Understanding
Trigonometry
This guide explains the why behind every formula — not just what to memorize, but how to think about it. Work through each section in order, try every checkpoint question, then check your answer.
You've been measuring angles in degrees your whole life. A full turn is 360°, a right angle is 90°. But why 360? It's largely historical — the Babylonians used a base-60 number system, and 360 is conveniently divisible by many numbers. It's a practical choice, not a mathematical one.
Mathematicians prefer a different unit: the radian. And once you understand what a radian actually is, you'll see why it makes everything cleaner.
Imagine taking the radius of a circle and bending it along the edge of that circle. The angle that arc creates at the centre is exactly 1 radian. This isn't an arbitrary choice — it's the angle you get when the arc length equals the radius. This natural definition makes every formula involving circles simpler: arc length becomes s = rθ with no conversion factor needed.
Since the full circumference is 2πr, and one radian uses an arc of r, you need exactly 2π radians to go all the way around. That gives us the golden bridge:
To convert, think of it as a proportion. Cross-multiply:
Minutes and Seconds
Degrees can be broken into smaller parts the same way hours are:
- 1° = 60 minutes (written 60′)
- 1′ = 60 seconds (written 60″)
- So 2° 5′ 30″ = 2 degrees, 5 minutes, 30 seconds
Use the relationship π = 180°. Write it as a conversion factor to multiply:
= 300π / 180
= 5π / 3
The π in the numerator and π in the denominator cancel:
= 1260 / 4
= 315°
One full turn = 360°. So 3/5 of a turn:
(b) Convert 13π/4 radians to degrees.
(c) How many degrees is 0.125 turns?
(a) 480 × π/180 = 480π/180 = 8π/3 radians
(b) (13π/4) × (180/π) = 13 × 45 = 585°
(c) 0.125 × 360° = 45°
An angle is in standard position when its vertex is at the origin and its initial arm lies along the positive x-axis. You measure the angle by rotating counter-clockwise (positive) or clockwise (negative) to reach the terminal arm.
Reference Angles — the Key Tool
A reference angle is the acute angle (always between 0° and 90°) between the terminal arm and the x-axis (not y-axis). It's your shortcut to finding trig values in any quadrant, because the trig function value is the same magnitude as the equivalent reference angle — you just adjust the sign using CAST.
| Quadrant | Angle range | Reference angle formula | Example |
|---|---|---|---|
| I | 0° to 90° | ref = θ | θ = 40° → ref = 40° |
| II | 90° to 180° | ref = 180° − θ | θ = 150° → ref = 30° |
| III | 180° to 270° | ref = θ − 180° | θ = 220° → ref = 40° |
| IV | 270° to 360° | ref = 360° − θ | θ = 310° → ref = 50° |
Reflections of Angles
If you reflect an angle across an axis, the terminal arm moves but the reference angle stays the same:
- Reflect 130° in the y-axis: stay in the same "row," flip the sign of x → 50° (QI)
- Reflect 130° in the x-axis: stay in the same "column," flip sign of y → 230° (QIII)
- Reflect in both axes: flip both → 310° (QIV)
In Quadrant III: standard angle = 180° + reference angle
For a circle of radius r at angle θ: P = (r·cos θ, r·sin θ)
150° is in QII. Reference angle = 180° − 150° = 30°
sin(150°) = +sin(30°) = 1/2
y = 48 × (1/2) = 24
(b) The reference angle is 25° and the terminal arm is in QIV. Find θ.
(c) Reflect 30° in the x-axis. What is the new angle in standard position?
(a) 255° is between 180° and 270° → Quadrant III. Reference angle = 255° − 180° = 75°
(b) QIV: θ = 360° − 25° = 335°
(c) Reflecting 30° (QI) in the x-axis: the y-values flip sign → terminal arm moves to QIV → θ = 360° − 30° = 330°
The unit circle is a circle with radius = 1 centred at the origin. It's not just a memorization chart — it's a tool that connects angles, coordinates, and trig values all in one picture.
For any angle θ, draw the terminal arm from the origin. Where it crosses the unit circle, the x-coordinate of that point is cos(θ) and the y-coordinate is sin(θ). This isn't a definition we impose — it falls naturally from SOH-CAH-TOA: in the right triangle formed, the hypotenuse = 1 (the radius), so cos = adjacent/1 = x, and sin = opposite/1 = y.
The CAST Rule — Which Signs Where?
Because x can be negative (left half of circle) and y can be negative (bottom half), the signs of cos and sin depend on the quadrant. "All Students Take Calculus":
The Key Values — Where They Come From
The "magic" values at 30°, 45°, and 60° come from two special triangles you should know cold:
• 45-45-90 triangle: sides 1, 1, √2. Hypotenuse = √2. So sin 45° = cos 45° = 1/√2 = √2/2.
• 30-60-90 triangle: sides 1, √3, 2. So sin 30° = 1/2, cos 30° = √3/2, sin 60° = √3/2, cos 60° = 1/2.
Once you know Q1, CAST handles the rest — just flip signs based on which quadrant you're in.
| Angle (°) | Radians | sin θ | cos θ | tan θ |
|---|---|---|---|---|
| 0° | 0 | 0 | 1 | 0 |
| 30° | π/6 | 1/2 | √3/2 | √3/3 |
| 45° | π/4 | √2/2 | √2/2 | 1 |
| 60° | π/3 | √3/2 | 1/2 | √3 |
| 90° | π/2 | 1 | 0 | undefined |
| 120° | 2π/3 | √3/2 | −1/2 | −√3 |
| 135° | 3π/4 | √2/2 | −√2/2 | −1 |
| 150° | 5π/6 | 1/2 | −√3/2 | −√3/3 |
| 180° | π | 0 | −1 | 0 |
| 210° | 7π/6 | −1/2 | −√3/2 | √3/3 |
| 225° | 5π/4 | −√2/2 | −√2/2 | 1 |
| 240° | 4π/3 | −√3/2 | −1/2 | √3 |
| 270° | 3π/2 | −1 | 0 | undefined |
| 300° | 5π/3 | −√3/2 | 1/2 | −√3 |
| 330° | 11π/6 | −1/2 | √3/2 | −√3/3 |
| 360° | 2π | 0 | 1 | 0 |
Periodic Values — Beyond 360°
Trig functions repeat every full cycle. For sin and cos that's 360° (or 2π), for tan it's 180° (or π):
- sin(600°) = sin(600° − 360°) = sin(240°) = −√3/2
- cos(−135°) = cos(360° − 135°) = cos(225°) = −√2/2
- sin(−270°) = sin(360° − 270°) = sin(90°) = 1
Subtract 2π repeatedly (or once with enough multiples):
4π/3 is in [0, 2π] ✓
4π/3 = 240°, which is in QIII.
sin(π/3) = √3/2. In QIII, sine is negative.
The restriction 180° ≤ t ≤ 270° places t in QIII. In QIII, cosine is negative ✓ — this is consistent with cos(t) = −12/13.
The reference angle has the same cosine value but positive: cos(ref) = 12/13.
In QIII: angle = 180° + reference angle
(b) Find the exact value of sin(−390°).
(c) Given that sin(t) = −7/25 and 270° ≤ t ≤ 360°, find t.
(a) 225° is in QIII, reference angle = 45°. Cosine is negative in QIII.
(b) −390° + 360° = −30°. Then −30° + 360° = 330°. That's QIV, reference = 30°. Sine is negative in QIV.
(c) QIV. Reference angle = arcsin(7/25) ≈ 16.3°. Angle = 360° − 16.3° ≈ 343.7°
Here's where radians pay off. The arc length of a sector is the portion of the circumference cut out by the angle. Since the full circumference is 2πr and a full angle is 2π radians:
The swing goes from 240° to 300°, so the angle covered is:
8 equal arcs means each arc = 1/8 of the full circle.
∠AOB = π/4 radians (45°)
In a right triangle, the three primary trig functions relate angles to side ratios. The three reciprocal functions flip those ratios:
| Function | Right triangle definition | Unit circle | Reciprocal |
|---|---|---|---|
| sin(θ) | opposite / hypotenuse | y-coordinate | csc(θ) = 1/sin(θ) |
| cos(θ) | adjacent / hypotenuse | x-coordinate | sec(θ) = 1/cos(θ) |
| tan(θ) | opposite / adjacent | sin(θ)/cos(θ) | cot(θ) = 1/tan(θ) |
Finding All Six Values from a Point on the Circle
If you're given a point P(t) on the unit circle, you can read off all six values directly. The key insight: the unit circle has hypotenuse = 1, so the right-triangle definitions simplify beautifully.
On the unit circle, x = cos(t) and y = sin(t):
sec(t) = 1/cos(t) = −13/5
cot(t) = 1/tan(t) = −5/12
sin(t) = −7/25. The restriction 270° ≤ t ≤ 360° places us in QIV. In QIV: sin is negative ✓ and cos is positive.
(−7/25)² + cos²(t) = 1
49/625 + cos²(t) = 1
cos²(t) = 576/625
cos(t) = ±24/25
Since we're in QIV, cos is positive: cos(t) = +24/25
sec(t) = −5/3 → cos(t) = −3/5. We're in QIII, so sin is also negative.
tan(t) = sin/cos = (−4/5)/(−3/5) = 4/3 → cot(t) = 1/tan = 3/4
The basic sine function y = sin(x) is a wave that goes from −1 to 1 with a period of 2π. The four parameters a, b, h, and k each transform this wave in a specific way. The key is understanding what each one actually does — not just memorizing formulas.
Parameter by Parameter
| Parameter | Name | What it actually does | Formula |
|---|---|---|---|
| a | Amplitude factor | Stretches the wave vertically. If a < 0, flips it upside-down. The wave now reaches |a| above and below the midline. | Amplitude = |a| |
| b | Frequency factor | Compresses or stretches the wave horizontally. Larger |b| = more waves in the same space = shorter period. | Period = 2π/|b| |
| h | Phase shift | Slides the entire wave left or right. The starting point moves. Positive h = shift RIGHT. | Shift = h units |
| k | Vertical shift | Raises or lowers the whole wave. Sets the midline of the wave at y = k. | Midline: y = k |
The basic wave sin(x) completes one full cycle in 2π. When we replace x with bx, the input reaches 2π (one full cycle) when bx = 2π, i.e. when x = 2π/b. So the cycle length (period) shrinks by a factor of b. Larger b = shorter period = more cycles packed in.
Period = 2π/1 = 2π
Frequency = 1/(2π)
Maximum = 0 + 2 = 2
Minimum = 0 − 2 = −2
Zeros in [0, 2π]: x = 0, π, 2π
Initial value = 2·sin(0) = 0
Period = 2π / 0.5 = 4π
Frequency = 0.5/(2π) = 1/(4π)
Phase shift = π to the RIGHT
Vertical shift = 1 DOWN (k = −1, midline at y = −1)
Maximum = −1 + 3 = 2
Minimum = −1 − 3 = −4
= 3·sin(−π/2) − 1
= 3·(−1) − 1 = −4
So the function starts at its minimum when x = 0.
The argument is 0.5πx + 2π. Factor out b = 0.5π:
Now parameters are clear: a = 2.5, b = 0.5π, h = −4, k = 1.5
Period = 2π/(0.5π) = 4
Phase shift = 4 to the LEFT (h = −4)
Maximum = 1.5 + 2.5 = 4
Minimum = 1.5 − 2.5 = −1
State the amplitude, period, phase shift, vertical shift, maximum, minimum, and initial value.
a=3, b=0.5, h=2, k=−1
Period = 2π/0.5 = 4π
Phase shift = 2 units RIGHT
Vertical shift = 1 DOWN (midline y = −1)
Maximum = −1 + 3 = 2
Minimum = −1 − 3 = −4
Initial value: f(0) = 3·sin(0.5(0−2))−1 = 3·sin(−1)−1 ≈ 3(−0.841)−1 ≈ −3.52
An identity is an equation that is true for all valid values of the variable — not just specific ones. Proving an identity means showing algebraically that one side simplifies to match the other.
The Pythagorean Identities — From First Principles
On the unit circle, any point is (cos θ, sin θ) and lies on the circle x² + y² = 1 (since the radius = 1). Substituting: cos²θ + sin²θ = 1. That's it — the Pythagorean theorem applied to the unit circle. It's not arbitrary at all.
The three Pythagorean identities come from one
- sin²θ = 1 − cos²θ = (1 − cosθ)(1 + cosθ)
- cos²θ = 1 − sin²θ = (1 − sinθ)(1 + sinθ)
- tan²θ = sec²θ − 1
- cot²θ = csc²θ − 1
The Reciprocal Identities
- csc θ = 1/sin θ sec θ = 1/cos θ cot θ = 1/tan θ
- tan θ = sin θ / cos θ cot θ = cos θ / sin θ
Strategy for Proving Identities
In order of what to try:
- Convert to sin and cos — replace csc, sec, tan, cot with their sin/cos equivalents
- Find common denominators — if adding/subtracting fractions
- Spot Pythagorean substitutions — see sin²+cos²? Replace with 1. See 1−cos²? Replace with sin².
- Factor — especially difference of squares: (1−cosθ)(1+cosθ) = sin²θ
- Multiply by a conjugate — if the denominator has (1+sinθ), multiply top and bottom by (1−sinθ)
Replace csc θ with its reciprocal identity:
1 − sin²θ looks like part of the Pythagorean identity. Replace it:
= cos²θ · (1/cos²θ)
= 1 = RS ✓
Expand (sin θ + cos θ)² using (a+b)² = a²+ 2ab + b²:
= 1 + 2sinθcosθ − 1
= 2sinθcosθ = RS ✓
Work on the left side. Factor out cos²θ:
Apply the Pythagorean identity tan²θ + 1 = sec²θ:
A trig equation asks: for which angles does this equation hold? Unlike algebraic equations, trig equations usually have infinitely many solutions — because sin and cos repeat. In Sec 5, you'll typically be asked to find solutions in [0°, 360°] or [0, 2π].
The Method — 4 Steps
- Step 1: Isolate the trig function on one side.
- Step 2: Use your calculator or unit circle to find the reference angle (always positive, always acute).
- Step 3: Use the value's sign and CAST to find which quadrant(s) the solution lives in.
- Step 4: Convert the reference angle to the actual angle(s)
using the quadrant formula.
sin is negative. The magnitude is 1/2. We know sin(30°) = 1/2.
Sine is negative in QIII and QIV.
QIV: t = 360° − 30° = 330°
cos(π/3) = 1/2, so the reference angle is π/3.
QIII: t = π + π/3 = 4π/3
cos(t) = −√3/2
cos(π/6) = √3/2, so reference angle = π/6.
QIII: t = π + π/6 = 7π/6
Use the absolute value: sin(ref) = 0.2
QIV: t = 2π − 0.201 ≈ 6.082 rad
6.082 ✓ (between 0 and 6.283)
(b) Given that cos(t) = −4/5 and π ≤ t ≤ 3π/2, find the exact value of t, and also find sin(t) and tan(t).
(c) Solve: 2sin(t) + 1 = 0 for 0 ≤ t ≤ 2π. Express answers in exact form.
(a) Reference angle = 45°. Cosine negative → QII and QIII.
(b) π ≤ t ≤ 3π/2 is QIII. cos(t) = −4/5. Use Pythagorean identity:
tan(t) = sin/cos = (−3/5)/(−4/5) = 3/4
t = π + arccos(4/5) ≈ π + 0.6435 ≈ 3.785 rad
(c) 2sin(t) = −1 → sin(t) = −1/2. Reference = π/6. Negative sine → QIII and QIV.