Graphs of trigonometric functions — CSCA (China)

In this lesson you will learn

Describe the graphs of y = sin x, y = cos x and y = tan x: period, parity, axes and centres of symmetry, monotonic intervals and extreme values.
Find the period, extreme values, monotonic intervals and symmetry of y = A sin(ωx + φ) + b by treating ωx + φ as one variable.
Read A, ω, φ and b from a graph and obtain the graph from y = sin x by shifts and stretches in either order.
Solve simple trigonometric equations and inequalities on a given interval with the graph or the unit circle.

Tides, alternating current, a point on a turning wheel: everything that repeats is described by a sine wave. In Chinese textbooks the topic is called 三角函数的图像与性质 (graphs and properties of trigonometric functions), and CSCA-style papers test it in several items at once: the period, where a function increases, which line is an axis of symmetry, how a graph was shifted. Here you learn to answer such questions from the graph, almost without calculations. The values of special angles, the identities and the formula a sin x + b cos x = R sin(x + φ) are in the lesson “Trigonometric functions and identities”; general shifts and stretches of graphs are in “Functions and their properties”.

The graphs of sin x, cos x and tan x

Definition

Periodic function and the smallest positive period

A function f is periodic with period T ≠ 0 if f(x + T) = f(x) for every x of its domain. The smallest positive period (最小正周期) is what CSCA questions simply call “the period”: 2π for sin x and cos x, π for tan x.

Unwrap the unit circle: the point for the angle x has the coordinates (cos x, sin x), so while x runs from 0 to 2π the height sin x goes 0 → 1 → 0 → −1 → 0. This is the sine curve (正弦曲线). Chinese students sketch one period by the five-point method (五点法): x = 0, π/2, π, 3π/2, 2π with the values 0, 1, 0, −1, 0. The cosine curve is the same wave moved left by π/2, because cos x = sin(x + π/2); its five values are 1, 0, −1, 0, 1. The tangent curve is different: it is unbounded and breaks at x = π/2 + kπ, where cos x = 0; these lines are its vertical asymptotes.

Property	y = sin x	y = cos x	y = tan x
Domain	ℝ	ℝ	x ≠ π/2 + kπ
Range	[−1, 1]	[−1, 1]	ℝ
Period	2π	2π	π
Parity	odd	even	odd
Maximum 1 at	x = π/2 + 2kπ	x = 2kπ	—
Minimum −1 at	x = −π/2 + 2kπ	x = π + 2kπ	—
Increasing on	[−π/2 + 2kπ, π/2 + 2kπ]	[−π + 2kπ, 2kπ]	(−π/2 + kπ, π/2 + kπ)
Decreasing on	[π/2 + 2kπ, 3π/2 + 2kπ]	[2kπ, π + 2kπ]	—
Axes of symmetry	x = π/2 + kπ	x = kπ	none
Centres of symmetry	(kπ, 0)	(π/2 + kπ, 0)	(kπ/2, 0)

Everywhere k ∈ Z. Monotonic intervals are written with k in answers: “the increasing intervals are [−π/2 + 2kπ, π/2 + 2kπ], k ∈ Z”.

axis of symmetry: x = x₀ ⇔ f(x₀) = ±1 centre of symmetry: (x₀, 0) ⇔ f(x₀) = 0

where:

x₀the x-coordinate of a point of the curve y = sin x or y = cos x (or of any sine wave)
±1the largest or the smallest value: axes pass through the peaks and the troughs
0the value on the midline: centres of symmetry are the zeros of the curve

For sin x the axes are x = π/2 + kπ and the centres (kπ, 0); for cos x the axes are x = kπ and the centres (π/2 + kπ, 0). Neighbouring axes are T/2 apart; an axis and the nearest centre are T/4 apart. The tangent curve has no axes, and its centres are (kπ/2, 0).

Worked examples: reading the basic graphs

1) On [0, 2π], where does y = sin x increase, and where does y = cos x decrease?
2) Put sin 1, sin 2 and sin 3 in increasing order (the angles are in radians).
3) Is y = tan x increasing on its whole domain? Is x = π/4 an axis of symmetry of y = sin x?

Show solution

1) sin x increases on [0, π/2] (from 0 to 1) and on [3π/2, 2π] (from −1 back to 0); cos x decreases on [0, π], from 1 to −1.
2) Move every angle into [0, π/2], where the sine increases: sin 2 = sin(π − 2) ≈ sin 1.14 and sin 3 = sin(π − 3) ≈ sin 0.14. Since 0.14 < 1 < 1.14: sin 3 < sin 1 < sin 2.
3) No: tan(π/4) = 1 but tan(3π/4) = −1, although 3π/4 > π/4. The tangent increases only on each interval (−π/2 + kπ, π/2 + kπ). And sin(π/4) = √2/2 is neither 1 nor −1, so x = π/4 is not an axis; the nearest axis is x = π/2.

y = A sin(ωx + φ) + b: amplitude, period, phase

Definition

Amplitude, period and phase

For y = A sin(ωx + φ) + b with A > 0 and ω > 0: A is the amplitude (振幅), T = 2π/ω the period, f = 1/T the frequency (频率), ωx + φ the phase (相位) and φ the initial phase (初相), the phase at x = 0. The line y = b is the midline: the graph oscillates between b − A and b + A.

T = 2π/ω y_max = b + A y_min = b − AT = 2π/ω y_max = b + A y_min = b − A

where:

Aamplitude (A > 0); for A < 0 use |A| — the maximum and the minimum swap places
ωangular frequency (ω > 0); it alone sets the period
φinitial phase; it moves the graph sideways but does not change T
bvertical shift: the midline is y = b

In general T = 2π/|ω|. The same formulas hold for y = A cos(ωx + φ) + b; for y = A tan(ωx + φ) the period is π/|ω|.

Worked examples: period and extreme values

1) y = 3 sin(2x − π/3) + 1: find the period, the largest and the smallest value and where they are reached.
2) y = −2 cos(x/2): find the period and the points of maximum.
3) y = tan(3x + π/4): find the period and the domain.

Show solution

1) T = 2π/2 = π. The maximum 1 + 3 = 4 is reached when 2x − π/3 = π/2 + 2kπ, i.e. x = 5π/12 + kπ; the minimum 1 − 3 = −2 when 2x − π/3 = −π/2 + 2kπ, i.e. x = −π/12 + kπ.
2) T = 2π ÷ (1/2) = 4π. Because A = −2 < 0, the maximum 2 is reached where cos(x/2) = −1: x/2 = π + 2kπ, x = 2π + 4kπ.
3) T = π/3. The tangent needs 3x + π/4 ≠ π/2 + kπ, so 3x ≠ π/4 + kπ and x ≠ π/12 + kπ/3.

To find the monotonic intervals, axes and centres of y = A sin(ωx + φ), Chinese textbooks use substitution as a whole (整体代换): call t = ωx + φ, write the known condition for sin t from the table, and solve it for x.

increasing: −π/2 + 2kπ ≤ ωx + φ ≤ π/2 + 2kπ axis: ωx + φ = π/2 + kπ centre: ωx + φ = kπincreasing: −π/2 + 2kπ ≤ ωx + φ ≤ π/2 + 2kπ axis: ωx + φ = π/2 + kπ centre: ωx + φ = kπ

where:

ωx + φthe phase, treated as one variable t
kany integer (k ∈ Z)

For A > 0, ω > 0. Decreasing: π/2 + 2kπ ≤ ωx + φ ≤ 3π/2 + 2kπ. For a cosine use the cosine conditions: increasing −π + 2kπ ≤ ωx + φ ≤ 2kπ, axis ωx + φ = kπ, centre ωx + φ = π/2 + kπ.

Worked examples: substitution as a whole

1) Find the increasing intervals of y = 2 sin(2x + π/6).
2) Find the axes and the centres of symmetry of y = sin(2x − π/4).
3) Find the range of y = 2 sin(2x + π/6) for x ∈ [0, π/2].

Show solution

1) −π/2 + 2kπ ≤ 2x + π/6 ≤ π/2 + 2kπ ⇒ −2π/3 + 2kπ ≤ 2x ≤ π/3 + 2kπ ⇒ [−π/3 + kπ, π/6 + kπ], k ∈ Z.
2) Axes: 2x − π/4 = π/2 + kπ ⇒ x = 3π/8 + kπ/2. Centres: 2x − π/4 = kπ ⇒ x = π/8 + kπ/2, the points (π/8 + kπ/2, 0).
3) For x ∈ [0, π/2] the phase t = 2x + π/6 runs over [π/6, 7π/6]. There sin t rises from 1/2 to 1 (at t = π/2) and falls to −1/2, so sin t ∈ [−1/2, 1] and y ∈ [−1, 2]. The end values alone, y(0) = 1 and y(π/2) = −1, would miss the maximum inside.

CSCA-style item

f(x) = 2 cos(2x − π/3). Which statement is correct?
A) The smallest positive period of f is 2π.
B) The graph of f is symmetric about the line x = π/6.
C) f is decreasing on [0, π/2].
D) The point (π/6, 0) is a centre of symmetry of the graph of f.

Show solution

f(π/6) = 2 cos 0 = 2 is the maximum value, so x = π/6 is an axis: B. A: T = 2π/2 = π. C: for x ∈ [0, π/2] the phase runs over [−π/3, 2π/3] and passes 0, where the cosine has its maximum, so f first increases and then decreases. D is the classic confusion of an axis with a centre: a centre must lie on the midline, but f(π/6) = 2.

Reading a graph and moving it

1
A and b from the heights
A = (max − min)/2 and b = (max + min)/2.
2
ω from horizontal distances
From a maximum to the next minimum is T/2; from a point on the midline to the next maximum is T/4. Then ω = 2π/T.
3
φ from one known point
Best a highest point (x₁, max): ωx₁ + φ = π/2 + 2kπ. Or a midline point where the graph goes up: ωx₀ + φ = 2kπ. A midline point where the graph goes down gives ωx₀ + φ = π + 2kπ — easy to mix up, so check the direction.
4
Choose k by the condition on φ
Take the k that puts φ in the given range (for example |φ| < π/2) and check one more point of the graph.

A highest point and the next lowest point are half a period apart; the midline lies halfway between them.

Worked examples: from the graph to the formula

1) The graph of y = A sin(ωx + φ) + b (A > 0, ω > 0, |φ| < π/2) has a highest point (π/6, 3), and the next lowest point is (2π/3, −1) (see the figure). Find the function.
2) The graph of y = 2 sin(ωx + φ) (ω > 0, |φ| < π/2) goes up through the point (π/8, 0), and the next highest point is (3π/8, 2). Find ω and φ.

Show solution

1) A = (3 − (−1))/2 = 2, b = (3 + (−1))/2 = 1. T/2 = 2π/3 − π/6 = π/2, so T = π and ω = 2. At the highest point 2 · π/6 + φ = π/2 + 2kπ ⇒ φ = π/6 + 2kπ, and |φ| < π/2 gives φ = π/6: y = 2 sin(2x + π/6) + 1. Check: at x = 2π/3 the phase is 3π/2 and y = 2 · (−1) + 1 = −1 ✓.
2) T/4 = 3π/8 − π/8 = π/4, so T = π and ω = 2. The graph goes up through (π/8, 0): 2 · π/8 + φ = 2kπ ⇒ φ = −π/4. So y = 2 sin(2x − π/4).

y = sin x → y = sin(x + φ) → y = sin(ωx + φ) → y = A sin(ωx + φ) + b

where:

x + φshift left by φ if φ > 0, right by |φ| if φ < 0: “left add, right subtract” (左加右减)
ωxthe x-coordinates are multiplied by 1/ω (y unchanged): a compression if ω > 1, a stretch if 0 < ω < 1
A, bthe y-coordinates are multiplied by A, then the graph moves up by b: “up add, down subtract” (上加下减)

The other order — stretch first, shift second — also works, but then the shift is φ/ω, because sin(ωx + φ) = sin ω(x + φ/ω): a shift always acts on x itself, never on ωx.

Worked examples: shifts and stretches

1) How must the graph of y = sin 3x be moved to get the graph of y = sin(3x − π/4)?
2) Obtain y = sin(2x + π/3) from y = sin x in both orders.
3) The graph of y = cos 2x is shifted right by π/8. Write the new function.

Show solution

1) sin(3x − π/4) = sin 3(x − π/12): shift right by π/12, not by π/4.
2) Shift first: left by π/3 gives y = sin(x + π/3); halving the x-coordinates gives y = sin(2x + π/3). Stretch first: halving gives y = sin 2x; then shift left by π/6: sin 2(x + π/6) = sin(2x + π/3).
3) Replace x by x − π/8: y = cos 2(x − π/8) = cos(2x − π/4).

CSCA-style item

How can the graph of y = cos 2x be shifted to obtain the graph of y = sin 2x?
A) Left by π/4 B) Right by π/4 C) Right by π/2 D) Left by π/2

Show solution

cos 2(x − π/4) = cos(2x − π/2) = sin 2x, so the shift is right by π/4: B. C forgets to divide by ω = 2: right by π/2 is the shift from y = cos x to y = sin x. A moves the wrong way: cos 2(x + π/4) = cos(2x + π/2) = −sin 2x. D gives cos(2x + π) = −cos 2x.

Interactive

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A is the amplitude, w plays the role of ω, p of φ, and b moves the midline. With A = 2, w = 2, p ≈ 0.52 (≈ π/6) and b = 1 you get the graph of the example above. Double w: the period halves. Change only p: the graph slides by p/w, not by p.

Simple equations and inequalities on an interval

Because the functions are periodic, a trigonometric equation has infinitely many roots. CSCA questions usually ask for the roots on an interval or for their number: find the roots on one period first, then add periods.

sin x = a: x = arcsin a + 2kπ or x = π − arcsin a + 2kπ cos x = a: x = ±arccos a + 2kπ tan x = a: x = arctan a + kπ

where:

athe given number; for sine and cosine a root exists only if |a| ≤ 1
arcsin athe angle in [−π/2, π/2] whose sine is a (arccos a ∈ [0, π], arctan a ∈ (−π/2, π/2))
kany integer (k ∈ Z)

For |a| < 1, sin x = a has two roots per period, symmetric about an axis x = π/2 + 2kπ, and cos x = a has two roots symmetric about x = 2kπ; tan x = a has one root in every period π.

Worked examples: equations on an interval

1) Solve sin x = √3/2 on [0, 2π].
2) Solve 2 cos x + 1 = 0 on [0, 2π].
3) Solve tan x = −1 on (0, 2π).
4) Solve sin(2x − π/6) = 1/2 on [0, π].

Show solution

1) arcsin(√3/2) = π/3, and the second root is π − π/3: x = π/3 or 2π/3.
2) cos x = −1/2: x = ±2π/3 + 2kπ; in [0, 2π]: x = 2π/3 or 4π/3.
3) x = −π/4 + kπ; in (0, 2π): x = 3π/4 or 7π/4 — one root in every interval of length π.
4) Put t = 2x − π/6. For x ∈ [0, π], t ∈ [−π/6, 11π/6]. sin t = 1/2 gives t = π/6 or 5π/6 (13π/6 is too large), so 2x = π/3 or π: x = π/6 or π/2.

1
Substitute
If the argument is ωx + φ, put t = ωx + φ and find the interval that t runs over.
2
Mark the boundary
Solve sin t = a (or cos t = a) on that interval: these points split it into pieces.
3
Read the graph
Keep the pieces where the curve lies above (or below) the line y = a. On the unit circle: above the horizontal line y = a for sine, to the right of the vertical line x = a for cosine.
4
Return to x
Solve for x and check the ends: ≥ keeps them, > drops them.

Worked examples: inequalities

Solve on the given interval:
1) sin x ≥ 1/2 on [0, 2π];
2) cos x < √2/2 on [0, 2π];
3) tan x > 1 on (−π/2, π/2).

Show solution

1) The sine curve is at or above the level 1/2 between its two crossings π/6 and 5π/6: [π/6, 5π/6].
2) cos x = √2/2 at π/4 and 7π/4; between them the cosine curve is below this level: (π/4, 7π/4).
3) tan x increases on (−π/2, π/2) and tan(π/4) = 1: (π/4, π/2).

CSCA-style item

What is the solution set of cos x ≥ 1/2 on [0, 2π]?
A) [0, π/3] B) [π/3, 5π/3] C) [0, π/3] ∪ [5π/3, 2π] D) [0, π/6] ∪ [11π/6, 2π]

Show solution

cos x = 1/2 at π/3 and 5π/3. The cosine curve starts at 1, drops below 1/2 after π/3 and climbs above it again after 5π/3: C. A forgets the piece at the end of the period, B is the solution of cos x ≤ 1/2, and D solves cos x ≥ √3/2.

How CSCA asks about this

Graph questions are frequent, very often in the form “which statement is correct”. Learn the key terms in English and Chinese:

Term	中文	Pinyin
sine / cosine / tangent function	正弦 / 余弦 / 正切函数	zhèngxián / yúxián / zhèngqiē hánshù
smallest positive period	最小正周期	zuìxiǎo zhèng zhōuqī
amplitude	振幅	zhènfú
frequency	频率	pínlǜ
phase / initial phase	相位 / 初相	xiàngwèi / chūxiàng
axis of symmetry	对称轴	duìchènzhóu
centre of symmetry	对称中心	duìchèn zhōngxīn
increasing interval	单调递增区间	dāndiào dìzēng qūjiān
maximum / minimum value	最大值 / 最小值	zuìdàzhí / zuìxiǎozhí
zero of a function	零点	língdiǎn
translation (shift)	平移	píngyí
stretch / compression	伸缩	shēnsuō
five-point method	五点法	wǔdiǎnfǎ
trigonometric equation	三角方程	sānjiǎo fāngchéng

Typical question patterns and time-savers (about 75 s per item):

Period and extreme values: T = 2π/|ω| (π/|ω| for tan); max = b + |A|, min = b − |A|; never multiply by ω.
“Which statement is correct”: test a proposed axis or centre by computing f(x₀): b ± A means an axis, b means a centre.
Monotonic intervals: substitute t = ωx + φ into the intervals of the table; if ω < 0, rewrite the function first.
Shifts: factor out ω — y = sin(ωx + φ) is y = sin ωx shifted by φ/ω; make the function names equal before comparing.
Graph → formula: A and b from the heights, T from max-to-min (T/2) or midline-to-max (T/4), φ from a highest point.
Equations and inequalities on an interval: find the interval of t = ωx + φ first, then count where the curve meets the line y = a.

Key points

sin x and cos x: period 2π, values in [−1, 1]; sin is odd, cos is even; tan x: period π, odd, increasing on each (−π/2 + kπ, π/2 + kπ).
Axes of symmetry pass through the highest and lowest points, centres of symmetry through the points on the midline.
y = A sin(ωx + φ) + b (A, ω > 0): T = 2π/ω, values from b − A to b + A.
Monotonic intervals, axes and centres: substitute t = ωx + φ into the conditions for sin t.
y = sin(ωx + φ) is y = sin ωx shifted by φ/ω (left if φ > 0): factor out ω first.
On an interval, find the range of t = ωx + φ first; sin x = a has the second root π − arcsin a, cos x = a has −arccos a.

Check yourself

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What is the smallest positive period of y = tan x?

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