Grade 11 · Pure Mathematics 1 · Cambridge A-Level 9709 · Age 16–17
Welcome to Trigonometry Fundamentals!
Trigonometry is the mathematics of angles and triangles, extended through the unit circle to a powerful tool for modelling waves, oscillations, and periodic phenomena. From finding missing sides in triangles to solving equations that arise in physics and engineering, trigonometry underpins much of A-Level mathematics and beyond.
sin²θ + cos²θ = 1 | sin(A+B) = sinA cosB + cosA sinB | R sin(θ+α) form
Learning Objectives
Define sin, cos, tan using SOH CAH TOA and the unit circle
Apply the CAST diagram to determine signs in all four quadrants
Convert between degrees and radians fluently
Recall exact values for 0°, 30°, 45°, 60°, 90° and beyond
Sketch and interpret graphs of sin, cos, and tan with period, amplitude, and asymptotes
Use and derive the Pythagorean identities sin²θ+cos²θ=1, 1+tan²θ=sec²θ, 1+cot²θ=cosec²θ
Solve equations of the form sinθ=k, cosθ=k, tanθ=k in given intervals using CAST
Solve equations reducible to quadratic form in trigonometry
Apply compound angle and double angle formulae
Express a sinθ + b cosθ in the harmonic form R sin(θ+α)
Trig Ratios
SOH CAH TOA, unit circle, CAST diagram, quadrant signs
Exact Values
0°–180° table, surds, graphs with period and amplitude
Principal value + CAST, quadratic trig, step-by-step method
Compound Angles
sin/cos/tan of A±B, double angle, R sin(θ+α) harmonic form
Applications
Max/min of trig expressions, modelling, equation solving
Learn 1 — Trig Ratios & Unit Circle
SOH CAH TOA
For a right-angled triangle with angle θ, the three primary trigonometric ratios are defined as:
sin θ = Opposite / Hypotenuse cos θ = Adjacent / Hypotenuse tan θ = Opposite / Adjacent
Memory aid: SOH CAH TOA Sin = Opposite / Hypotenuse
Cos = Adjacent / Hypotenuse
Tan = Opposite / Adjacent
Also note: tan θ = sin θ / cos θ (follows directly from the definitions above)
The Unit Circle
The unit circle has centre (0,0) and radius 1. For any angle θ measured anticlockwise from the positive x-axis, the point on the unit circle is (cos θ, sin θ).
A point P on the unit circle at angle θ satisfies: x-coordinate = cos θ y-coordinate = sin θ tan θ = y/x = sin θ / cos θ (undefined when x = 0, i.e. θ = 90°, 270°, …)
Since the radius = 1, by Pythagoras: x² + y² = 1, giving the fundamental identity sin²θ + cos²θ = 1.
The CAST Diagram
The CAST diagram tells you which trig ratios are positive in each quadrant:
S Sin positive (2nd quadrant) 90°–180°
A All positive (1st quadrant) 0°–90°
T Tan positive (3rd quadrant) 180°–270°
C Cos positive (4th quadrant) 270°–360°
Reading the CAST diagram:
1st quadrant (0° to 90°): sin, cos, tan all positive
2nd quadrant (90° to 180°): sin positive; cos, tan negative
3rd quadrant (180° to 270°): tan positive; sin, cos negative
4th quadrant (270° to 360°): cos positive; sin, tan negative
Negative Angles and Angles Beyond 360°
Negative angles: Measured clockwise from the positive x-axis.
sin(−θ) = −sin θ cos(−θ) = cos θ tan(−θ) = −tan θ
Angles beyond 360°: Trig functions are periodic, so you can subtract multiples of 360° (or 2π for radians).
sin(θ + 360°) = sin θ cos(θ + 360°) = cos θ tan(θ + 180°) = tan θ
Angles can be measured in degrees or radians. The full circle is 360° = 2π radians.
Degrees to radians: multiply by π/180 Radians to degrees: multiply by 180/π
Key conversions to memorise:
180° = π rad 90° = π/2 rad 60° = π/3 rad 45° = π/4 rad 30° = π/6 rad
Example: Convert 135° to radians: 135 × π/180 = 3π/4 Example: Convert 5π/6 to degrees: (5π/6) × (180/π) = 150°
In Cambridge A-Level, unless degrees are explicitly stated, work in radians. Always check the mode on your calculator before computing trig values.
Learn 2 — Exact Values & Graphs
Exact Values Table
You must know these exact values without a calculator. Derive them from the 30-60-90 and 45-45-90 triangles.
θ (degrees)
θ (radians)
sin θ
cos θ
tan θ
0°
0
0
1
0
30°
π/6
1/2
√3/2
1/√3 = √3/3
45°
π/4
1/√2 = √2/2
1/√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
1/√2
−1/√2
−1
150°
5π/6
1/2
−√3/2
−1/√3
180°
π
0
−1
0
For angles in the 2nd quadrant (90°–180°): use the related acute angle. e.g. sin 120° = sin 60° = √3/2 (sin positive in 2nd quadrant). cos 120° = −cos 60° = −1/2 (cos negative in 2nd quadrant).
Graph of y = sin θ
Period: 360° (or 2π rad) — the pattern repeats every 360° Amplitude: 1 — oscillates between −1 and +1 Key points: (0°,0), (90°,1), (180°,0), (270°,−1), (360°,0) Symmetry: sin(180°−θ) = sin θ sin(−θ) = −sin θ (odd function) No asymptotes — smooth continuous wave
Graph of y = cos θ
Period: 360° (or 2π rad) Amplitude: 1 — oscillates between −1 and +1 Key points: (0°,1), (90°,0), (180°,−1), (270°,0), (360°,1) Symmetry: cos(−θ) = cos θ (even function) cos(360°−θ) = cos θ Relationship to sin: cos θ = sin(90°−θ) — the cos graph is the sin graph shifted left by 90°
Graph of y = tan θ
Period: 180° (or π rad) — repeats every 180° Amplitude: none — range is all real numbers (−∞, +∞) Key points: (0°,0), (45°,1), (135°,−1), (180°,0) Asymptotes: vertical asymptotes at θ = 90°, 270°, −90°, … (i.e. θ = 90° + 180°n for integer n)
At these angles tan θ is undefined (cos θ = 0)
To remember which way the exact-value table goes: sin increases from 0 to 1 as θ goes from 0° to 90°, while cos decreases from 1 to 0. Think "sin starts at zero, cos starts at one".
Learn 3 — Trigonometric Identities
The Fundamental Pythagorean Identity
sin²θ + cos²θ = 1
This comes directly from the unit circle: a point (cos θ, sin θ) lies on the circle x² + y² = 1, so cos²θ + sin²θ = 1.
Derived Identities
Dividing sin²θ + cos²θ = 1 through by different terms gives two further identities:
Divide both sides by cos²θ:
sin²θ/cos²θ + 1 = 1/cos²θ tan²θ + 1 = sec²θ (equivalently: 1 + tan²θ = sec²θ)
(valid when cos θ ≠ 0)
Divide both sides by sin²θ:
1 + cos²θ/sin²θ = 1/sin²θ 1 + cot²θ = cosec²θ
(valid when sin θ ≠ 0)
tan θ = sin θ / cos θ cot θ = cos θ / sin θ sec θ = 1/cos θ cosec θ = 1/sin θ
Using Identities to Simplify Expressions
Example: Simplify (1 − sin²θ) / cos θ
Use sin²θ + cos²θ = 1, so 1 − sin²θ = cos²θ
= cos²θ / cos θ = cos θ
Example: Show that (tan θ + cot θ) sin θ cos θ ≡ 1
LHS = (sin θ/cos θ + cos θ/sin θ) sin θ cos θ
= (sin²θ/cos θ · cos θ / ... ) Let me expand directly:
= sin θ cos θ · sin θ/cos θ + sin θ cos θ · cos θ/sin θ
= sin²θ + cos²θ = 1 ✓
Using Identities to Prove Results
Always work on one side only (usually the more complicated side) and show it equals the other.
Prove: sin²θ/(1 + cos θ) ≡ 1 − cos θ
LHS = (1 − cos²θ)/(1 + cos θ) = (1 − cos θ)(1 + cos θ)/(1 + cos θ) = 1 − cos θ = RHS ✓
Strategy for Identity Proofs
1. Work on ONE side only — never cross-multiply or move terms across the equals sign.
2. Convert everything to sin and cos if stuck.
3. Look for chances to use sin²θ + cos²θ = 1 (or its rearrangements: sin²θ = 1 − cos²θ, cos²θ = 1 − sin²θ).
4. Factorise where possible (difference of two squares is common).
5. If you see sec or cosec, rewrite as 1/cos or 1/sin first.
Learn 4 — Solving Trigonometric Equations
The General Method
Step 1: Find the principal value (PV) using your calculator: θ = sin⁻¹(k), cos⁻¹(k), or tan⁻¹(k) Step 2: Use the CAST diagram (or symmetry of the graph) to find all solutions in the given interval Step 3: Check whether you need to add/subtract 360° (or 2π) to find more solutions within the interval
Solving sin θ = k
The second solution comes from the symmetry of the sin graph: sin(180° − θ) = sin θ
Never divide both sides by a trig function — you will lose solutions. For example, never divide by sin θ; instead move all terms to one side and factorise.
Learn 5 — Compound & Double Angle Formulae
Compound Angle Formulae
sin(A ± B) = sinA cosB ± cosA sinB
cos(A ± B) = cosA cosB ∓ sinA sinB
tan(A ± B) = (tanA ± tanB) / (1 ∓ tanA tanB)
Example: Find the exact value of sin 75°
sin 75° = sin(45° + 30°) = sin45°cos30° + cos45°sin30°
= (√2/2)(√3/2) + (√2/2)(1/2) = √6/4 + √2/4 = (√6 + √2)/4
Example: Find the exact value of cos 15°
cos 15° = cos(45° − 30°) = cos45°cos30° + sin45°sin30°
= (√2/2)(√3/2) + (√2/2)(1/2) = √6/4 + √2/4 = (√6 + √2)/4
Double Angle Formulae
Setting B = A in the compound angle formulae gives the double angle results:
sin 2A = 2 sinA cosA
cos 2A = cos²A − sin²A = 1 − 2sin²A = 2cos²A − 1
tan 2A = 2tanA / (1 − tan²A)
Three forms of cos 2A — when to use each:
cos 2A = cos²A − sin²A (when you have a mix of sin² and cos²)
cos 2A = 1 − 2sin²A → sin²A = (1 − cos2A)/2 (to eliminate sin²)
cos 2A = 2cos²A − 1 → cos²A = (1 + cos2A)/2 (to eliminate cos²)
The Harmonic Form: R sin(θ + α)
Any expression of the form a sinθ + b cosθ can be written as R sin(θ + α), where R > 0 and 0 < α < 90°.
a sinθ + b cosθ = R sin(θ + α) where R = √(a² + b²) and tan α = b/a
Derivation: Expand R sin(θ + α) = R sinθ cosα + R cosθ sinα
Comparing with a sinθ + b cosθ:
R cosα = a R sinα = b
R² = a² + b² → R = √(a² + b²)
tan α = (R sinα)/(R cosα) = b/a → α = tan⁻¹(b/a)
Example: Write 3sinθ + 4cosθ in the form Rsin(θ + α)
R = √(3² + 4²) = √(9 + 16) = √25 = 5
tan α = 4/3 → α = tan⁻¹(4/3) ≈ 53.13°
Answer: 5sin(θ + 53.13°)
Maximum value = 5 (when θ + 53.13° = 90°, i.e. θ ≈ 36.87°)
Minimum value = −5 (when θ + 53.13° = 270°, i.e. θ ≈ 216.87°)
Other forms: You may also need R sin(θ − α), R cos(θ + α), or R cos(θ − α).
R cos(θ − α) = R cosθ cosα + R sinθ sinα → R cosα = b, R sinα = a → same R, same method.
For the R form, always check the sign of both a and b carefully before computing α. If either coefficient is negative, determine the quadrant of α from the signs of R sinα and R cosα, not just from tan α.
Worked Examples
Eight fully worked examples covering all key trigonometry techniques for Cambridge A-Level 9709.
Example 1 — Exact Value Using CAST
Q: Without a calculator, find the exact value of sin 225° and cos(−60°).
B1: 225° = 180° + 45°, so it is in the 3rd quadrant. In the 3rd quadrant, sin is negative.
M1: Related acute angle = 45°. So sin 225° = −sin 45° = −1/√2 = −√2/2.
B1: cos(−60°) = cos(60°) because cos is an even function (cos(−θ) = cos θ).
A1: cos(−60°) = cos 60° = 1/2. sin 225° = −√2/2. 4 marks
Example 2 — CAST: All Solutions in an Interval
Q: Solve cos θ = −1/2 for 0° ≤ θ ≤ 360°.
M1: PV: cos⁻¹(1/2) = 60°. Since cos is negative, solutions are in the 2nd and 3rd quadrants.
These are the errors examiners see most often. Understanding why they are wrong helps you avoid them.
Mistake 1 — Missing Solutions in the Interval
✗ Solving sin θ = 0.5 for 0° ≤ θ ≤ 360°: only writing θ = 30°
✓ Always find BOTH solutions. Sin is positive in quadrants 1 and 2: θ = 30° and θ = 150°. Always draw or mentally use the CAST diagram.
Mistake 2 — Mixing Degrees and Radians
✗ Solving sin θ = 0.5 in radians but computing sin⁻¹(0.5) = 30 (degrees mode on calculator)
✓ Always check your calculator mode. In radians, sin⁻¹(0.5) = π/6 ≈ 0.5236. If the interval is given in radians (e.g. 0 ≤ θ ≤ 2π), work in radians throughout.
Mistake 3 — Dividing by a Trig Function and Losing Solutions
✗ sin 2θ = sin θ → 2sin θ cos θ = sin θ → divide by sin θ → 2cos θ = 1
✓ Never divide by a trig function. Instead, rearrange to zero and factorise: 2sinθ cosθ − sinθ = 0 → sinθ(2cosθ − 1) = 0. This gives sinθ = 0 (extra solutions!) AND cosθ = 1/2.
Mistake 4 — Wrong Sign in cos(A − B)
✗ cos(A − B) = cosA cosB − sinA sinB (copying the sin rule sign pattern)
✓ cos(A − B) = cosA cosB + sinA sinB. The signs for cos compound formulae are the OPPOSITE of what you might expect: cos(A+B) has a minus, cos(A−B) has a plus.
Mistake 5 — Using Wrong Form of cos 2A
✗ Solving 3cos 2θ + sinθ = 0 by using cos 2θ = 2cos²θ − 1 (leaves both sin and cos in equation)
✓ When the equation contains sinθ, use cos 2θ = 1 − 2sin²θ to get everything in terms of sinθ. Choose the form that eliminates one trig function.
Mistake 6 — Incorrect R Form: Wrong α Quadrant
✗ For −3sinθ + 4cosθ = Rsin(θ + α): computing α = tan⁻¹(4/3) ≈ 53.1° (ignoring the negative coefficient)
✓ After expanding Rsin(θ+α) = Rsinθ cosα + Rcosθ sinα, match: Rcosα = −3, Rsinα = 4. Since cosα < 0 and sinα > 0, α is in the 2nd quadrant: α = 180° − tan⁻¹(4/3).
Mistake 7 — Forgetting Periodicity When Extending the Interval
✗ Solving sin 2θ = 0.5 for 0° ≤ θ ≤ 360°: finding 2θ = 30° and 150° only (missing the next period)
✓ If you substitute u = 2θ, the interval for u is 0° ≤ u ≤ 720°. Find ALL solutions for u in this doubled interval: u = 30°, 150°, 390°, 510°. Then halve each: θ = 15°, 75°, 195°, 255°.
Mistake 8 — Identity Proof: Working on Both Sides
✗ In a proof, writing LHS = ... = RHS while simultaneously manipulating both sides
✓ Always work on ONE side only. Start from the more complex side and transform it step by step until it matches the other side. Never move terms across the ≡ sign during a proof.
These proofs form the foundation of trigonometry. Knowing them will help you answer "show that" questions and understand where the formulae come from.
Proof 1 — sin²θ + cos²θ = 1 from the Unit Circle
Consider a unit circle (radius = 1) centred at the origin.
For any angle θ, the point P on the circle is P = (cos θ, sin θ) by definition.
Since P lies on the circle x² + y² = 1, we substitute:
(cos θ)² + (sin θ)² = 1 ∴ sin²θ + cos²θ = 1 Q.E.D.
Proof 2 — 1 + tan²θ = sec²θ from the Pythagorean Identity
Start with sin²θ + cos²θ = 1.
Divide every term by cos²θ (valid when cos θ ≠ 0):
sin²θ/cos²θ + cos²θ/cos²θ = 1/cos²θ
(sinθ/cosθ)² + 1 = (1/cosθ)² tan²θ + 1 = sec²θ Q.E.D.
Proof 3 — sin(A+B) Geometric Derivation
Consider angle A+B. Draw a right triangle where the hypotenuse makes angle A+B with the horizontal.
Using area methods or construction of two nested triangles with angles A and B:
In the construction: project the hypotenuse onto the vertical axis.
The vertical component = sin(A+B) by definition.
Breaking the angle into A and B separately using rotation:
Horizontal component at angle A then rotated by B:
sin(A+B) = sinA cosB + cosA sinB (The full geometric proof uses a unit right triangle divided into two triangles sharing the diagonal; reading off opposite/hypotenuse for each sub-triangle and combining gives the result.) ∴ sin(A+B) = sinA cosB + cosA sinB Q.E.D.
Proof 4 — Double Angle: sin 2A = 2 sinA cosA
Use the compound angle formula for sin(A+B) with B = A:
sin(A + A) = sinA cosA + cosA sinA
sin 2A = 2 sinA cosA Q.E.D.
Proof 5 — cos 2A = 1 − 2sin²A
Start with cos 2A = cos²A − sin²A (from compound angle with B = A).
Replace cos²A using the identity cos²A = 1 − sin²A:
cos 2A = (1 − sin²A) − sin²A = 1 − 2sin²A ∴ cos 2A = 1 − 2sin²A Q.E.D.
Rearranging: sin²A = (1 − cos 2A)/2 — a very useful form for integration and equation-solving.
Interactive Unit Circle Visualiser
Drag the slider or type an angle to see the point on the unit circle and the values of sin, cos, and tan.
Express 8sinθ + 15cosθ in the form Rsin(θ + α), where R > 0 and 0 < α < 90°. Give α correct to 2 decimal places. Hence find the maximum value of 8sinθ + 15cosθ and state the value of θ in [0°, 360°] at which this maximum occurs.
Find the exact value of cos²(π/12) − sin²(π/12). [Hint: use a double angle formula.]
cos²A − sin²A = cos 2A [B1 recognising double angle]
With A = π/12: cos²(π/12) − sin²(π/12) = cos(2 × π/12) = cos(π/6) [M1 A1]
cos(π/6) = √3/2 [A1 A1]
Q8 [6 marks]
The function f(θ) = 5sinθ + 12cosθ is written in the form f(θ) = Rsin(θ + α). (i) Find R and α. (ii) Hence solve f(θ) = 6.5 for 0° ≤ θ ≤ 360°, giving solutions to 1 decimal place.
Express 3sinθ − 4cosθ in the form Rsin(θ − α), where R > 0 and 0 < α < π/2. Hence (i) find the greatest and least values of 3sinθ − 4cosθ, and (ii) solve 3sinθ − 4cosθ = 2 for 0 ≤ θ ≤ 2π.