The branch of mathematics that studies relationships between the sides and angles of triangles. Trigonometry extends far beyond triangles, providing essential tools for modeling waves, circular motion, and periodic phenomena throughout science and engineering.
Trigonometry — from the Greek words trigonon (triangle) and metron (measure) — is the study of the relationships between the angles and sides of triangles. It originated in the ancient world as a practical tool for astronomy and land surveying, but has grown into a vast discipline that underlies much of modern mathematics, physics, and engineering.
At its core, trigonometry answers a simple question: if you know some measurements of a triangle, can you determine the rest? The answer is yes, and the six trigonometric functions — sine, cosine, tangent, cosecant, secant, and cotangent — are the tools that make it possible.
Beyond triangles, trigonometric functions turn out to be the perfect mathematical language for describing anything that repeats in a cycle: sound waves, light waves, alternating current, the motion of a pendulum, the orbit of a planet, and countless other periodic phenomena.
Key areas where trigonometry is indispensable include:
The foundation of trigonometry is the right triangle — a triangle containing one 90° angle. In a right triangle, the side opposite the right angle is called the hypotenuse (always the longest side), and the other two sides are called legs. Relative to a given acute angle θ, the legs are labeled as the opposite side (across from θ) and the adjacent side (next to θ).
The three primary trigonometric ratios are defined using the sides of a right triangle relative to an acute angle θ:
The mnemonic SOH-CAH-TOA is the single most important memory aid in trigonometry. It tells you exactly how to set up each ratio.
The three reciprocal ratios are:
A right triangle has legs of length 3 and 4, and a hypotenuse of 5. Find all six trig ratios for the angle θ opposite the side of length 3.
Opposite = 3, Adjacent = 4, Hypotenuse = 5
sin(θ) = 3/5 = 0.6
cos(θ) = 4/5 = 0.8
tan(θ) = 3/4 = 0.75
csc(θ) = 5/3 ≈ 1.667
sec(θ) = 5/4 = 1.25
cot(θ) = 4/3 ≈ 1.333
A right triangle has a hypotenuse of 10 and an angle of 30°. Find the length of the side opposite the 30° angle.
Step 1: Identify the relationship: sin(30°) = Opposite / Hypotenuse
Step 2: Substitute: sin(30°) = x / 10
Step 3: Solve: x = 10 · sin(30°) = 10 · 0.5 = 5
Two right triangles appear so frequently in mathematics that their ratios should be memorized.
An isosceles right triangle where both legs are equal. If each leg has length 1, the hypotenuse is √2 (by the Pythagorean theorem).
Side ratio: 1 : 1 : √2
This triangle is half of an equilateral triangle. If the shortest side (opposite the 30° angle) has length 1, then the hypotenuse is 2 and the longer leg is √3.
Side ratio: 1 : √3 : 2
A ladder leans against a wall, forming a 60° angle with the ground. If the ladder is 12 feet long, how high up the wall does it reach?
Step 1: The wall height is the side opposite the 60° angle, and the ladder is the hypotenuse.
Step 2: sin(60°) = height / 12
Step 3: height = 12 · sin(60°) = 12 · (√3/2) = 6√3 ≈ 10.39 feet
The unit circle is a circle of radius 1 centered at the origin of a coordinate plane. It is one of the most powerful tools in trigonometry because it extends the definitions of sine and cosine beyond acute angles to all real numbers.
For any angle θ measured from the positive x-axis (counterclockwise is positive), the terminal side of the angle intersects the unit circle at a point (x, y). We define:
Since the radius is 1, this is consistent with SOH-CAH-TOA: cos(θ) = adjacent/hypotenuse = x/1 = x, and sin(θ) = opposite/hypotenuse = y/1 = y.
The following table lists the most important angles and their sine and cosine values:
A reference angle is the acute angle (between 0° and 90°) formed between the terminal side of an angle and the x-axis. Reference angles allow you to evaluate trig functions for any angle by relating them back to a first-quadrant angle.
Find the reference angle for θ = 240°.
Step 1: 240° is in Quadrant III (between 180° and 270°).
Step 2: Reference angle = 240° - 180° = 60°
So sin(240°) = -sin(60°) = -√3/2 and cos(240°) = -cos(60°) = -1/2 (both negative in Quadrant III).
The mnemonic "All Students Take Calculus" (ASTC) tells you which trig functions are positive in each quadrant:
Radians are the standard unit of angle measurement in mathematics. One radian is the angle subtended at the center of a circle by an arc equal in length to the radius.
Convert 150° to radians: 150 × (π/180) = 150π/180 = 5π/6
Convert 3π/4 to degrees: (3π/4) × (180/π) = 540/4 = 135°
When we treat the trig ratios as functions of a real variable, we unlock the full power of trigonometry. Each trigonometric function takes an angle as input and returns a ratio as output, and each has its own distinctive graph, domain, range, and period.
The sine function starts at the origin, rises to a maximum of 1 at π/2, returns to 0 at π, drops to a minimum of -1 at 3π/2, and returns to 0 at 2π, completing one full cycle.
The cosine function is identical to the sine function but shifted left by π/2. That is, cos(x) = sin(x + π/2).
The tangent function increases through each period, passing through zero at multiples of π and having vertical asymptotes where cosine equals zero.
The general sinusoidal function is:
Where:
Describe the graph of y = 3 sin(2x - π) + 1.
Amplitude: |A| = 3 (the graph oscillates 3 units above and below the midline)
Period: 2π/|B| = 2π/2 = π (one full cycle every π units)
Phase shift: C/B = π/2 (shifted π/2 units to the right)
Vertical shift: D = 1 (midline is y = 1 instead of y = 0)
The graph oscillates between y = 1 - 3 = -2 and y = 1 + 3 = 4.
Trigonometric identities are equations involving trig functions that are true for every value in their domain. They are essential for simplifying expressions, solving equations, and proving other mathematical results. Mastering these identities is critical for success in calculus.
These are derived directly from the Pythagorean theorem applied to the unit circle (x² + y² = 1):
The first identity is the most fundamental. The other two are obtained by dividing through by cos²(θ) and sin²(θ) respectively.
If sin(θ) = 3/5 and θ is in Quadrant I, find cos(θ).
Step 1: sin²(θ) + cos²(θ) = 1
Step 2: (3/5)² + cos²(θ) = 1
Step 3: 9/25 + cos²(θ) = 1
Step 4: cos²(θ) = 16/25
Step 5: cos(θ) = 4/5 (positive because θ is in Quadrant I)
These relate the trig functions of complementary angles (angles that add up to 90° or π/2):
These formulas allow you to find the exact trig values of angle sums and differences:
Find the exact value of sin(75°).
Step 1: Write 75° = 45° + 30°
Step 2: Apply the sum formula: sin(75°) = sin(45° + 30°)
= sin(45°)cos(30°) + cos(45°)sin(30°)
= (√2/2)(√3/2) + (√2/2)(1/2)
= √6/4 + √2/4
= (√6 + √2) / 4 ≈ 0.9659
Setting B = A in the sum formulas yields the double-angle identities:
If cos(θ) = 3/5, find sin(2θ) and cos(2θ), assuming θ is in Quadrant I.
Step 1: Find sin(θ) using the Pythagorean identity: sin(θ) = 4/5
Step 2: sin(2θ) = 2 sin(θ) cos(θ) = 2(4/5)(3/5) = 24/25
Step 3: cos(2θ) = cos²(θ) - sin²(θ) = (9/25) - (16/25) = -7/25
The ± sign is determined by the quadrant in which A/2 lies.
Find the exact value of cos(15°).
Step 1: 15° = 30°/2, so we use the half-angle formula with A = 30°.
Step 2: cos(15°) = √[(1 + cos 30°) / 2] = √[(1 + √3/2) / 2]
= √[(2 + √3) / 4] = √(2 + √3) / 2 ≈ 0.9659
(Positive because 15° is in Quadrant I.)
The inverse trigonometric functions answer the question: "What angle has this trig value?" Since trig functions are periodic and not one-to-one, we must restrict their domains to create proper inverse functions.
Find the exact value of each:
arcsin(1/2): What angle in [-π/2, π/2] has sin = 1/2? Answer: π/6 (30°)
arccos(-√2/2): What angle in [0, π] has cos = -√2/2? Answer: 3π/4 (135°)
arctan(1): What angle in (-π/2, π/2) has tan = 1? Answer: π/4 (45°)
arcsin(-1): What angle in [-π/2, π/2] has sin = -1? Answer: -π/2 (-90°)
When composing trig functions with their inverses, be careful about domain restrictions:
Find cos(arcsin(3/5)).
Step 1: Let θ = arcsin(3/5). Then sin(θ) = 3/5 with θ in [-π/2, π/2].
Step 2: We need cos(θ). Using sin²(θ) + cos²(θ) = 1:
cos²(θ) = 1 - 9/25 = 16/25
Step 3: cos(θ) = 4/5 (positive because θ is in [-π/2, π/2])
Therefore, cos(arcsin(3/5)) = 4/5
A trigonometric equation is an equation that involves trigonometric functions of unknown angles. Unlike identities (which are true for all values), trig equations are true only for specific values. Because trig functions are periodic, these equations typically have infinitely many solutions.
Solve: 2 sin(x) - 1 = 0
Step 1: Isolate sin(x): sin(x) = 1/2
Step 2: Reference angle: arcsin(1/2) = π/6
Step 3: sin is positive in Quadrants I and II:
x = π/6 and x = π - π/6 = 5π/6
Step 4: General solution: x = π/6 + 2nπ or x = 5π/6 + 2nπ, where n is any integer.
Solve: 2cos²(x) - cos(x) - 1 = 0 on [0, 2π)
Step 1: This is a quadratic in cos(x). Let u = cos(x): 2u² - u - 1 = 0
Step 2: Factor: (2u + 1)(u - 1) = 0
Step 3: u = -1/2 or u = 1, so cos(x) = -1/2 or cos(x) = 1
Step 4: cos(x) = -1/2: x = 2π/3, 4π/3
cos(x) = 1: x = 0
Solutions: x = 0, 2π/3, 4π/3
Solve: sin(2x) = cos(x) on [0, 2π)
Step 1: Use the double-angle identity: 2sin(x)cos(x) = cos(x)
Step 2: Rearrange: 2sin(x)cos(x) - cos(x) = 0
Step 3: Factor: cos(x)[2sin(x) - 1] = 0
Step 4: cos(x) = 0: x = π/2, 3π/2
sin(x) = 1/2: x = π/6, 5π/6
Solutions: x = π/6, π/2, 5π/6, 3π/2
Solve: tan(3x) = 1 on [0, 2π)
Step 1: Reference angle: arctan(1) = π/4
Step 2: General solution for 3x: 3x = π/4 + nπ
Step 3: Solve for x: x = π/12 + nπ/3
Step 4: Find values in [0, 2π) by substituting n = 0, 1, 2, 3, 4, 5:
x = π/12, 5π/12, 3π/4, 13π/12, 17π/12, 7π/4
Solutions: x = π/12, 5π/12, 3π/4, 13π/12, 17π/12, 7π/4
While SOH-CAH-TOA only works for right triangles, the Law of Sines and Law of Cosines allow us to solve any triangle — including oblique triangles (those without a right angle). These are arguably the two most powerful formulas in triangle trigonometry.
In any triangle with sides a, b, c opposite angles A, B, C respectively:
Equivalently: sin(A)/a = sin(B)/b = sin(C)/c
Use the Law of Sines when you know:
In triangle ABC, A = 42°, B = 73°, and a = 15. Find side b.
Step 1: Find C: C = 180° - 42° - 73° = 65°
Step 2: Apply the Law of Sines: a/sin(A) = b/sin(B)
15/sin(42°) = b/sin(73°)
Step 3: Solve: b = 15 · sin(73°)/sin(42°) = 15 · 0.9563/0.6691 ≈ 21.44
When given two sides and an angle opposite one of them, there may be zero, one, or two possible triangles. This is because the sine function gives the same value for supplementary angles (sin(θ) = sin(180° - θ)).
In triangle ABC, a = 8, b = 12, and A = 30°. How many triangles are possible?
Step 1: sin(B)/b = sin(A)/a → sin(B) = b · sin(A)/a = 12 · sin(30°)/8 = 12 · 0.5/8 = 0.75
Step 2: B = arcsin(0.75) ≈ 48.59° or B = 180° - 48.59° = 131.41°
Step 3: Check both: A + B = 30° + 48.59° = 78.59° < 180° ✓
A + B = 30° + 131.41° = 161.41° < 180° ✓
Result: Two triangles are possible.
The Law of Cosines is a generalization of the Pythagorean theorem to all triangles:
Notice: when C = 90°, cos(C) = 0, and the formula reduces to c² = a² + b² — the Pythagorean theorem!
Use the Law of Cosines when you know:
In triangle ABC, a = 7, b = 10, and C = 51°. Find side c.
Step 1: Apply the formula: c² = a² + b² - 2ab cos(C)
c² = 49 + 100 - 2(7)(10)cos(51°)
c² = 149 - 140 · 0.6293
c² = 149 - 88.10 = 60.90
Step 2: c = √60.90 ≈ 7.80
A triangle has sides a = 5, b = 8, c = 11. Find angle C.
Step 1: Rearrange: cos(C) = (a² + b² - c²) / (2ab)
cos(C) = (25 + 64 - 121) / (2 · 5 · 8) = -32/80 = -0.4
Step 2: C = arccos(-0.4) ≈ 113.58°
When you know two sides and the included angle, the area is:
When you know all three sides, use Heron's formula:
Find the area of a triangle with sides a = 9, b = 14, and included angle C = 62°.
Area = ½ · 9 · 14 · sin(62°) = 63 · 0.8829 ≈ 55.62 square units
Polar coordinates provide an alternative to the Cartesian (rectangular) coordinate system. Instead of locating a point by horizontal and vertical distances (x, y), polar coordinates use a distance from the origin (r) and an angle from the positive x-axis (θ).
When converting to polar, be sure to adjust θ for the correct quadrant (arctan only gives values in Quadrants I and IV).
Convert (5, π/3) to rectangular coordinates.
x = 5 cos(π/3) = 5 · (1/2) = 5/2
y = 5 sin(π/3) = 5 · (√3/2) = 5√3/2
Rectangular coordinates: (5/2, 5√3/2) ≈ (2.5, 4.33)
Convert (-3, 3) to polar coordinates.
Step 1: r = √((-3)² + 3²) = √(9 + 9) = √18 = 3√2
Step 2: θ = arctan(3/(-3)) = arctan(-1) = -π/4
Step 3: The point (-3, 3) is in Quadrant II, so adjust: θ = π - π/4 = 3π/4
Polar coordinates: (3√2, 3π/4)
Many beautiful curves have simple equations in polar coordinates:
Describe the polar curve r = 4 cos(3θ).
This is a rose curve with a = 4 and n = 3. Since n = 3 is odd, the rose has 3 petals, each with length 4.
The petals are centered at θ = 0, θ = 2π/3, and θ = 4π/3 (equally spaced at 120° intervals).
You can convert equations between rectangular and polar form using the substitution relationships.
Convert x² + y² = 9 to polar form.
Since x² + y² = r²: r² = 9, so r = 3
(A circle of radius 3 centered at the origin.)
Convert r = 4 sin(θ) to rectangular form.
Step 1: Multiply both sides by r: r² = 4r sin(θ)
Step 2: Substitute r² = x² + y² and r sin(θ) = y:
x² + y² = 4y
Step 3: Rearrange: x² + y² - 4y = 0 → x² + (y - 2)² = 4
(A circle of radius 2 centered at (0, 2).)
Trigonometry is one of the most applied branches of mathematics. Its reach extends from ancient navigation to cutting-edge signal processing. Here are some of the most important real-world applications.
Trigonometry has been used for navigation since ancient times. By measuring angles to known landmarks or celestial objects, navigators can determine their position with remarkable accuracy.
From a point 200 meters from the base of a building, the angle of elevation to the top is 35°. How tall is the building?
Setup: tan(35°) = height / 200
height = 200 · tan(35°) = 200 · 0.7002 ≈ 140.0 meters
A ship sails 50 km on a bearing of N 30° E, then turns and sails 80 km on a bearing of S 60° E. How far is the ship from its starting point?
Step 1: Convert to standard position and find displacement components:
First leg: x₁ = 50 sin(30°) = 25 km east, y₁ = 50 cos(30°) = 43.3 km north
Second leg: x₂ = 80 sin(60°) = 69.3 km east, y₂ = -80 cos(60°) = -40 km south
Step 2: Total displacement: x = 25 + 69.3 = 94.3 km, y = 43.3 - 40 = 3.3 km
Step 3: Distance = √(94.3² + 3.3²) = √(8892.5 + 10.9) ≈ 94.4 km
Sine and cosine functions are the natural language of waves. Any wave — whether it's a sound wave, light wave, radio wave, or water wave — can be described using sinusoidal functions.
The general equation of a simple harmonic wave is:
Where:
Middle C on a piano has a frequency of 261.63 Hz. Write the equation for this sound wave with amplitude 0.02 and no phase shift.
y(t) = 0.02 sin(2π · 261.63 · t)
y(t) = 0.02 sin(1643.8t)
The period is T = 1/261.63 ≈ 0.00382 seconds (about 3.82 milliseconds per cycle).
Objects on springs, pendulums, and many other physical systems exhibit simple harmonic motion (SHM), which is described by trigonometric functions:
Where ω = 2πf is the angular frequency (in radians per second).
A mass on a spring oscillates with amplitude 0.1 m and period 2 seconds. Find the position and velocity at t = 0.5 s if the mass starts at maximum displacement.
Step 1: ω = 2π/T = 2π/2 = π rad/s
Step 2: Position: x(t) = 0.1 cos(πt)
x(0.5) = 0.1 cos(0.5π) = 0.1 · 0 = 0 m (at equilibrium)
Step 3: Velocity: v(t) = -0.1π sin(πt)
v(0.5) = -0.1π sin(0.5π) = -0.1π · 1 = -0.314 m/s
Trigonometry is essential for decomposing vectors into components and for computing resultant vectors — fundamental operations in physics and engineering.
A vector v with magnitude |v| at angle θ from the positive x-axis has components:
A force of 100 N acts at an angle of 40° above the horizontal. Find its horizontal and vertical components.
Fₓ = 100 cos(40°) = 100 · 0.7660 ≈ 76.6 N
Fᵧ = 100 sin(40°) = 100 · 0.6428 ≈ 64.3 N
Two forces act on an object: F₁ = 50 N at 30° and F₂ = 80 N at 120°. Find the magnitude and direction of the resultant force.
Step 1: Find components:
F₁ₓ = 50 cos(30°) = 43.3 N, F₁ᵧ = 50 sin(30°) = 25 N
F₂ₓ = 80 cos(120°) = -40 N, F₂ᵧ = 80 sin(120°) = 69.3 N
Step 2: Sum components: Rₓ = 43.3 + (-40) = 3.3 N, Rᵧ = 25 + 69.3 = 94.3 N
Step 3: Magnitude: |R| = √(3.3² + 94.3²) = √(8903.8) ≈ 94.4 N
Step 4: Direction: θ = arctan(94.3/3.3) ≈ 88.0° from the positive x-axis
Alternating current (AC) voltage and current are sinusoidal, making trigonometry the essential mathematical language for electrical engineering:
In the United States, household electricity uses V₀ ≈ 170 V and f = 60 Hz, giving the familiar "120 V AC" (the root-mean-square value is V₀/√2 ≈ 120 V).
Rotations in 2D computer graphics rely on trigonometry. To rotate a point (x, y) by angle θ around the origin:
This rotation matrix is the foundation of every 2D graphics engine, and its 3D extension (using Euler angles or quaternions) powers video games, CAD software, and CGI in films.
Rotate the point (3, 4) by 90° counterclockwise around the origin.
x' = 3 cos(90°) - 4 sin(90°) = 3(0) - 4(1) = -4
y' = 3 sin(90°) + 4 cos(90°) = 3(1) + 4(0) = 3
New point: (-4, 3)