Further Trigonometry

This lesson extends trigonometry with addition formulae, double angle formulae, and the R cos/sin form — powerful tools for simplifying expressions and solving equations.

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Addition Formulae

Formula
sin(A + B) = sin A cos B + cos A sin B
sin(A − B) = sin A cos B − cos A sin B
cos(A + B) = cos A cos B − sin A sin B
cos(A − B) = cos A cos B + sin A sin B
tan(A + B) = (tan A + tan B)/(1 − tan A tan B)
tan(A − B) = (tan A − tan B)/(1 + tan A tan B)

Worked Example 1

Find the exact value of sin 75°.

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

Worked Example 2

Given sin A = 3/5 (A acute) and cos B = 5/13 (B acute), find cos(A − B).

cos A = 4/5 (from Pythagoras), sin B = 12/13.

cos(A − B) = cos A cos B + sin A sin B = (4/5)(5/13) + (3/5)(12/13) = 20/65 + 36/65 = 56/65

Answer: 56/65

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Double Angle Formulae

Formula
sin 2A = 2 sin A cos A
cos 2A = cos²A − sin²A = 2cos²A − 1 = 1 − 2sin²A
tan 2A = 2 tan A/(1 − tan²A)

Worked Example 3

Given cos θ = 3/5 (θ acute), find sin 2θ and cos 2θ.

sin θ = 4/5 (Pythagoras)

sin 2θ = 2(4/5)(3/5) = 24/25

cos 2θ = 2(3/5)² − 1 = 18/25 − 1 = −7/25

Answers: sin 2θ = 24/25, cos 2θ = −7/25

Half-Angle Rearrangements

From cos 2A = 2cos²A − 1:

cos²A = (1 + cos 2A)/2

From cos 2A = 1 − 2sin²A:

sin²A = (1 − cos 2A)/2

These are useful for integration (e.g. ∫cos²x dx).

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The R cos/sin Form

Expressions of the form a cos θ + b sin θ can be written as:

R cos(θ − α) or R sin(θ + α)

where R = √(a² + b²) and α is found from tan α = b/a (or a/b, depending on the form).

Why This Is Useful

• Solve equations like 3 cos θ + 4 sin θ = 2 more easily.
• Find maximum and minimum values of trigonometric expressions.
• Simplify trigonometric integrals.

Worked Example 4

Express 3 cos θ + 4 sin θ in the form R cos(θ − α).

R cos(θ − α) = R cos θ cos α + R sin θ sin α

Comparing: R cos α = 3 and R sin α = 4

R = √(9 + 16) = 5

tan α = 4/3 → α = arctan(4/3) ≈ 53.13°

So: 3 cos θ + 4 sin θ = 5 cos(θ − 53.13°)

Finding Maximum and Minimum Values

Since −1 ≤ cos(θ − α) ≤ 1:

Maximum value = R (when θ − α = 0, i.e. θ = α) Minimum value = −R (when θ − α = π, i.e. θ = α + π)

Worked Example 5

Find the maximum value of 3 cos θ + 4 sin θ and the value of θ at which it occurs (0 ≤ θ ≤ 2π).

From above: 3 cos θ + 4 sin θ = 5 cos(θ − 53.13°)

Maximum = 5, occurring when θ = 53.13° ≈ 0.927 radians.

Worked Example 6

Solve 3 cos θ + 4 sin θ = 2 for 0° ≤ θ ≤ 360°.

5 cos(θ − 53.13°) = 2 → cos(θ − 53.13°) = 0.4

θ − 53.13° = ±66.42°

θ = 53.13° + 66.42° = 119.55° or θ = 53.13° − 66.42° = −13.29° → 346.71°

Solutions: θ ≈ 119.6° or θ ≈ 346.7°

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Proving Trigonometric Identities

Worked Example 7

Prove that (sin 2θ)/(1 + cos 2θ) ≡ tan θ.

LHS = 2 sin θ cos θ / (1 + 2cos²θ − 1) = 2 sin θ cos θ / (2cos²θ) = sin θ / cos θ = tan θ = RHS ✓

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Exam Tips

• Memorise all addition and double angle formulae — they are not given in the formula book for all exam boards.
• When using R cos(θ − α), always state R and α clearly.
• For solving equations using R cos/sin form, find all solutions within the given range.
• The double angle formulae for cos 2A have three equivalent forms — choose the one that simplifies your working.
• When proving identities, work on one side only and transform it to the other.

A-Level Deep Dive: Further Trigonometry

Spec mapping

AQA 7357 specification, Paper 2 — Pure Mathematics, Section E (Trigonometry), Year 2 sub-strands: "Understand and use the addition formulae; understand and use double-angle formulae; understand and use expressions of the form $a\cos\theta + b\sin\theta$acosθ+bsinθ in the equivalent forms $R\cos(\theta \pm \alpha)$Rcos(θ±α) or $R\sin(\theta \pm \alpha)$Rsin(θ±α); understand and use the reciprocal trigonometric functions $\sec\theta$secθ, $\csc\theta$cscθ and $\cot\theta$cotθ, and the identities $1 + \tan^2\theta \equiv \sec^2\theta$1+tan2θ≡sec2θ and $1 + \cot^2\theta \equiv \csc^2\theta$1+cot2θ≡csc2θ." This material sits explicitly in the Year 2 portion of Pure and is examined across both Paper 1 (Pure) and Paper 2 (Pure), with synoptic appearances in Paper 3 (Mechanics) wherever simple harmonic motion or oscillating systems appear.

A point that catches many students: the compound-angle formulae ($\sin(A \pm B)$sin(A±B), $\cos(A \pm B)$cos(A±B), $\tan(A \pm B)$tan(A±B)), the double-angle formulae ($\sin 2\theta$sin2θ, $\cos 2\theta$cos2θ, $\tan 2\theta$tan2θ), and the harmonic-form identity (the relationship between $R$R, $\alpha$α, $a$a, $b$b) are included in the AQA formula booklet for 7357. The reciprocal-function identities ($1 + \tan^2\theta \equiv \sec^2\theta$1+tan2θ≡sec2θ and $1 + \cot^2\theta \equiv \csc^2\theta$1+cot2θ≡csc2θ) are likewise listed. The Pythagorean identity $\sin^2\theta + \cos^2\theta \equiv 1$sin2θ+cos2θ≡1 is not listed and must be memorised — an oversight that has cost candidates marks when they assume it appears in the booklet alongside the others.

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Worked example with full mark scheme

Question (8 marks): Solve the equation $3\sin\theta - 4\cos\theta = 2.5$3sinθ−4cosθ=2.5 for $0 \leq \theta \leq 2\pi$0≤θ≤2π, giving your answers to 3 significant figures.

Solution with mark scheme:

Step 1 — express the LHS in harmonic form $R\sin(\theta - \alpha)$Rsin(θ−α).

We seek $R$R and $\alpha$α (with $R > 0$R>0, $0 < \alpha < \pi/2$0<α<π/2) such that

$3\sin\theta - 4\cos\theta \equiv R\sin(\theta - \alpha) = R\sin\theta\cos\alpha - R\cos\theta\sin\alpha$3sinθ−4cosθ≡Rsin(θ−α)=Rsinθcosα−Rcosθsinα

Comparing coefficients: $R\cos\alpha = 3$Rcosα=3 and $R\sin\alpha = 4$Rsinα=4.

M1 — choosing the correct harmonic form to match the LHS pattern (positive $\sin\theta$sinθ coefficient, negative $\cos\theta$cosθ coefficient $\Rightarrow R\sin(\theta - \alpha)$⇒Rsin(θ−α)). Picking the wrong form (e.g. $R\cos(\theta - \alpha)$Rcos(θ−α)) typically loses both M1s downstream.

Step 2 — find $R$R.

$R^2 = R^2\cos^2\alpha + R^2\sin^2\alpha = 3^2 + 4^2 = 25 \implies R = 5$R2=R2cos2α+R2sin2α=32+42=25⟹R=5

A1 — $R = 5$R=5 exactly. Note: $R$R is always taken positive, so we never need to reject $R = -5$R=−5.

Step 3 — find $\alpha$α.

$\tan\alpha = \dfrac{R\sin\alpha}{R\cos\alpha} = \dfrac{4}{3} \implies \alpha = \arctan(4/3) \approx 0.9273 \text{ rad}$tanα=RcosαRsinα​=34​⟹α=arctan(4/3)≈0.9273 rad

M1 A1 — correct ratio for $\tan\alpha$tanα (M1) and acute angle in radians (A1). A frequent slip: students compute $\arctan(3/4) \approx 0.6435$arctan(3/4)≈0.6435, having reversed the ratio. The convention $\tan\alpha = b/a$tanα=b/a matches the coefficient assignment chosen in Step 1.

Step 4 — rewrite and solve.

$5\sin(\theta - \alpha) = 2.5 \implies \sin(\theta - \alpha) = 0.5$5sin(θ−α)=2.5⟹sin(θ−α)=0.5

M1 — correct division and reduction to a standard $\sin$sin equation.

The principal solutions of $\sin\phi = 0.5$sinϕ=0.5 are $\phi = \pi/6$ϕ=π/6 and $\phi = \pi - \pi/6 = 5\pi/6$ϕ=π−π/6=5π/6. Adding multiples of $2\pi$2π generates further solutions.

So $\theta - \alpha = \pi/6 + 2k\pi$θ−α=π/6+2kπ or $\theta - \alpha = 5\pi/6 + 2k\pi$θ−α=5π/6+2kπ.

M1 — using both branches of $\arcsin$arcsin (the $\pi - \phi$π−ϕ branch is the one most often forgotten).

Step 5 — recover $\theta$θ in the required interval.

$\theta = \alpha + \pi/6 \approx 0.9273 + 0.5236 \approx 1.451$θ=α+π/6≈0.9273+0.5236≈1.451. $\theta = \alpha + 5\pi/6 \approx 0.9273 + 2.6180 \approx 3.545$θ=α+5π/6≈0.9273+2.6180≈3.545.

Adding $2\pi$2π to either pushes $\theta$θ outside $[0, 2\pi]$[0,2π], but we should also check $\theta = \alpha + \pi/6 - 2\pi < 0$θ=α+π/6−2π<0 (reject) — the two values found are the complete solution set in range.

A1 A1 — both correct values to 3 s.f.: $\theta \approx 1.45$θ≈1.45 and $\theta \approx 3.55$θ≈3.55.

Total: 8 marks (M4 A4). A* candidates write a one-line interval check at the end ("both values lie in $[0, 2\pi]$[0,2π]; adding $2\pi$2π takes us outside the range") to secure the final A1 cleanly.

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Specimen question modelled on the AQA 7357 Paper 2 format

Question (6 marks):

(a) Show that $\cos 2\theta + \sin\theta = 0$cos2θ+sinθ=0 can be rewritten as $2\sin^2\theta - \sin\theta - 1 = 0$2sin2θ−sinθ−1=0. (2)

(b) Hence solve $\cos 2\theta + \sin\theta = 0$cos2θ+sinθ=0 for $0 \leq \theta \leq 2\pi$0≤θ≤2π, giving exact answers. (4)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — using the double-angle identity $\cos 2\theta = 1 - 2\sin^2\theta$cos2θ=1−2sin2θ to eliminate $\cos 2\theta$cos2θ.
• A1 (AO1.1b) — substituting and simplifying: $1 - 2\sin^2\theta + \sin\theta = 0$1−2sin2θ+sinθ=0, then negating to obtain the printed form $2\sin^2\theta - \sin\theta - 1 = 0$2sin2θ−sinθ−1=0.

(b)

• M1 (AO1.1b) — factorising the quadratic in $\sin\theta$sinθ: $(2\sin\theta + 1)(\sin\theta - 1) = 0$(2sinθ+1)(sinθ−1)=0.
• A1 (AO1.1b) — extracting roots $\sin\theta = -1/2$sinθ=−1/2 or $\sin\theta = 1$sinθ=1.
• M1 (AO2.4) — solving each within the interval: $\sin\theta = -1/2 \Rightarrow \theta = 7\pi/6$sinθ=−1/2⇒θ=7π/6 or $\theta = 11\pi/6$θ=11π/6; $\sin\theta = 1 \Rightarrow \theta = \pi/2$sinθ=1⇒θ=π/2.
• A1 (AO2.5) — full solution set $\theta \in \{\pi/2, 7\pi/6, 11\pi/6\}$θ∈{π/2,7π/6,11π/6} presented as exact multiples of $\pi$π.

Total: 6 marks split AO1 = 4, AO2 = 2. This is the canonical AQA structure: a "show that" reduces the problem to a routine quadratic, then the second part rewards interval discipline.

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Synoptic links

Connects to:

1. Pure 1 trig basics (Year 1): the unit-circle definitions of $\sin$sin, $\cos$cos, $\tan$tan, the small-angle approximations, and $\sin^2\theta + \cos^2\theta \equiv 1$sin2θ+cos2θ≡1 are all assumed knowledge. Year 2 trigonometry layers compound and double-angle identities on top of Year 1 fluency — without instant recall of $\sin(\pi/6)$sin(π/6), $\cos(\pi/3)$cos(π/3) and the quadrant-sign diagram, harmonic-form questions become impossible to land cleanly.

2. Differentiation of trigonometric functions: $\frac{d}{dx}\sin x = \cos x$dxd​sinx=cosx and $\frac{d}{dx}\tan x = \sec^2 x$dxd​tanx=sec2x are listed in the formula booklet. Many max/min problems require expanding $\sin(2x + \pi/3)$sin(2x+π/3) using compound-angle formulae before differentiating. Reciprocal-trig identities reappear when differentiating $\sec x = (\cos x)^{-1}$secx=(cosx)−1 via the chain rule, giving $\sec x \tan x$secxtanx.

3. Integration of trigonometric functions: integrals like $\int \sin^2 x \, dx$∫sin2xdx are intractable until you apply the double-angle identity $\sin^2 x = \tfrac{1}{2}(1 - \cos 2x)$sin2x=21​(1−cos2x). The integrals $\int \sec^2 x \, dx = \tan x + C$∫sec2xdx=tanx+C and $\int \sec x \tan x \, dx = \sec x + C$∫secxtanxdx=secx+C rely on knowing the reciprocal-trig derivative pairs in reverse.

4. Mechanics — simple harmonic motion (Paper 3): the general SHM solution $x(t) = A\cos(\omega t) + B\sin(\omega t)$x(t)=Acos(ωt)+Bsin(ωt) converts via harmonic form into $x(t) = R\cos(\omega t - \alpha)$x(t)=Rcos(ωt−α), which is the form physicists use to read off amplitude $R$R and phase shift $\alpha$α directly. Every harmonic-form practice problem in Pure trains this Mechanics skill.

5. Complex numbers (Further Maths, but the bridge sits here): the addition formulae $\cos(A + B) = \cos A \cos B - \sin A \sin B$cos(A+B)=cosAcosB−sinAsinB and $\sin(A + B) = \sin A \cos B + \cos A \sin B$sin(A+B)=sinAcosB+cosAsinB are exactly the real and imaginary parts of $e^{iA} \cdot e^{iB} = e^{i(A + B)}$eiA⋅eiB=ei(A+B). A* students who plan on Further Maths see the addition formulae as Euler's identity in disguise.

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Mark-scheme literacy

Further-trig questions on AQA 7357 typically split AO marks roughly as:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–65%	Recalling and applying compound, double-angle and reciprocal identities; converting to harmonic form; standard solve-in-interval routines
AO2 (reasoning / interpretation)	25–35%	Choosing which identity unlocks an equation; justifying $R > 0$R>0 and the acute-$\alpha$α convention; "show that" steps that demand explicit working
AO3 (problem-solving)	10–20%	Modelling oscillating systems; max/min problems where harmonic form reveals amplitude; multi-step proofs combining several identities

Examiner-rewarded phrasing: "applying the identity $\cos 2\theta \equiv 1 - 2\sin^2\theta$cos2θ≡1−2sin2θ to eliminate $\cos 2\theta$cos2θ"; "writing $3\sin\theta - 4\cos\theta$3sinθ−4cosθ in the form $R\sin(\theta - \alpha)$Rsin(θ−α) with $R > 0$R>0 and $0 < \alpha < \pi/2$0<α<π/2"; "the second solution comes from the identity $\sin(\pi - \phi) \equiv \sin\phi$sin(π−ϕ)≡sinϕ". Phrases that lose marks: "by inspection" (no method shown); failing to state the interval check at the end; giving decimal answers when the question asks for exact answers.

A specific AQA pattern: in "Hence" questions, marks are reserved for using the previous part's result. Re-deriving from scratch in part (b) typically loses the M1 awarded for using the printed identity — the command word is binding.

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Grade-band model answers

3-mark question

Question: Express $\sin 2x$sin2x in terms of $\tan x$tanx.

Grade C response (~140 words):