BTEC Level 3 · Further Engineering MathsTrig Identities
🏠 Trigonometric Identities
An identity is a relationship that's true for every value of the angle involved — not just a solution to one equation. These identities let you rewrite trig expressions into simpler or more useful forms, which is exactly what's needed to combine waveforms, analyse oscillations, and solve equations that a calculator alone can't handle directly.
An identity is true for all angles; an equation is only true for specific angles
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Everything here builds from just one starting identity: sin²θ + cos²θ = 1
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Compound and double angle formulae let you expand or combine angles
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Engineers use these to add waveforms of the same frequency into one single wave
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Key Words You'll Meet Throughout
IdentityTrue for every value of the angle
Compound AngleAn angle made of two added/subtracted angles
Double AngleAn angle that's exactly double another
AmplitudeThe maximum size of a wave
Phase AngleA shift in the starting point of a wave
🗺️ How these topics connect
Pythagorean identities — the foundation. Everything else in this resource can be derived from sin²θ + cos²θ = 1.
Compound angle formulae — expand sin(A ± B), cos(A ± B), tan(A ± B) — essential whenever an angle is made of two parts added or subtracted.
Double angle formulae — a special case of compound angles where A = B, giving formulae for sin2A, cos2A, tan2A.
Solving trig equations — using these identities to simplify an equation into a form you can actually solve.
Engineering applications — combining sinusoidal waveforms of the same frequency into a single R sin(θ + α) wave, essential in AC theory and mechanical vibration.
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Pythagorean Identities
Three identities, all derived from a single starting point — the relationship between sin and cos on the unit circle.
📋 Pythagorean Identities — Quick Reference
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sin²θ + cos²θ = 1 — the starting identity
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Divide by cos²θ: 1 + tan²θ = sec²θ
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Divide by sin²θ: 1 + cot²θ = cosec²θ
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Key Words — What Do They Mean?
sec θ1 / cos θ
cosec θ1 / sin θ
cot θ1 / tan θ
Where sin²θ + cos²θ = 1 comes from
On a right-angled triangle inside a unit circle (hypotenuse = 1), the opposite and adjacent sides are exactly sinθ and cosθ. Pythagoras' theorem then gives directly:
sin²θ + cos²θ = 1
This single identity is the seed for everything else in this section — and, eventually, for the compound and double angle formulae too.
Deriving the other two forms
✏️ Worked derivation
Start: sin²θ + cos²θ = 1
Divide every term by cos²θ: sin²θ/cos²θ + cos²θ/cos²θ = 1/cos²θ
Simplify: tan²θ + 1 = sec²θ
Answer:1 + tan²θ = sec²θ
Dividing the original identity by sin²θ instead gives the third form, 1 + cot²θ = cosec²θ, by exactly the same method.
Enter an angle, then click Verify to check all three Pythagorean identities numerically.
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Compound Angle Formulae
How to expand sin, cos and tan of an angle that's really two angles added or subtracted together.
📋 Compound Angle Formulae — Quick Reference
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sin(A±B) = sinA cosB ± cosA sinB
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cos(A±B) = cosA cosB ∓ sinA sinB (signs swap!)
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tan(A±B) = (tanA ± tanB) / (1 ∓ tanA tanB)
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Key Words — What Do They Mean?
Compound AngleAn angle written as A + B or A − B
ExpandWrite out in terms of the individual angles
The three 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)
📌 Notice cos(A±B) flips the sign compared to what you might expect — a "+" in the angle becomes a "−" in the formula, and vice versa. This trips people up constantly, so double check it every time.
Using a compound angle formula
✏️ Worked example
Problem: Find the exact value of sin75°, using sin75° = sin(45°+30°).
Apply the formula: sin(45°+30°) = sin45°cos30° + cos45°sin30°
Pick a function and two angles, then click Verify — compares the compound angle formula against direct calculation.
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Double Angle Formulae
A special case of the compound angle formulae, where B = A — giving compact expressions for sin2A, cos2A and tan2A.
📋 Double Angle Formulae — Quick Reference
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sin2A = 2 sinA cosA
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cos2A has three equivalent forms
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tan2A = 2tanA / (1 − tan²A)
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Key Words — What Do They Mean?
Double Angle2A — exactly twice angle A
Deriving sin2A
✏️ Worked derivation
Start from the compound angle formula: sin(A+B) = sinAcosB + cosAsinB
Set B = A: sin(A+A) = sinAcosA + cosAsinA
Answer:sin2A = 2sinAcosA
The three forms of cos2A
Starting from cos(A+A) = cosAcosA − sinAsinA = cos²A − sin²A, and then substituting the Pythagorean identity sin²A + cos²A = 1 in two different directions, gives three equivalent forms — useful for different situations:
cos2A = cos²A − sin²A
cos2A = 2cos²A − 1 (using sin²A = 1−cos²A)
cos2A = 1 − 2sin²A (using cos²A = 1−sin²A)
📌 The 2nd and 3rd forms are especially useful for solving equations, since each is written in terms of just one trig function instead of two.
Using double angle formulae
✏️ Worked example
Problem: Given sinA = 0.6 and A is acute, find sin2A and cos2A.
Find cosA using Pythagorean identity: cosA = √(1−0.6²) = √0.64 = 0.8
Rearrange to quadratic form (let x=cosθ): −2x² + x + 1 = 0 → 2x² − x − 1 = 0
Factorise: (2x + 1)(x − 1) = 0 → x = −0.5 or x = 1
So cosθ = −0.5 or cosθ = 1: θ = 120°, 240° (from cosθ=−0.5) and θ = 0° (from cosθ=1)
Answer: θ = 0°, 120°, 240°
💡 The pattern in both examples is the same: spot which identity turns the equation into one trig function, solve that as an ordinary equation (sometimes even a quadratic), then convert back to angles — checking every solution in the required range.
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Engineering Applications — Combining Waveforms
Adding two sine waves of the same frequency together always produces another single sine wave — trig identities are exactly how you find its amplitude and phase.
📋 Waveform Combination — Quick Reference
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a sinθ + b cosθ = R sin(θ + α)
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Amplitude: R = √(a² + b²)
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Phase angle: α = tan⁻¹(b/a)
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Key Words — What Do They Mean?
R sin(θ + α) formA single wave equivalent to two combined waves
ResultantThe combined single wave
Why this works
Expand R sin(θ + α) using the compound angle formula:
R sin(θ + α) = R sinθ cosα + R cosθ sinα
Comparing this to a sinθ + b cosθ term by term shows that a = R cosα and b = R sinα. Squaring and adding these two equations eliminates α and gives R; dividing them eliminates R and gives α:
R = √(a² + b²) α = tan⁻¹(b/a)
Combining two AC voltages of the same frequency
✏️ Worked example
Problem: Two voltages v₁ = 8 sinθ and v₂ = 6 cosθ (same frequency) are combined. Express the resultant as R sin(θ + α).
Here a = 8, b = 6.
Amplitude: R = √(8²+6²) = √(64+36) = √100
Phase angle: α = tan⁻¹(6/8) = 36.87°
Answer: v₁ + v₂ = 10 sin(θ + 36.87°)
This is exactly how engineers combine phase-shifted signals (e.g. in AC circuit analysis or vibration analysis) into one wave with a single peak size and a single timing offset, rather than working with two separate waves.
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Waveform Combiner (a sinθ + b cosθ → R sin(θ+α))
ab
Enter a and b, then click Combine.
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Practice
A mixed set of problems across all the topics — working from straightforward substitution through to multi-step engineering scenarios.
💡 Show your working for every problem. Click "Show answer" to check once you've had a genuine attempt — the value is in the attempt, not the answer.
Q1
Simplify (1 − sin²θ) using a Pythagorean identity.
Two displacement signals x₁ = 5sinθ and x₂ = 12cosθ (same frequency) act on a component simultaneously. Express the resultant displacement as R sin(θ + α), and state the maximum displacement.
a=5, b=12. R = √(5²+12²) = √(25+144) = √169 = 13. α = tan⁻¹(12/5) = 67.38°. Resultant: x = 13sin(θ+67.38°). The maximum displacement is the amplitude: 13 units, occurring when sin(θ+67.38°)=1.
Q9
Explain why cos(A+B) uses a minus sign, even though the angle itself is a sum. Use the formula to justify your answer.
cos(A+B) = cosAcosB − sinAsinB. Even though the angle A+B is a sum, the formula's sign pattern is opposite to the sin(A+B) formula (which uses a plus). This is simply how the identity works out algebraically when derived — it is not intuitive from the angle sign alone, which is exactly why it needs to be memorised or re-derived carefully rather than guessed.
Q10
A vibrating system has two components of motion, y₁ = 3cosθ and y₂ = 4sinθ. A technician claims the combined motion has maximum amplitude 7 (simply adding 3+4). Explain why this is incorrect, and find the correct maximum amplitude.
Simply adding the amplitudes is only valid if the two waves peak at exactly the same instant — but sinθ and cosθ are 90° out of phase, so they never peak simultaneously. The correct combined amplitude uses R = √(a²+b²), treating the two components as perpendicular contributions rather than directly additive ones: R = √(4²+3²) = √25 = 5, not 7.