Trigonometric Functions

This lesson covers the graphs, properties and transformations of the sine, cosine and tangent functions. Understanding these graphs is essential for solving equations and modelling periodic phenomena at A-Level (9MA0).

---

The Sine Function: y = sin(x)

Key Properties

Property	Value
Period	2pi (360 degrees)
Amplitude	1
Range	-1 <= y <= 1
Symmetry	Odd function: sin(-x) = -sin(x)
Key points	sin(0) = 0, sin(pi/2) = 1, sin(pi) = 0, sin(3pi/2) = -1
Zeros	x = n x pi, where n is any integer

The sine curve starts at the origin, rises to 1 at pi/2, returns to 0 at pi, drops to -1 at 3pi/2, and returns to 0 at 2pi.

---

The Cosine Function: y = cos(x)

Key Properties

Property	Value
Period	2pi (360 degrees)
Amplitude	1
Range	-1 <= y <= 1
Symmetry	Even function: cos(-x) = cos(x)
Key points	cos(0) = 1, cos(pi/2) = 0, cos(pi) = -1, cos(3pi/2) = 0
Zeros	x = pi/2 + n x pi, where n is any integer

The cosine curve is the same shape as the sine curve but shifted left by pi/2:

cos(x) = sin(x + pi/2)

---

The Tangent Function: y = tan(x)

Key Properties

Property	Value
Period	pi (180 degrees)
Range	All real numbers (-infinity to infinity)
Asymptotes	x = pi/2 + n x pi, where n is any integer
Symmetry	Odd function: tan(-x) = -tan(x)
Key points	tan(0) = 0, tan(pi/4) = 1, tan(pi) = 0
Zeros	x = n x pi, where n is any integer

The tangent function has vertical asymptotes where cos(x) = 0, since tan(x) = sin(x)/cos(x).

---

Transformations of Trigonometric Graphs

All standard graph transformations apply to trig functions.

y = A sin(x) — Vertical Stretch

• Amplitude becomes |A|
• If A < 0, the graph is reflected in the x-axis

y = sin(Bx) — Horizontal Compression

• Period becomes 2pi/B
• The graph is compressed horizontally by factor 1/B

y = sin(x) + C — Vertical Translation

• The entire graph shifts up by C units

y = sin(x - D) — Horizontal Translation

• The graph shifts right by D units

Worked Example 1

Sketch y = 3sin(2x) for 0 <= x <= 2pi.

• Amplitude: 3
• Period: 2pi/2 = pi
• The graph completes two full cycles in the interval [0, 2pi]
• Maximum value: 3 at x = pi/4, 5pi/4
• Minimum value: -3 at x = 3pi/4, 7pi/4

Worked Example 2

Describe the transformation that maps y = cos(x) onto y = cos(x - pi/4) + 2.

• Translation by (pi/4, 2): shift right by pi/4 and up by 2
• The range becomes 1 <= y <= 3

Worked Example 3

The function f(x) = 5sin(3x) + 1 models the height of a wave.

(a) State the amplitude: 5 (b) State the period: 2pi/3 (c) State the maximum height: 5 + 1 = 6 (d) State the minimum height: -5 + 1 = -4

---

Comparing sin, cos and tan

Feature	sin(x)	cos(x)	tan(x)
Period	2pi	2pi	pi
Amplitude	1	1	N/A
Range	[-1, 1]	[-1, 1]	All reals
Odd/Even	Odd	Even	Odd
Asymptotes	None	None	x = pi/2 + npi
Relationship	cos(x) = sin(x + pi/2)	sin(x) = cos(x - pi/2)	tan(x) = sin(x)/cos(x)

---

Solving Simple Trig Equations from Graphs

Worked Example 4

Using the graph of y = sin(x), find all solutions of sin(x) = 0.5 in the range 0 <= x <= 2pi.

The principal value is x = pi/6.

Using the symmetry of the sine curve: the second solution in [0, 2pi] is x = pi - pi/6 = 5pi/6.

Solutions: x = pi/6 and x = 5pi/6

Worked Example 5

Find all solutions of cos(x) = -0.5 in the range 0 <= x <= 2pi.

Principal value: cos^(-1)(-0.5) = 2pi/3

Using symmetry of cosine (even function): second solution is 2pi - 2pi/3 = 4pi/3.

Solutions: x = 2pi/3 and x = 4pi/3

---

Exam Tips

Tip	Detail
Sketch first	Always sketch the relevant trig graph before solving equations
Period formula	For y = sin(Bx), period = 2pi/B; for tan(Bx), period = pi/B
Transformations	Apply transformations in the order: horizontal stretch, horizontal shift, vertical stretch, vertical shift
Asymptotes	Mark asymptotes clearly when sketching tan graphs
Calculator mode	Always check radian vs degree mode

---

Practice Problems

1. State the period and amplitude of y = 4cos(3x). [Answer: period = 2pi/3, amplitude = 4]
2. Describe the transformation from y = sin(x) to y = 2sin(x - pi/6). [Answer: horizontal shift right pi/6, vertical stretch factor 2]
3. Find all solutions of tan(x) = 1 for 0 <= x < 2pi. [Answer: pi/4, 5pi/4]
4. A function y = A sin(Bx) + C has maximum 7, minimum 1, and period pi. Find A, B and C. [Answer: A = 3, B = 2, C = 4]
5. Sketch y = -cos(x) for 0 <= x <= 2pi and state its range. [Answer: range [-1, 1], reflected cosine]

---

Summary

• sin(x) and cos(x) have period 2pi, amplitude 1, and range [-1, 1].
• tan(x) has period pi, no amplitude, and range all reals, with vertical asymptotes.
• cos(x) = sin(x + pi/2); the two functions are phase-shifted versions of each other.
• Transformations: y = A sin(B(x - D)) + C changes amplitude, period, phase and vertical position.
• Always sketch graphs to visualise solutions.

A-Level Deep Dive: Trigonometric Functions

Spec mapping

Edexcel 9MA0 specification section 5 — Trigonometry covers the sine, cosine and tangent functions; their graphs, symmetries and periodicity. Know and use exact values of $\sin$, $\cos$ and $\tan$ for $0$, $\pi/6$, $\pi/4$, $\pi/3$, $\pi/2$ and multiples thereof. Solve trigonometric equations within a given interval, including those involving $\sin(kx)$, $\cos(kx)$ and $\tan(kx)$ and combinations thereof (refer to the official specification document for exact wording). Trigonometric graph work appears principally on Paper 2 — Pure Mathematics, but synoptic crossover is heavy. Specifically: section 2 (Algebra and functions — transformations of graphs) is the geometric language of every trig graph question; section 7 (Differentiation) requires $\frac{d}{dx}\sin x = \cos x$, which only makes sense in radians; section 8 (Integration) depends on the same convention; section 5 (Trigonometry — harmonic form $R\sin(x + \alpha)$) at Year 2 is a direct generalisation of "amplitude–phase" sketching; and section 9 (Modelling) uses sinusoids for tides, sound, mechanical oscillation. The Edexcel formula booklet does list compound-angle and double-angle identities but does not list exact values at standard angles or the period/amplitude rules — these must be memorised.

---

Worked example with full mark scheme

Question (8 marks):

(a) Sketch the graph of $y = 3\sin!\left(2x - \dfrac{\pi}{3}\right) + 1$ for $0 \leq x \leq 2\pi$, stating clearly the amplitude, period, vertical shift, and the coordinates of the first maximum in the interval. (5)

(b) Hence, or otherwise, find all solutions of $3\sin!\left(2x - \dfrac{\pi}{3}\right) + 1 = 1$ in $0 \leq x \leq 2\pi$, giving exact values. (3)

Solution with mark scheme:

(a) Step 1 — rewrite in standard transformation form.

$y = 3\sin\!\left(2\!\left(x - \dfrac{\pi}{6}\right)\right) + 1$y=3sin(2(x−6π​))+1

The factor of 2 has been pulled outside the bracket, exposing the genuine horizontal translation $\pi/6$. This is the key A* move — the raw expression $\sin(2x - \pi/3)$ tempts candidates to read the phase shift as $\pi/3$, but factorising shows it is $\pi/6$.

M1 — correct factorisation of the inner bracket. Without this step, the phase shift is reported as $\pi/3$ and any sketch that follows is misaligned.

Step 2 — read off the four transformation parameters.

• Amplitude $A = 3$.
• Angular frequency $B = 2$, so period $T = \dfrac{2\pi}{B} = \pi$.
• Phase shift $D = \dfrac{\pi}{6}$ to the right.
• Vertical shift $C = 1$ upward, so the centre line is $y = 1$ and the range is $[1 - 3,, 1 + 3] = [-2,, 4]$.

A1 — all four parameters identified correctly.

Step 3 — locate the first maximum.

The parent function $\sin\theta$ peaks at $\theta = \pi/2$. Setting the inner argument to $\pi/2$:

$2x - \dfrac{\pi}{3} = \dfrac{\pi}{2} \implies 2x = \dfrac{5\pi}{6} \implies x = \dfrac{5\pi}{12}$2x−3π​=2π​⟹2x=65π​⟹x=125π​

At this $x$, $y = 3(1) + 1 = 4$. So the first maximum has coordinates $\left(\dfrac{5\pi}{12},, 4\right)$.

M1 — setting inner argument equal to $\pi/2$ to locate the maximum.

A1 — correct exact coordinates.

Step 4 — produce the sketch.

The curve oscillates between $y = -2$ and $y = 4$, with two complete cycles in $[0, 2\pi]$ (since the period is $\pi$). The starting value at $x = 0$ is $y = 3\sin(-\pi/3) + 1 = 3 \cdot (-\sqrt{3}/2) + 1 = 1 - \dfrac{3\sqrt{3}}{2} \approx -1.60$. Axes must be labelled with $x$ in radians and the centre line $y = 1$ marked. Maxima at $x = 5\pi/12$ and $x = 5\pi/12 + \pi = 17\pi/12$; minima at $x = 5\pi/12 + \pi/2 = 11\pi/12$ and $x = 11\pi/12 + \pi = 23\pi/12$.

A1 — sketch with correctly labelled axes, period, amplitude, centre line, and at least one labelled maximum.

(b) Step 1 — reduce.

$3\sin(2x - \pi/3) + 1 = 1$ gives $\sin(2x - \pi/3) = 0$.

M1 — correct isolation of the sine term.

Step 2 — solve over the extended interval for the inner argument.

If $0 \leq x \leq 2\pi$, then $-\pi/3 \leq 2x - \pi/3 \leq 11\pi/3$. Within this interval, $\sin\theta = 0$ at $\theta = 0,, \pi,, 2\pi,, 3\pi$.

M1 — extending the search interval to match the inner argument range. This is the most-missed mark on transformed-trig solving questions.

Step 3 — back-substitute.

$2x - \pi/3 = 0 \Rightarrow x = \pi/6$; $2x - \pi/3 = \pi \Rightarrow x = 2\pi/3$; $2x - \pi/3 = 2\pi \Rightarrow x = 7\pi/6$; $2x - \pi/3 = 3\pi \Rightarrow x = 5\pi/3$.

A1 — all four exact solutions: $x = \dfrac{\pi}{6},, \dfrac{2\pi}{3},, \dfrac{7\pi}{6},, \dfrac{5\pi}{3}$.

Total: 8 marks (M4 A4, split as shown).

---

Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Question (6 marks): The function $f$ is defined by $f(x) = 2\cos!\left(\dfrac{x}{2}\right) - 1$ for $0 \leq x \leq 4\pi$.

(a) Sketch the graph of $y = f(x)$, marking the coordinates of the points where the curve crosses the axes. (4)

(b) State the range of $f$. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — identifying amplitude 2, period $2\pi/(1/2) = 4\pi$, vertical shift $-1$.
• M1 (AO1.1b) — finding $y$-intercept: $f(0) = 2\cos 0 - 1 = 1$.
• M1 (AO1.1b) — solving $2\cos(x/2) = 1$: $\cos(x/2) = 1/2$ gives $x/2 = \pi/3$ or $x/2 = 5\pi/3$, so $x = 2\pi/3$ or $x = 10\pi/3$.
• A1 (AO2.5) — sketch with labelled centre line $y = -1$, intercepts at $(0, 1)$, $(2\pi/3, 0)$, $(10\pi/3, 0)$, and exactly one full cycle visible.

(b)

• M1 (AO1.1b) — recognising max at $y = 2(1) - 1 = 1$ and min at $y = 2(-1) - 1 = -3$.
• A1 (AO2.5) — range $-3 \leq f(x) \leq 1$ stated in interval notation or inequality form.

Total: 6 marks split AO1 = 4, AO2 = 2. Paper 2 trig-graph questions reward clean axis labelling and explicit centre-line marking — the AO2 marks are presentation marks, awarded for communication rather than calculation.

---

Synoptic links

Connects to:

1. Section 2 — Algebra and functions (transformations of graphs): every trig-graph question is a transformation question wearing trigonometric clothing. The mapping $f(x) \to A f(B(x - D)) + C$ is universal; sine/cosine/tangent are simply the parent functions on which it acts. A candidate fluent in the four-parameter transformation framework can read off period, amplitude, phase shift and centre line on inspection.
2. Section 7 — Differentiation of trigonometric functions: $\dfrac{d}{dx}\sin x = \cos x$ and $\dfrac{d}{dx}\cos x = -\sin x$ hold only when $x$ is in radians. The proof rests on the small-angle limit $\lim_{h \to 0} \dfrac{\sin h}{h} = 1$, which itself depends on radian measure. Stationary-point questions on $y = \sin(2x - \pi/3)$ require chain-rule application: $\dfrac{dy}{dx} = 2\cos(2x - \pi/3)$.
3. Section 8 — Integration of trigonometric functions: $\int \sin(ax + b),dx = -\dfrac{1}{a}\cos(ax + b) + C$. Areas under one full cycle of a sinusoid are zero — a fact used in the modelling of net displacement of an oscillating body and a frequent Paper 2 setup.
4. Section 5 — Harmonic form $R\sin(x + \alpha)$ (Year 2): writing $a\sin x + b\cos x$ as $R\sin(x + \alpha)$ where $R = \sqrt{a^2 + b^2}$ and $\tan\alpha = b/a$ is the algebraic recipe for combining two phase-shifted sinusoids into a single transformed sine. Sketching transformed sines fluently in Year 1 is the prerequisite for handling harmonic form in Year 2.
5. Section 9 / Mechanics — modelling oscillations: simple harmonic motion produces displacement $x(t) = A\cos(\omega t + \phi)$. Reading amplitude, angular frequency $\omega$, period $2\pi/\omega$ and phase $\phi$ from a graph is the bridge between pure trigonometry and 9MA0-03 mechanics work on SHM and waves.

---

Mark-scheme literacy

Trig-graph questions on 9MA0 Paper 2 split AO marks as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	55–65%	Identifying amplitude/period/phase/shift; computing exact intercepts; back-substituting solutions through the inner argument
AO2 (reasoning / interpretation)	25–35%	Labelled axes with radians; correct centre-line marking; exact-form answers; selecting the correct interval for the inner argument
AO3 (problem-solving)	5–15%	Modelling oscillations; combining transformations in non-obvious order; harmonic-form questions with a sketching component

Examiner-rewarded phrasing: "the curve has labelled axes with $x$ in radians"; "centre line $y = c$ shown with a dashed horizontal"; "amplitude $|A|$ measured from the centre line, not from the $x$-axis"; "period $2\pi/|B|$ for sine and cosine, $\pi/|B|$ for tangent"; "exact coordinates given in the form $(x, y)$ where $x$ is a rational multiple of $\pi$". Phrases that lose marks: writing decimals where exact values are demanded ("3.14" instead of $\pi$); omitting the centre line on a vertically-shifted graph; labelling the axis as "degrees" when the question is in radians; failing to extend the search interval when solving $\sin(kx + \alpha) = c$ — solving over $[0, 2\pi]$ instead of $[\alpha, 2k\pi + \alpha]$ silently drops solutions.

A specific Edexcel pattern: the "first maximum / first minimum / first axis crossing" command. Candidates often list all extrema in the interval; the question asks only for the first. The A1 is reserved for the candidate who reads the question precisely.

---

Grade-band model answers

3-mark question

Question: State the period and amplitude of the curve $y = 4\sin!\left(\dfrac{x}{3}\right) - 2$.

Grade C response (~210 words):