  • sin(A) = 3/8, cos(A) < 0, tan(B) = 6/5, sin(B) > 0, cos(B) > 0. Find the exact value of sin(a + b), cos(a – b), tan(2a), sin(b/2). Determine in which quadrant it is located, then find sin(A) [3/8 (given)], cos(A) [-√55/8 (find with x² + y² = r²)], and tan(A) [-3/√55 (find with x² + y² = r²)]. Also, find sin(B), cos(B), and tan(B) using the same method. No fractions inside the fractions.
  • f(x) = 4cos(3x/π + 5π/2) – 3: Period: 2π²/3, Amplitude: 4, Phase Shift: -5π²/6, y-coordinates of peak: 1 (4 above -3), y-coordinates of troughs: -7 (4 below -3)
  • Convert the rectangular point [-3√3, -3] to polar coordinates, with θ in degrees. → Determine the quadrant → [-3√3]² + (-3)² = r² → 27 + 9 = r² → 36 = r² → 6 = r, then tan(θ) = y/x → -3/-3√3 → 1/√3. Answer: (6, 210°)
  • v = -3√3i – 3j. What are the magnitude and direction of v? Determine in which quadrant it is located. Magnitude = 6 (√((-3√3)²) + (-3)²) → 6, Direction = 210° (cos(θ) = [-3√3]/6 → -√3/2 → 210°)
  • w has magnitude 8 and direction 100°. Express w in terms of its i and j components = (8cos(100°))i + (8sin(100°))j
  • Convert the complex number -3√3 – 3i to its trigonometric form rcis(θ): r = 6 (√((-3√3)²) + (-3)²) → Find degree: -3√3 = 6cos(θ) → -√3/2 = cos(θ) → Answer: 6(cos(210°) + isin(210°))
  • Convert 8(cos(100°) + isin(100°)) to the form a + bi = Answer: 8cos(100°) + (8sin(100°))i
  • Given the information tan(θ) = 11/15, 180° < θ < 270°, find the exact value of sin(θ), cos(θ), and tan(θ): Determine which quadrant, then use x² + y² = r² (to find any side), then use the Law of Cosines. Answers: (sin = -11/√346, cos = -15/√346, tan = 11/15)
  • Using Heron’s formula: A piece of commercial real estate is priced at $4.50 per square foot. Find the cost, to the nearest dollar, of a triangular lot measuring 320 feet by 510 feet by 410 feet. → s = (320 + 510 + 410)/2 = 620 → √(620(620 – 320)(620 – 510)(620 – 410)) * 4.50 → $294,968.00
  • sec(cos-1(1/x)) → sec(θ) = 1/cos(θ) → 1/(1/x) → Answer: x