○ The Unit Circle & Trig Graphs

The unit circle (radius = 1) is the foundation of trigonometry. As an angle θ sweeps round the circle, the x-coordinate traces out cos θ and the y-coordinate traces out sin θ. The resulting waves are the basis of all engineering oscillations.

Unit circle — radius 1 sin θ = y-coordinate cos θ = x-coordinate tan θ = sin θ / cos θ 360° = full cycle

Drag the point or click anywhere on the circle

What the unit circle shows

For any angle θ, the point P on the unit circle has coordinates (cos θ, sin θ). This means:

cos θ = horizontal distance from centre to P

sin θ = vertical distance from centre to P

tan θ = sin θ / cos θ = vertical/horizontal

💡 The unit circle explains why trig values repeat every 360° — after a full rotation, P is back where it started.

Exact values — memorise these

sin 0°=0 sin 30°=½ sin 45°=√2/2 sin 60°=√3/2 sin 90°=1

cos 0°=1 cos 30°=√3/2 cos 45°=√2/2 cos 60°=½ cos 90°=0

tan 0°=0 tan 30°=1/√3 tan 45°=1 tan 60°=√3 tan 90°=∞

📌 Memory trick: sin values for 0°,30°,45°,60°,90° are √0/2, √1/2, √2/2, √3/2, √4/2 = 0, 0.5, 0.71, 0.87, 1

📦

Quadrants & Sign Rules

The sign (+ or −) of sin, cos, and tan depends on which quadrant the angle is in. Knowing this lets you work out values for angles above 90° without a calculator.

Reference angle method

Every angle has a reference angle — the acute angle between the terminal side and the x-axis:

Q1 (0°–90°): ref = θ

Q2 (90°–180°): ref = 180° − θ

Q3 (180°–270°): ref = θ − 180°

Q4 (270°–360°): ref = 360° − θ

💡 Calculate trig of the reference angle, then apply the correct sign from the CAST rule.

Worked examples

sin 210°: Q3, ref=30°, sin negative → sin210° = −sin30° = −0.5

cos 135°: Q2, ref=45°, cos negative → cos135° = −cos45° = −0.707

tan 300°: Q4, ref=60°, tan negative → tan300° = −tan60° = −1.732

sin 330°: Q4, ref=30°, sin negative → sin330° = −sin30° = −0.5

∿

The Sine Wave — y = sin θ

The sine wave starts at 0, rises to +1 at 90°, returns to 0 at 180°, falls to −1 at 270°, and returns to 0 at 360°. It is the most fundamental waveform in engineering.

Drag the point on the circle · watch the sine wave trace out

Key properties of y = sin θ

Period = 360° (repeats every full turn)

Amplitude = 1 (peaks at +1 and −1)

Starts at 0, crosses zero at 0°, 180°, 360°

Maximum +1 at θ = 90°

Minimum −1 at θ = 270°

💡 Engineering form: y = A·sin(ωt + φ)
A = amplitude, ω = angular frequency, φ = phase shift

Reading values from the graph

The graph shows the value of sin θ at a glance:

∿

The Cosine Wave — y = cos θ

The cosine wave starts at +1 at 0°, falls to 0 at 90°, reaches −1 at 180°, rises back to 0 at 270°, and returns to +1 at 360°. It is the sine wave shifted 90° to the left.

Drag the point on the circle · watch the cosine wave trace out

θ	sin θ	cos θ	tan θ
0°	0	+1	0
30°	+½	+√3/2	+1/√3
45°	+√2/2	+√2/2	+1
60°	+√3/2	+½	+√3
90°	+1	0	—
120°	+√3/2	−½	−√3
135°	+√2/2	−√2/2	−1
150°	+½	−√3/2	−1/√3
180°	0	−1	0
210°	−½	−√3/2	+1/√3
225°	−√2/2	−√2/2	+1
240°	−√3/2	−½	+√3
270°	−1	0	—
300°	−√3/2	+½	−√3
315°	−√2/2	+√2/2	−1
330°	−½	+√3/2	−1/√3
360°	0	+1	0

Key properties of y = cos θ

Period = 360°

Amplitude = 1

Starts at MAXIMUM (+1) at θ = 0°

Crosses zero at 90° and 270°

cos θ = sin(θ + 90°) — same shape, 90° phase shift

💡 Key identity: sin²θ + cos²θ = 1 (always true — comes from Pythagoras on the unit circle!)

Sine vs Cosine comparison

θ	sin θ	cos θ
0°	0	1
90°	1	0
180°	0	−1
270°	−1	0
360°	0	1

📌 When sin is at its peak (90°), cos is zero — and vice versa at 0°/360°. They are always 90° apart.

∿

The Tangent Wave — y = tan θ

Tan is very different from sin and cos. It has asymptotes at 90° and 270° where the value shoots off to ±infinity. Its period is only 180°.

Drag the point on the circle · asymptotes at 90° and 270°

Key properties of y = tan θ

Period = 180° (half that of sin and cos)

No maximum or minimum — goes to ±∞

Asymptotes (undefined) at 90°, 270°, 450°…

Crosses zero at 0°, 180°, 360°…

tan θ = sin θ / cos θ — undefined when cos θ = 0

⚠️ tan 90° is UNDEFINED — there is no answer. Your calculator gives an error or a very large number.

Why tan has asymptotes

tan θ = sin θ / cos θ. At 90°, cos 90° = 0. You cannot divide by zero — so tan 90° is undefined.

As θ → 90° from below: tan → +∞

As θ → 90° from above: tan → −∞

💡 In engineering, tan θ appears in the impedance triangle (tan φ = X/R), pipe slopes, and cutting tool angles. It is always used for angles well away from 90°.

θ	tan θ
0°	0
45°	1
89°	57.3 (very large)
90°	UNDEFINED
135°	−1
180°	0

📊

Comparing Sin, Cos and Tan

Sine (red), Cosine (navy) and Tan (purple, clipped) overlaid on one graph. See how they relate at every angle.

Key relationships

sin θ = cos(90° − θ) — complementary angles

cos θ = sin(90° + θ) — cos leads sin by 90°

tan θ = sin θ / cos θ

sin²θ + cos²θ = 1 — Pythagorean identity

sin and cos cross at 45° and 225° (value = ±√2/2 = ±0.707)

What to notice on the graph

🔴 Sin starts at 0, rises first
🔵 Cos starts at +1 (already at its peak)
🟣 Tan starts at 0 but rises much faster than sin
⚪ Where sin=0, tan=0 (at 0°, 180°, 360°)
⚪ Where cos=0, tan is undefined (at 90°, 270°)
⚪ Tan = 1 when sin = cos (at 45°, 225°)

⚙️

Applications

Where this topic is used in engineering, manufacturing, maintenance and daily life.

⚡Electrical: Mains AC: V=325sin(100πt). 50 Hz, 230 V RMS.
🔊Acoustics: Sound: P=P₀sin(2πft). Read amplitude & phase from graph.
⚙️Mechanical: Piston: x=rcos(ωt). Simple harmonic motion.
📡Communications: Phase shift φ in V=V₀sin(ωt+φ). Read from graph.

🎮

Quick Fire Quiz

Test your knowledge — 10 questions, instant feedback.

⚡ Trig Graphs Blitz

Score: 0 Streak: 0 🔥 Q 1/10

Loading...

✏

Practice Questions

Use the CAST rule and your knowledge of the graphs. Always state which quadrant.

🟢 Tier 1 — Reading Values

Q1

State the value of sin 90°, cos 180°, and tan 45° without a calculator.

[3]

sin 90° = 1 (peak of sine wave)
cos 180° = −1 (minimum of cosine wave)
tan 45° = 1 (sin45°/cos45° = 0.707/0.707 = 1)

Q2

Using the CAST rule, state the sign (+ or −) of: sin 150°, cos 210°, tan 320°.

[3]

sin 150°: Q2, only SIN positive → +
cos 210°: Q3, only TAN positive → cos is −
tan 320°: Q4, only COS positive → tan is −

Q3

Find the exact value of sin 210° and cos 315° using the reference angle method.

[4]

sin 210°: Q3, ref = 210°−180° = 30°, sin negative → sin210° = −0.5
cos 315°: Q4, ref = 360°−315° = 45°, cos positive → cos315° = +0.707

Q4

At what TWO angles between 0° and 360° does sin θ = 0.5?

[2]

sin is positive in Q1 and Q2

Primary angle: θ = sin⁻¹(0.5) = 30°
Q2 angle: 180° − 30° = 150°
θ = 30° and 150°

Q5

Find all angles between 0° and 360° where cos θ = −0.866.

[3]

cos is negative in Q2 and Q3.
Reference: cos⁻¹(0.866) = 30°
Q2: 180° − 30° = 150°
Q3: 180° + 30° = 210°
θ = 150° and 210°

🟡 Tier 2 — Graph Properties & Engineering

Q6

An AC voltage is given by v = 325·sin(314t) V. Find: (a) peak voltage, (b) frequency, (c) value at t = 0.005 s.

[4]

(a) Peak = 325 V
(b) ω=314 rad/s → f=314/(2π) = 50 Hz
(c) 314×0.005=1.57 rad=90° → v=325·sin90° = 325 V

Q7

A crank of radius 80 mm rotates at 1200 rpm. At θ = 120° from TDC, find the horizontal and vertical displacement of the crank pin from centre.

[3]

x = 80·cos120° = 80×(−0.5) = −40 mm
y = 80·sin120° = 80×0.866 = +69.3 mm
The negative x means the crank has passed centre to the left.

Q8

A structure has natural frequency f_n = 4 Hz and mass 50 kg. Find the stiffness k using ω_n = √(k/m).

[3]

ω_n = 2π×4 = 25.13 rad/s
k = m×ω_n² = 50×(25.13)² = 50×632 = 31,600 N/m

🔴 Tier 3 — T Level Challenge

Q9

Two AC signals: V₁ = 10·sin(ωt) and V₂ = 10·sin(ωt + 60°). Find the resultant V₁ + V₂ at ωt = 30° and ωt = 90°.

[4]

At ωt=30°: V₁=10·sin30°=5.0V V₂=10·sin90°=10.0V Total=15.0 V
At ωt=90°: V₁=10·sin90°=10.0V V₂=10·sin150°=5.0V Total=15.0 V

Q10

A cam follower has SHM profile y=(30/2)·(1−cos(πθ/90°)) mm for 0°≤θ≤90°. Find the displacement at θ=30° and θ=60°, and state where the follower is at half lift.

[5]

At θ=30°: y=15·(1−cos(π×30/90))=15·(1−cos60°)=15·(1−0.5)=7.5mm
At θ=60°: y=15·(1−cos120°)=15·(1−(−0.5))=15×1.5=22.5mm
Half lift (y=15mm): 1−cosα=1 → cosα=0 → α=90° → θ=90°×(90°/π)... wait:
y=15 → 1−cos(πθ/90)=1 → cos(πθ/90)=0 → πθ/90=90° → θ=45°

🎯 Score: Q1–5 reading/CAST (15 marks) · Q6–8 engineering (10 marks) · Q9–10 challenge (9 marks)

👉 Next topics: Sequences & Series — arithmetic, geometric, and their engineering uses.