△ Trigonometry & Trig Calculus
Circles, Triangles, Angles — and the Calculus That Moves Them
Complete reference: rigorous proofs, derivatives & integrals of every trig function, interactive explorations, cryptographic applications & GNU Octave laboratory.
§01 Angles & Measurement
1.1 — Degrees, Radians & Gradians
An angle measures rotation between two rays sharing a common vertex. The radian is the natural unit: the angle subtended by an arc equal in length to the radius.
θ (radians) = arc length / radius = s / r
360° = 2π rad = 400 grad ⟹ 1 rad = 180°/π ≈ 57.2958°
1.2 — Conversion Formulas
| From | To Radians | To Degrees |
|---|---|---|
| Degrees d | d · π/180 | — |
| Radians r | — | r · 180/π |
| Turns t | t · 2π | t · 360 |
1.3 — Special Angles
| Degrees | Radians | sin | cos | tan |
|---|---|---|---|---|
| 0° | 0 | 0 | 1 | 0 |
| 30° | π/6 | 1/2 | √3/2 | 1/√3 |
| 45° | π/4 | √2/2 | √2/2 | 1 |
| 60° | π/3 | √3/2 | 1/2 | √3 |
| 90° | π/2 | 1 | 0 | undefined |
| 180° | π | 0 | −1 | 0 |
| 270° | 3π/2 | −1 | 0 | undefined |
1.4 — Arc Length & Sector Area
Arc length: s = rθ Sector area: A = ½r²θ (θ in radians)
PROOF — Sector area A = ½r²θ
A full circle has area πr² and subtends 2π radians.
A sector of angle θ occupies fraction θ/(2π) of the circle.
A = πr² · θ/(2π) = ½r²θ. ■
§02 The Unit Circle
2.1 — Definition
The unit circle is the circle x² + y² = 1 centered at the origin. Every point on it can be written as (cos θ, sin θ) where θ is the angle measured counterclockwise from the positive x-axis.
2.2 — Interactive Unit Circle
↑ Move cursor to explore angles. Shows (cos θ, sin θ), all six trig functions, and the reference triangle.
2.3 — Quadrant Signs (ASTC Rule)
| Quadrant | Angle Range | sin | cos | tan | Mnemonic |
|---|---|---|---|---|---|
| I | 0° – 90° | + | + | + | All positive |
| II | 90° – 180° | + | − | − | Sin positive |
| III | 180° – 270° | − | − | + | Tan positive |
| IV | 270° – 360° | − | + | − | Cos positive |
2.4 — Reference Angles
The reference angle θ′ is the acute angle between the terminal side and the x-axis. For any θ: trig(θ) = ±trig(θ′), with the sign determined by the quadrant (ASTC rule).
§03 The Six Trigonometric Functions
3.1 — Definitions on the Unit Circle
sin θ = y cos θ = x tan θ = y/x = sin θ / cos θ
csc θ = 1/sin θ sec θ = 1/cos θ cot θ = 1/tan θ = cos θ / sin θ
3.2 — Domains & Ranges
| Function | Domain | Range | Period |
|---|---|---|---|
| sin θ | All ℝ | [−1, 1] | 2π |
| cos θ | All ℝ | [−1, 1] | 2π |
| tan θ | θ ≠ π/2 + nπ | (−∞, ∞) | π |
| csc θ | θ ≠ nπ | (−∞,−1]∪[1,∞) | 2π |
| sec θ | θ ≠ π/2 + nπ | (−∞,−1]∪[1,∞) | 2π |
| cot θ | θ ≠ nπ | (−∞, ∞) | π |
3.3 — Even/Odd Properties
cos(−θ) = cos θ (even) sin(−θ) = −sin θ (odd) tan(−θ) = −tan θ (odd)
PROOF — sin is odd, cos is even
On the unit circle, reflecting angle θ across the x-axis gives angle −θ.
Point (cos θ, sin θ) maps to (cos θ, −sin θ).
The x-coordinate is unchanged: cos(−θ) = cos θ. The y-coordinate negates: sin(−θ) = −sin θ. ■
§04 Right Triangles & SOH-CAH-TOA
4.1 — Definitions via Right Triangle
In a right triangle with acute angle θ, hypotenuse h, opposite side o, adjacent side a:
sin θ = o/h cos θ = a/h tan θ = o/a
csc θ = h/o sec θ = h/a cot θ = a/o
Mnemonic: SOH-CAH-TOA — Sin = Opposite/Hypotenuse, Cos = Adjacent/Hypotenuse, Tan = Opposite/Adjacent.
4.2 — The Pythagorean Theorem
a² + b² = c² (where c is the hypotenuse)
PROOF — Pythagorean Theorem (Euclid's proof via similar triangles)
Drop an altitude h from the right angle to the hypotenuse c, creating two sub-triangles.
Both sub-triangles are similar to the original (AA similarity).
From the first: a/c = (segment₁)/a ⟹ a² = c · segment₁.
From the second: b/c = (segment₂)/b ⟹ b² = c · segment₂.
Adding: a² + b² = c(segment₁ + segment₂) = c · c = c². ■
4.3 — Special Right Triangles
| Triangle | Angles | Side Ratios |
|---|---|---|
| 45-45-90 | π/4, π/4, π/2 | 1 : 1 : √2 |
| 30-60-90 | π/6, π/3, π/2 | 1 : √3 : 2 |
PROOF — 30-60-90 side ratios
Start with an equilateral triangle of side 2. All angles are 60°.
Drop an altitude: it bisects the base, creating two 30-60-90 triangles.
Each has hypotenuse 2, short side 1, and altitude = √(2²−1²) = √3. ■
4.4 — Interactive Triangle Solver
↑ Adjust angle A, angle B, and side c. Triangle auto-solves using Law of Sines. C = 180° − A − B.
§05 Oblique Triangles
5.1 — Law of Sines
a/sin A = b/sin B = c/sin C = 2R (R = circumradius)
PROOF — Law of Sines
Drop altitude h from vertex C to side c. Then h = a sin B = b sin A.
So a sin B = b sin A ⟹ a/sin A = b/sin B.
Repeat with altitude from A: b/sin B = c/sin C.
For the circumradius: inscribe the triangle in a circle of radius R. The inscribed angle theorem gives c = 2R sin C, so c/sin C = 2R. ■
5.2 — Law of Cosines
c² = a² + b² − 2ab cos C
PROOF — Law of Cosines (coordinate method)
Place C at origin, B at (a, 0), A at (b cos C, b sin C).
c² = ‖A − B‖² = (b cos C − a)² + (b sin C)²
= b² cos²C − 2ab cos C + a² + b² sin²C
= a² + b²(cos²C + sin²C) − 2ab cos C = a² + b² − 2ab cos C. ■
5.3 — Law of Tangents
(a − b)/(a + b) = tan((A−B)/2) / tan((A+B)/2)
5.4 — Heron's Formula
Area = √[s(s−a)(s−b)(s−c)] where s = (a+b+c)/2
PROOF — Heron's Formula from the Law of Cosines
Area = ½ab sin C. From the law of cosines: cos C = (a²+b²−c²)/(2ab).
sin²C = 1 − cos²C = [4a²b² − (a²+b²−c²)²] / (4a²b²).
Factor the numerator as a difference of squares: [2ab − (a²+b²−c²)][2ab + (a²+b²−c²)].
= [c² − (a−b)²][(a+b)² − c²] = (c−a+b)(c+a−b)(a+b−c)(a+b+c).
Substituting s = (a+b+c)/2: Area² = (½ab)² · sin²C = s(s−a)(s−b)(s−c). ■
5.5 — Area Formulas Summary
| Method | Formula |
|---|---|
| Base × Height | A = ½bh |
| Two sides + included angle | A = ½ab sin C |
| Heron's (three sides) | A = √[s(s−a)(s−b)(s−c)] |
| Circumradius | A = abc / (4R) |
| Inradius | A = rs (r = inradius) |
§06 Circles — Geometry & Trigonometry
6.1 — Circle Equations
Standard: (x−h)² + (y−k)² = r² Center (h,k), radius r
Parametric: x = h + r cos t, y = k + r sin t, t ∈ [0, 2π)
6.2 — Arc Length & Sector Area
Arc: s = rθ Sector: A = ½r²θ Segment: A = ½r²(θ − sin θ)
6.3 — Inscribed Angle Theorem
THEOREM & PROOF — Inscribed angle = ½ central angle
An inscribed angle is formed by two chords meeting at a point on the circle.
Case 1 (one side is a diameter): Let the inscribed angle be α at point A, subtending arc BC where AB is a diameter. Triangle OAC is isosceles (OA = OC = r). The exterior angle at O = 2α. So the central angle = 2α, giving α = ½(central angle).
General case: Draw diameter AD. Apply Case 1 to angles BAD and DAC separately, then add or subtract. ■
6.4 — Thales' Theorem
An angle inscribed in a semicircle is always 90°. This is the inscribed angle theorem with central angle 180° (π), giving inscribed angle = π/2.
6.5 — Chord Length & Tangent Lines
Chord length: ℓ = 2r sin(θ/2) (θ = central angle)
Tangent-Radius: tangent ⊥ radius at point of tangency
Power of a Point: PA · PB = PC · PD (intersecting chords/secants)
6.6 — Circles & Trigonometry in Navigation
The Haversine formula computes great-circle distances on a sphere using trig:
a = sin²(Δφ/2) + cos φ₁ cos φ₂ sin²(Δλ/2)
d = 2R · arctan2(√a, √(1−a))
where φ = latitude, λ = longitude, R = Earth's radius. Every GPS device uses this formula.
§07 Trigonometric Identities
7.1 — Pythagorean Identities
sin²θ + cos²θ = 1 1 + tan²θ = sec²θ 1 + cot²θ = csc²θ
PROOF — sin²θ + cos²θ = 1
On the unit circle, (cos θ, sin θ) satisfies x² + y² = 1.
Therefore cos²θ + sin²θ = 1. Divide by cos²θ: 1 + tan²θ = sec²θ. Divide by sin²θ: cot²θ + 1 = csc²θ. ■
7.2 — Sum & Difference Formulas
sin(α ± β) = sin α cos β ± cos α sin β
cos(α ± β) = cos α cos β ∓ sin α sin β
tan(α ± β) = (tan α ± tan β) / (1 ∓ tan α tan β)
PROOF — cos(α − β) via distance formula on the unit circle
Points A = (cos α, sin α) and B = (cos β, sin β) lie on the unit circle.
Distance² = (cos α − cos β)² + (sin α − sin β)²
= cos²α − 2cos α cos β + cos²β + sin²α − 2sin α sin β + sin²β
= 2 − 2(cos α cos β + sin α sin β).
But also, by the law of cosines on the unit circle triangle: Distance² = 2 − 2cos(α − β).
Equating: cos(α − β) = cos α cos β + sin α sin β. ■
Replace β with −β: cos(α + β) = cos α cos β − sin α sin β.
Use sin θ = cos(π/2 − θ) to derive the sine formulas.
7.3 — Double-Angle Formulas
sin 2θ = 2 sin θ cos θ
cos 2θ = cos²θ − sin²θ = 2cos²θ − 1 = 1 − 2sin²θ
tan 2θ = 2tan θ / (1 − tan²θ)
7.4 — Half-Angle Formulas
sin(θ/2) = ±√[(1 − cos θ)/2] cos(θ/2) = ±√[(1 + cos θ)/2]
tan(θ/2) = sin θ/(1 + cos θ) = (1 − cos θ)/sin θ
7.5 — Product-to-Sum & Sum-to-Product
sin α cos β = ½[sin(α+β) + sin(α−β)]
cos α cos β = ½[cos(α−β) + cos(α+β)]
sin α + sin β = 2 sin((α+β)/2) cos((α−β)/2)
cos α + cos β = 2 cos((α+β)/2) cos((α−β)/2)
7.6 — Power-Reduction Formulas
sin²θ = (1 − cos 2θ)/2 cos²θ = (1 + cos 2θ)/2
7.7 — Auxiliary Angle Method
PROOF — a sin θ + b cos θ = R sin(θ + φ)
Let R = √(a² + b²) and φ = arctan(b/a).
R sin(θ + φ) = R[sin θ cos φ + cos θ sin φ] = R cos φ · sin θ + R sin φ · cos θ.
Set R cos φ = a and R sin φ = b: these give R² = a² + b² and tan φ = b/a. ✓ ■
§08 Inverse Trigonometric Functions
8.1 — Definitions & Ranges
| Function | Domain | Range (Principal Value) |
|---|---|---|
| arcsin x = sin⁻¹ x | [−1, 1] | [−π/2, π/2] |
| arccos x = cos⁻¹ x | [−1, 1] | [0, π] |
| arctan x = tan⁻¹ x | (−∞, ∞) | (−π/2, π/2) |
| arccsc x | |x| ≥ 1 | [−π/2, 0) ∪ (0, π/2] |
| arcsec x | |x| ≥ 1 | [0, π/2) ∪ (π/2, π] |
| arccot x | (−∞, ∞) | (0, π) |
8.2 — Key Identities
arcsin x + arccos x = π/2 arctan x + arccot x = π/2
arctan x + arctan(1/x) = π/2 (x > 0)
8.3 — atan2: The Full-Circle Arctangent
atan2(y, x) = angle of (x,y) from the positive x-axis, range (−π, π]
Unlike arctan(y/x), atan2 uses both coordinates to determine the correct quadrant. Essential in computer graphics, robotics, and signal processing.
8.4 — Compositions
sin(arccos x) = √(1−x²) cos(arcsin x) = √(1−x²) tan(arcsin x) = x/√(1−x²)
PROOF — sin(arccos x) = √(1−x²)
Let θ = arccos x, so cos θ = x and θ ∈ [0,π].
sin²θ = 1 − cos²θ = 1 − x². Since θ ∈ [0,π], sin θ ≥ 0.
Therefore sin(arccos x) = √(1−x²). ■
§09 Graphs, Transformations & Waves
9.1 — General Sinusoid
y = A sin(Bx + C) + D
| Parameter | Effect | Formula |
|---|---|---|
| A | Amplitude | |A| |
| B | Frequency | Period = 2π/|B| |
| C | Phase shift | Shift = −C/B |
| D | Vertical shift | Midline y = D |
9.2 — Interactive Wave Explorer
↑ y = A·sin(Bx + C) + D — adjust sliders to see amplitude, frequency, phase & vertical shift in real time.
9.3 — Graphs of All Six Functions
| Function | Shape | Period | Asymptotes |
|---|---|---|---|
| sin x | Smooth wave, through origin | 2π | None |
| cos x | Smooth wave, max at x=0 | 2π | None |
| tan x | Increasing through origin, S-curves | π | x = π/2 + nπ |
| csc x | U/∩ curves between asymptotes | 2π | x = nπ |
| sec x | U/∩ curves between asymptotes | 2π | x = π/2 + nπ |
| cot x | Decreasing S-curves | π | x = nπ |
9.4 — Beats & Interference
sin(ω₁t) + sin(ω₂t) = 2 cos((ω₁−ω₂)t/2) sin((ω₁+ω₂)t/2)
When ω₁ ≈ ω₂, the cosine factor creates a slow envelope — the phenomenon of beats in acoustics. The sum-to-product identity is the mathematical engine.
§10 Derivatives of Trigonometric Functions
10.1 — The Foundational Limits
PROOF — limθ→0 sin θ / θ = 1 (Squeeze Theorem)
For 0 < θ < π/2, consider a unit circle sector. By area comparison:
Area(inner triangle) ≤ Area(sector) ≤ Area(outer triangle)
½ sin θ ≤ ½θ ≤ ½ tan θ
Divide by ½ sin θ > 0: 1 ≤ θ/sin θ ≤ 1/cos θ.
Invert: cos θ ≤ sin θ/θ ≤ 1.
As θ → 0⁺, cos θ → 1, so by the Squeeze Theorem, sin θ/θ → 1. By the even/odd argument, the limit from the left also equals 1. ■
PROOF — limθ→0 (cos θ − 1)/θ = 0
Multiply by (cos θ + 1)/(cos θ + 1): (cos²θ − 1)/(θ(cos θ + 1)) = −sin²θ/(θ(cos θ + 1)).
= −(sin θ/θ) · sin θ/(cos θ + 1) → −1 · 0/(1+1) = 0. ■
10.2 — d/dx [sin x] = cos x
PROOF — Derivative of sine from the definition
d/dx[sin x] = limh→0 [sin(x+h) − sin x]/h
= limh→0 [sin x cos h + cos x sin h − sin x]/h
= sin x · lim[(cos h − 1)/h] + cos x · lim[sin h / h]
= sin x · 0 + cos x · 1 = cos x. ■
10.3 — d/dx [cos x] = −sin x
PROOF — Derivative of cosine
d/dx[cos x] = d/dx[sin(π/2 − x)] = cos(π/2 − x) · (−1) = −sin x.
Alternatively, from the definition: limh→0 [cos(x+h)−cos x]/h = lim[cos x(cos h−1) − sin x sin h]/h = −sin x. ■
10.4 — d/dx [tan x] = sec² x
PROOF — Derivative of tangent via quotient rule
d/dx[tan x] = d/dx[sin x / cos x] = (cos x · cos x − sin x · (−sin x)) / cos²x
= (cos²x + sin²x) / cos²x = 1/cos²x = sec²x. ■
10.5 — Complete Derivative Table
| f(x) | f′(x) | f(x) | f′(x) |
|---|---|---|---|
| sin x | cos x | arcsin x | 1/√(1−x²) |
| cos x | −sin x | arccos x | −1/√(1−x²) |
| tan x | sec²x | arctan x | 1/(1+x²) |
| csc x | −csc x cot x | arccsc x | −1/(|x|√(x²−1)) |
| sec x | sec x tan x | arcsec x | 1/(|x|√(x²−1)) |
| cot x | −csc²x | arccot x | −1/(1+x²) |
10.6 — Inverse Trig Derivative Proofs
PROOF — d/dx [arcsin x] = 1/√(1−x²)
Let y = arcsin x, so sin y = x and y ∈ [−π/2, π/2].
Differentiate implicitly: cos y · dy/dx = 1 ⟹ dy/dx = 1/cos y.
Since cos y = √(1 − sin²y) = √(1 − x²) (cos y ≥ 0 on [−π/2, π/2]):
d/dx[arcsin x] = 1/√(1 − x²). ■
PROOF — d/dx [arctan x] = 1/(1+x²)
Let y = arctan x, so tan y = x.
Differentiate: sec²y · dy/dx = 1 ⟹ dy/dx = 1/sec²y = 1/(1 + tan²y) = 1/(1 + x²). ■
10.7 — The nth Derivative Cycle
dⁿ/dxⁿ [sin x] = sin(x + nπ/2) dⁿ/dxⁿ [cos x] = cos(x + nπ/2)
The derivatives cycle with period 4: sin → cos → −sin → −cos → sin → ⋯
§11 Integration of Trigonometric Functions
11.1 — Basic Antiderivatives
| ∫ f(x) dx | Result + C |
|---|---|
| ∫ sin x dx | −cos x |
| ∫ cos x dx | sin x |
| ∫ tan x dx | −ln|cos x| = ln|sec x| |
| ∫ cot x dx | ln|sin x| |
| ∫ sec x dx | ln|sec x + tan x| |
| ∫ csc x dx | −ln|csc x + cot x| = ln|csc x − cot x| |
| ∫ sec²x dx | tan x |
| ∫ csc²x dx | −cot x |
| ∫ sec x tan x dx | sec x |
| ∫ csc x cot x dx | −csc x |
11.2 — Proofs of Non-Obvious Integrals
PROOF — ∫ tan x dx = ln|sec x| + C
∫ tan x dx = ∫ sin x / cos x dx. Let u = cos x, du = −sin x dx.
= −∫ du/u = −ln|u| + C = −ln|cos x| + C = ln|sec x| + C. ■
PROOF — ∫ sec x dx = ln|sec x + tan x| + C
Multiply by (sec x + tan x)/(sec x + tan x):
∫ sec x dx = ∫ [sec²x + sec x tan x] / [sec x + tan x] dx.
Let u = sec x + tan x, du = (sec x tan x + sec²x) dx = numerator dx.
= ∫ du/u = ln|u| + C = ln|sec x + tan x| + C. ■
11.3 — Powers of Trig Functions
∫ sinⁿx dx and ∫ cosⁿx dx
Odd power: Strip one factor, convert the rest using sin²x = 1 − cos²x, substitute u = cos x (or vice versa).
Even power: Use power-reduction: sin²x = (1 − cos 2x)/2, cos²x = (1 + cos 2x)/2.
PROOF — Reduction formula: ∫ sinⁿx dx = −(sinⁿ⁻¹x cos x)/n + (n−1)/n · ∫ sinⁿ⁻²x dx
Write ∫ sinⁿx dx = ∫ sinⁿ⁻¹x · sin x dx. Integration by parts: u = sinⁿ⁻¹x, dv = sin x dx.
du = (n−1)sinⁿ⁻²x cos x dx, v = −cos x.
= −sinⁿ⁻¹x cos x + (n−1)∫ sinⁿ⁻²x cos²x dx
= −sinⁿ⁻¹x cos x + (n−1)∫ sinⁿ⁻²x (1 − sin²x) dx
= −sinⁿ⁻¹x cos x + (n−1)∫ sinⁿ⁻²x dx − (n−1)∫ sinⁿx dx.
Move the last term to the left: n∫ sinⁿx dx = −sinⁿ⁻¹x cos x + (n−1)∫ sinⁿ⁻²x dx. ÷n. ■
11.4 — Trigonometric Substitution
| Expression | Substitution | Identity Used |
|---|---|---|
| √(a² − x²) | x = a sin θ | 1 − sin²θ = cos²θ |
| √(a² + x²) | x = a tan θ | 1 + tan²θ = sec²θ |
| √(x² − a²) | x = a sec θ | sec²θ − 1 = tan²θ |
EXAMPLE — ∫ dx/√(1−x²) = arcsin x + C
Let x = sin θ, dx = cos θ dθ. Then √(1−x²) = cos θ.
∫ cos θ dθ / cos θ = ∫ dθ = θ + C = arcsin x + C. ■
11.5 — The Weierstrass Substitution
t = tan(x/2) ⟹ sin x = 2t/(1+t²), cos x = (1−t²)/(1+t²), dx = 2dt/(1+t²)
PROOF — Weierstrass substitution formulas
Let t = tan(x/2). Then sin x = 2sin(x/2)cos(x/2).
Divide by cos²(x/2) + sin²(x/2) = 1: sin x = 2tan(x/2)/sec²(x/2) = 2t/(1+t²).
Similarly cos x = (cos²(x/2) − sin²(x/2))/(cos²(x/2) + sin²(x/2)) = (1−t²)/(1+t²).
For dx: x = 2 arctan t, so dx = 2/(1+t²) dt. ■
This converts any rational function of sin and cos into a rational function of t, solvable by partial fractions.
11.6 — The Gaussian Integral via Trig
PROOF — ∫−∞∞ e−x² dx = √π (polar coordinates method)
Let I = ∫−∞∞ e−x² dx. Then I² = ∫∫ e−(x²+y²) dx dy.
Convert to polar: x = r cos θ, y = r sin θ, dx dy = r dr dθ.
I² = ∫₀2π ∫₀∞ e−r² r dr dθ = 2π · [−½e−r²]₀∞ = 2π · ½ = π.
Therefore I = √π. This integral underpins Gaussian distributions in probability and cryptographic lattice sampling. ■
§12 Polar Coordinates & Complex Numbers
12.1 — Polar Coordinates
x = r cos θ y = r sin θ r = √(x²+y²) θ = atan2(y,x)
12.2 — Common Polar Curves
| Curve | Equation | Shape |
|---|---|---|
| Circle | r = a | Circle radius a |
| Cardioid | r = a(1 + cos θ) | Heart shape |
| Rose | r = a cos(nθ) | n petals (n odd) or 2n petals (n even) |
| Lemniscate | r² = a² cos 2θ | Figure-8 |
| Spiral | r = aθ | Archimedean spiral |
12.3 — Area in Polar Coordinates
A = ½ ∫αβ r(θ)² dθ
PROOF — Polar area formula
Divide the region into thin sectors of angle Δθ. Each has area ≈ ½r²Δθ (sector area formula).
Summing: A ≈ Σ ½r(θᵢ)²Δθ → ½∫ r²dθ as Δθ → 0. ■
12.4 — Complex Numbers & Euler's Formula
z = a + bi = r(cos θ + i sin θ) = reiθ
PROOF — Euler's formula eiθ = cos θ + i sin θ (via Taylor series)
eiθ = Σ (iθ)ⁿ/n! = 1 + iθ − θ²/2! − iθ³/3! + θ⁴/4! + ⋯
Separate real and imaginary: = (1 − θ²/2! + θ⁴/4! − ⋯) + i(θ − θ³/3! + θ⁵/5! − ⋯)
= cos θ + i sin θ. ■
12.5 — De Moivre's Theorem
(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ
PROOF — De Moivre's Theorem
(eiθ)ⁿ = einθ. By Euler's formula:
(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ. ■
12.6 — Roots of Unity
zn = 1 ⟹ zk = e2πik/n = cos(2πk/n) + i sin(2πk/n), k = 0,1,…,n−1
The n-th roots of unity form a regular n-gon on the unit circle. They are the foundation of the DFT (§13) and NTT (§14, lattice cryptography).
12.7 — Euler's Identity
eiπ + 1 = 0
Five fundamental constants — e, i, π, 1, 0 — united by trigonometry and the exponential function.
§13 Fourier Series & Applications
13.1 — Fourier Series
Any periodic function f(x) with period 2π can be expressed as:
f(x) = a₀/2 + Σn=1∞ [aₙ cos(nx) + bₙ sin(nx)]
aₙ = (1/π) ∫−ππ f(x) cos(nx) dx bₙ = (1/π) ∫−ππ f(x) sin(nx) dx
PROOF — Orthogonality of {sin(nx), cos(mx)}
∫−ππ sin(nx)cos(mx) dx = 0 for all n,m (product of odd × even function over symmetric interval).
∫−ππ cos(nx)cos(mx) dx = πδnm (use product-to-sum: ½∫[cos(n−m)x + cos(n+m)x]dx).
∫−ππ sin(nx)sin(mx) dx = πδnm (similarly).
Multiply f(x) by cos(mx) and integrate, using orthogonality: only the aₘ term survives. ■
13.2 — Discrete Fourier Transform (DFT)
X[k] = Σn=0N−1 x[n] · e−2πikn/N (analysis: decompose into frequencies)
x[n] = (1/N) Σk=0N−1 X[k] · e2πikn/N (synthesis: reconstruct from frequencies)
13.3 — The FFT Algorithm
The Fast Fourier Transform computes the DFT in O(N log N) instead of O(N²) by exploiting the symmetry of roots of unity (the Cooley-Tukey butterfly). This makes audio processing, image compression (JPEG uses DCT, a real cousin of DFT), and polynomial multiplication practical.
13.4 — Parseval's Theorem
(1/2π) ∫−ππ |f(x)|² dx = |a₀|²/4 + ½Σ(aₙ² + bₙ²)
Energy in the time domain equals energy in the frequency domain — a conservation law derived from trig orthogonality.
13.5 — Application: The Basel Problem
PROOF — Σ 1/n² = π²/6 via Fourier series
Let f(x) = x on [−π,π]. Its Fourier coefficients: a₀ = 0, aₙ = 0, bₙ = (−1)ⁿ⁺¹ · 2/n.
Parseval: (1/2π)∫|x|²dx = ½Σ(4/n²). Left side: (1/2π)·(2π³/3) = π²/3.
So π²/3 = 2 Σ 1/n² ⟹ Σ 1/n² = π²/6. ■
§14 Cryptography — Trigonometry in Security
14.1 — Why Trig in Crypto?
Trigonometry appears throughout cryptography: from the periodicity that makes modular arithmetic work, to the Fourier transforms at the heart of lattice-based post-quantum schemes, to the geometry of elliptic curves.
14.2 — The Number Theoretic Transform (NTT)
The NTT is a DFT over a finite field 𝔽q instead of ℂ. It replaces e−2πi/N with a primitive N-th root of unity ω in 𝔽q (where ωN ≡ 1 mod q). The NTT enables fast polynomial multiplication in lattice cryptography (CRYSTALS-Kyber, Dilithium) without floating-point errors.
CONNECTION — Trig roots → NTT roots
In ℂ: the N-th roots of unity are e2πik/N = cos(2πk/N) + i sin(2πk/N).
In 𝔽q: we need ω with ωN ≡ 1 (mod q) and ωk ≢ 1 for 0 < k < N.
This exists when N | (q−1). The FFT butterfly structure is identical in both settings.
The trig orthogonality relations become: Σn ωkn ≡ 0 (mod q) for k ≢ 0. ■
14.3 — Gaussian Sampling for Lattice Crypto
Lattice schemes sample noise from a discrete Gaussian ρσ(x) = exp(−π‖x‖²/σ²). The normalization uses the Gaussian integral (§11.6). The smoothing parameter ηε is computed via the Poisson summation formula, which is the Fourier transform (a trig integral) applied to lattices.
14.4 — Elliptic Curve Trigonometric Parameterization
Elliptic curves over ℂ can be parameterized by the Weierstrass ℘-function, which is doubly-periodic (two independent "periods" in the complex plane, generalizing the single period 2π of sin/cos). The group law on E corresponds to the addition theorem for ℘ — a direct generalization of trig addition formulas.
14.5 — CORDIC Algorithm
The CORDIC (COordinate Rotation DIgital Computer) algorithm computes sin, cos, arctan using only shifts and adds — no multiplication. It rotates a vector by successively smaller angles αᵢ = arctan(2⁻ⁱ):
xi+1 = xi − σᵢ 2⁻ⁱ yi yi+1 = yi + σᵢ 2⁻ⁱ xi
where σᵢ = ±1 chooses the rotation direction. After n iterations, (xn, yn) ≈ K(cos θ, sin θ). CORDIC is used in hardware crypto accelerators where multipliers are expensive.
14.6 — Diffie-Hellman via Trig Analogy
The Diffie-Hellman key exchange: Alice picks secret a, publishes ga mod p. Bob picks secret b, publishes gb mod p. Shared secret: gab mod p.
Trig analogy: ga is like rotating by angle a on a "discrete unit circle" (the cyclic group ℤp*). Knowing the rotation angle from the final position is the discrete logarithm problem — as hard as un-rotating a circle when you can only see the endpoint.
14.7 — RSA & Euler's Totient
CONNECTION — Euler's formula and Euler's theorem
Euler's formula: eiθ has period 2π, so ei(θ+2π) = eiθ.
Euler's theorem: aφ(n) ≡ 1 (mod n) — exponentiation has period φ(n) in ℤn*.
Both are periodicity: one in the continuous world (trig), one in the discrete world (number theory).
RSA exploits this: e·d ≡ 1 mod φ(n), so med = m1+kφ(n) = m · (mφ(n))k ≡ m. ■
GNU Octave — Crypto Trig
% --- NTT roots of unity in F_q --- q = 7681; N = 256; % q ≡ 1 mod 256 g = 17; % generator of F_q* omega = mod(power_mod(g, (q-1)/N, q), q); function r = power_mod(b,e,m) r=1;b=mod(b,m);while e>0;if mod(e,2)==1,r=mod(r*b,m);end;e=floor(e/2);b=mod(b*b,m);end end printf("NTT: ω=%d, ω^N mod q = %d (should be 1)\n", omega, power_mod(omega, N, q)); % --- CORDIC sine/cosine --- angles = atan(2.^-(0:29)); K = prod(1./sqrt(1 + 2.^(-2*(0:29)))); target = pi/5; x = K; y = 0; z = target; for i = 0:29 s = sign(z) + (z==0); xn = x - s*y*2^(-i); yn = y + s*x*2^(-i); z = z - s*angles(i+1); x = xn; y = yn; end printf("CORDIC: cos(π/5)=%.10f (exact %.10f)\n", x, cos(target)); printf("CORDIC: sin(π/5)=%.10f (exact %.10f)\n", y, sin(target));
§15 GNU Octave — Trigonometry & Trig Calculus Laboratory
15.1 — Verifying Trig Identities
GNU Octave — Identity Verification
% Verify fundamental identities at 1000 random angles th = 2*pi*rand(1,1000); e1 = max(abs(sin(th).^2 + cos(th).^2 - 1)); e2 = max(abs(sin(2*th) - 2*sin(th).*cos(th))); e3 = max(abs(cos(2*th) - (cos(th).^2 - sin(th).^2))); a = rand(1,1000)*pi; b = rand(1,1000)*pi; e4 = max(abs(sin(a+b) - (sin(a).*cos(b) + cos(a).*sin(b)))); printf("Identity errors (should be ≈0):\n"); printf(" sin²+cos²=1: %.2e\n sin 2θ: %.2e\n", e1, e2); printf(" cos 2θ: %.2e\n sin(a+b): %.2e\n", e3, e4);
15.2 — Numerical Derivatives of Trig Functions
GNU Octave — Complex-Step Derivatives
% Verify all 6 trig derivatives with complex-step method h = 1e-20; x = 0.7; deriv = @(f,x) imag(f(x + 1i*h)) / h; funcs = {@sin, @cos, @tan, @csc, @sec, @cot}; exact_d = {cos(x), -sin(x), sec(x)^2, -csc(x)*cot(x), sec(x)*tan(x), -csc(x)^2}; names = {'sin','cos','tan','csc','sec','cot'}; printf("Trig derivatives at x = %.1f:\n", x); for k = 1:6 d = deriv(funcs{k}, x); printf(" d/dx[%s] = %12.8f (exact %12.8f, err %.1e)\n", ... names{k}, d, exact_d{k}, abs(d - exact_d{k})); end
15.3 — Numerical Integration of Trig Functions
GNU Octave — Trig Integrals via Quadrature
% Verify antiderivatives numerically printf("Trig integrals:\n"); I1 = quad(@sin, 0, pi); printf(" ∫₀^π sin x dx = %.10f (exact 2)\n", I1); I2 = quad(@(x) sec(x).^2, 0, pi/4); printf(" ∫₀^(π/4) sec²x dx = %.10f (exact tan(π/4)=1)\n", I2); I3 = quad(@(x) 1./(1+x.^2), 0, 1); printf(" ∫₀¹ 1/(1+x²) dx = %.10f (exact π/4=%.10f)\n", I3, pi/4); I4 = quad(@(x) 1./sqrt(1-x.^2), 0, 0.5); printf(" ∫₀^0.5 1/√(1-x²) dx = %.10f (exact arcsin(0.5)=%.10f)\n", I4, asin(0.5)); % Reduction formula: ∫₀^(π/2) sin⁶x dx = 5π/32 I5 = quad(@(x) sin(x).^6, 0, pi/2); printf(" ∫₀^(π/2) sin⁶x dx = %.10f (exact 5π/32=%.10f)\n", I5, 5*pi/32);
15.4 — Fourier Synthesis & Analysis
GNU Octave — Fourier Series of Square Wave
% Square wave: f(x) = (4/π) Σ sin((2k-1)x)/(2k-1) x = linspace(-pi, pi, 1000); for N = [1, 5, 20, 100] f = zeros(size(x)); for k = 1:N n = 2*k - 1; f = f + sin(n*x)/n; end f = f * 4/pi; err = mean(abs(f - sign(sin(x)))); printf("Fourier N=%3d terms: mean error = %.6f\n", N, err); end
15.5 — Haversine Navigation
GNU Octave — Great-Circle Distance
% Haversine formula: distance between two points on Earth function d = haversine(lat1, lon1, lat2, lon2) R = 6371; % Earth radius km p = pi/180; dlat = (lat2-lat1)*p; dlon = (lon2-lon1)*p; a = sin(dlat/2)^2 + cos(lat1*p)*cos(lat2*p)*sin(dlon/2)^2; d = 2*R*atan2(sqrt(a), sqrt(1-a)); end % New York → London d = haversine(40.7128, -74.0060, 51.5074, -0.1278); printf("NYC → London: %.1f km\n", d); % Tokyo → Sydney d = haversine(35.6762, 139.6503, -33.8688, 151.2093); printf("Tokyo → Sydney: %.1f km\n", d);
15.6 — Lissajous Figures & 3D Trig Surfaces
GNU Octave — Lissajous & Parametric Surfaces
% Lissajous: x = sin(at+δ), y = sin(bt) t = linspace(0, 2*pi, 10000); for params = [3,2,pi/4; 5,4,pi/3; 7,6,pi/6]' a = params(1); b = params(2); delta = params(3); x = sin(a*t + delta); y = sin(b*t); printf("Lissajous a=%d b=%d: bounding box [%.2f,%.2f]×[%.2f,%.2f]\n", ... a, b, min(x), max(x), min(y), max(y)); end % 3D Torus: (R+r cos v)cos u, (R+r cos v)sin u, r sin v R = 3; r = 1; [u,v] = meshgrid(linspace(0,2*pi,50)); X = (R+r*cos(v)).*cos(u); Y = (R+r*cos(v)).*sin(u); Z = r*sin(v); printf("Torus: R=%d, r=%d, surface area = %.4f (exact 4π²Rr = %.4f)\n", ... R, r, 4*pi^2*R*r, 4*pi^2*R*r);
15.7 — Comprehensive Verification
GNU Octave — Master Verification Suite
printf("=== TRIG & CALCULUS VERIFICATION ===\n\n"); % Euler's identity printf("|e^(iπ)+1| = %.2e\n", abs(exp(1i*pi)+1)); % De Moivre n=7 th = 0.3; n = 7; lhs = (cos(th)+1i*sin(th))^n; rhs = cos(n*th)+1i*sin(n*th); printf("De Moivre (n=%d): error = %.2e\n", n, abs(lhs-rhs)); % Machin's formula: π = 4(4·atan(1/5) - atan(1/239)) pi_machin = 4*(4*atan(1/5) - atan(1/239)); printf("Machin π = %.15f (error %.2e)\n", pi_machin, abs(pi_machin-pi)); % Wallis product: π/2 = Π(4n²/(4n²-1)) n = 1:100000; wallis = prod(4*n.^2 ./ (4*n.^2 - 1)); printf("Wallis π/2 = %.10f (exact %.10f)\n", wallis, pi/2); % Leibniz: π/4 = 1 - 1/3 + 1/5 - ... n = 0:1e6; leibniz = sum((-1).^n ./ (2*n+1)); printf("Leibniz π/4 = %.10f (exact %.10f)\n", leibniz, pi/4); % Basel: Σ1/n² = π²/6 n = 1:1e6; basel = sum(1./n.^2); printf("Basel π²/6 = %.10f (exact %.10f)\n", basel, pi^2/6); % Gaussian integral I = quad(@(x) exp(-x.^2), -20, 20); printf("∫exp(-x²) = %.10f (exact √π = %.10f)\n", I, sqrt(pi));
⚡ Quick Reference Card
Trig Values
| θ | 0 | π/6 | π/4 | π/3 | π/2 |
|---|---|---|---|---|---|
| sin | 0 | ½ | √2/2 | √3/2 | 1 |
| cos | 1 | √3/2 | √2/2 | ½ | 0 |
| tan | 0 | 1/√3 | 1 | √3 | ∞ |
Derivatives & Integrals
| f | f′ | ∫f dx |
|---|---|---|
| sin x | cos x | −cos x |
| cos x | −sin x | sin x |
| tan x | sec²x | ln|sec x| |
| sec x | sec x tan x | ln|sec x+tan x| |
| csc x | −csc x cot x | ln|csc x−cot x| |
| cot x | −csc²x | ln|sin x| |
Inverse Trig Derivatives
| f | f′ |
|---|---|
| arcsin x | 1/√(1−x²) |
| arccos x | −1/√(1−x²) |
| arctan x | 1/(1+x²) |
Key Identities
| Identity |
|---|
| sin²θ + cos²θ = 1 |
| sin(α±β) = sin α cos β ± cos α sin β |
| cos(α±β) = cos α cos β ∓ sin α sin β |
| sin 2θ = 2 sin θ cos θ |
| cos 2θ = cos²θ − sin²θ |
| eiθ = cos θ + i sin θ |
Triangle Laws
| Law | Formula |
|---|---|
| Sines | a/sin A = b/sin B = c/sin C |
| Cosines | c² = a² + b² − 2ab cos C |
| Heron | A = √[s(s−a)(s−b)(s−c)] |
Trig Substitution
| √(a²−x²) | √(a²+x²) | √(x²−a²) |
|---|---|---|
| x = a sin θ | x = a tan θ | x = a sec θ |