Trigonometric Graphs (Grade 11) — Worksheet & Memo

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Grade 11 · Trigonometry

Trigonometric Graphs (Grade 11)

MARKING GUIDELINE

Marks

Duration

1 hour 20 minutes

Questions

Name:

Class:

Date:

Mark

/ 52

Instructions and Information

Answer ALL the questions in this question paper.
Answer QUESTION 1 by circling the letter (A–D) in the answer grid at the end of that section.
Show ALL calculations clearly.
Show all units where applicable.
Number the answers correctly according to the numbering system used in this question paper.
A non-programmable calculator may be used, unless stated otherwise.
Write neatly and legibly.

Question 1

[10 MARKS]

Four options are given as possible answers to the following questions. Choose the correct answer and circle the letter (A–D) in the grid at the end of this section. If you want to change your choice, put a cross through the wrong letter and circle your new choice.

1.1
The PERIOD of y = sin 2x is …
(1)
A)180°✓
B)360°
C)720°
D)90°
Answer: A — The period of sin bx is 360° ÷ b.
- B — that is the period of y = sin x
- C — MULTIPLIED by 2 instead of dividing
- D — divided by 4 instead of by 2
1.2
The PERIOD of y = tan 3x is …
(1)
A)60°✓
B)120°
C)180°
D)540°
Answer: A — Tangent's basic period is 180°, so it becomes 180° ÷ 3 = 60°.
- B — used 360° ÷ 3, but tangent's period is 180°, not 360°
- C — that is the period of y = tan x
- D — MULTIPLIED by 3 instead of dividing
1.3
The AMPLITUDE of y = −4 cos x is …
(1)
A)4✓
B)−4
C)8
D)1
Answer: A — Amplitude is the SIZE of a, and is always positive.
- B — amplitude is never negative — the minus only flips the graph over
- C — that is the distance from minimum to maximum, twice the amplitude
- D — that is the amplitude of y = cos x
1.4
The RANGE of y = 3 sin x − 1 is …
(1)
A)−4 ≤ y ≤ 2✓
B)−3 ≤ y ≤ 3
C)−2 ≤ y ≤ 4
D)−1 ≤ y ≤ 3
Answer: A — 3 sin x runs from −3 to 3, and the −1 lowers both ends.
- B — forgot the vertical shift of −1
- C — ADDED the 1 instead of subtracting it
- D — used the shift as though it were the amplitude
1.5
Compared with y = sin x, the graph of y = sin(x − 30°) is …
(1)
A)shifted 30° to the right✓
B)shifted 30° to the left
C)shifted 30 units down
D)reflected in the x-axis
Answer: A — A minus INSIDE the bracket shifts the graph in the POSITIVE x-direction.
- B — a minus inside the bracket shifts to the RIGHT, not to the left
- C — the 30° sits inside the function, so the shift is horizontal
- D — a reflection needs a minus in FRONT of the function
1.6
The graph of y = cos(x + 60°) has a maximum at x = …
(1)
A)−60°✓
B)60°
C)0°
D)180°
Answer: A — Cosine peaks when its argument is 0°, so x + 60° = 0°.
- B — the shift is to the LEFT, so the peak moves to −60°
- C — 0° is where y = cos x peaks, before the shift
- D — 180° is where cosine is at its MINIMUM
1.7
y = 2 sin x and y = sin 2x differ because …
(1)
A)the outside 2 changes the AMPLITUDE; the inside 2 changes the PERIOD✓
B)both change the amplitude
C)both change the period
D)they are in fact the same graph
Answer: A — A factor in front stretches the graph vertically; a factor on x squeezes it horizontally.
- B — a factor ON x affects the period, not the amplitude
- C — a factor in FRONT affects the amplitude, not the period
- D — their maxima and their periods both differ
1.8
On [0° ; 360°], how many complete cycles does y = cos 3x complete?
(1)
A)3✓
B)1
C)6
D)13
Answer: A — The period is 120°, and 360° ÷ 120° = 3.
- B — that is the number of cycles of y = cos x
- C — doubled the correct count
- D — inverted the relationship
1.9
The MAXIMUM value of y = −2 cos x + 3 is …
(1)
A)5✓
B)1
C)3
D)−2
Answer: A — With a negative a, the maximum happens where cos x = −1, giving 2 + 3 = 5.
- B — that is the MINIMUM, where cos x = +1
- C — that is the vertical shift on its own
- D — that is the value of a
1.10
The graph of y = tan x has asymptotes wherever …
(1)
A)cos x = 0✓
B)sin x = 0
C)tan x = 0
D)x = 0
Answer: A — tan x = sin x ÷ cos x, so it is undefined where the denominator is zero.
- B — sin x = 0 gives tan x = 0, which is an x-INTERCEPT
- C — tan x = 0 is where the graph crosses the axis
- D — tan 0° = 0, so x = 0 is an intercept, not an asymptote

Answer grid — marking guideline

1.1	A	B	C	D
1.2	A	B	C	D
1.3	A	B	C	D
1.4	A	B	C	D
1.5	A	B	C	D
1.6	A	B	C	D
1.7	A	B	C	D
1.8	A	B	C	D
1.9	A	B	C	D
1.10	A	B	C	D

Question 2

[17 MARKS]

The graphs of f(x) = 2 cos x and g(x) = sin(x + 30°) are drawn below on the SAME set of axes for x ∈ [−180° ; 180°]. P and Q are the points of intersection of f and g.

2.1
Write down the range of f and the period of g, and state how many complete cycles f completes in the interval [−180° ; 180°].
(3)
Range of f: y ∈ [−2 ; 2] (1)
Period of g: 360° (1)
The interval is 360° wide and f also has a period of 360°, so f completes ONE complete cycle (1)
2.2
Write down the x-coordinates of the points at which g cuts the x-axis in the given interval, and show how you obtained them.
(3)
g(x) = 0 when sin(x + 30°) = 0 (1)
x + 30° = 0° or x + 30° = 180° (1)
x = −30° or x = 150° (1)
2.3
Write down the coordinates of P and Q, and hence use the graphs to determine the values of x for which f(x) > g(x).
(4)
P(−120° ; −1) (1)
Q(60° ; 1) (1)
f lies ABOVE g strictly between P and Q (1)
∴ −120° < x < 60° (1)
2.4
Determine, from the graphs, the values of x ∈ [−180° ; 180°] for which f(x) · g(x) ≤ 0.
(4)
A product is negative or zero where the two graphs have OPPOSITE signs, or where either one is zero (1)
f is zero at −90° and 90°; g is zero at −30° and 150° (1)
Between −90° and −30°: f is above the axis, g is below it (1)
∴ x ∈ [−90° ; −30°] or x ∈ [90° ; 150°] (1)
2.5
The graph of f is shifted 3 units DOWN to give a new graph m. Write down the range of m, and state, with a reason, whether m still cuts the x-axis.
(3)
m(x) = 2 cos x − 3, so the range is y ∈ [−5 ; −1] (2)
m does NOT cut the x-axis, because 0 does not lie in that range — the maximum value of m is −1 (1)

Question 3

[15 MARKS]

The graph of k(x) = a sin(x + p) is sketched below for x ∈ [−180° ; 180°]. The maximum turning point of k in this interval is (−30° ; 3), and 0° < p < 180°.

3.1
Show that a = 3 and p = 120°.
(4)
The maximum value of a sin(θ) is a, and the sketch gives a maximum of 3 (1)
∴ a = 3 (1)
A sine graph reaches its maximum when the angle equals 90°: −30° + p = 90° (1)
∴ p = 120° (1)
3.2
Write down the range of k, and determine the coordinates of the minimum turning point of k in the given interval.
(4)
Range: y ∈ [−3 ; 3] (1)
The minimum occurs when x + 120° = 270° (1)
x = 150° (1)
∴ the minimum turning point is (150° ; −3) (1)
3.3
Determine the x-intercepts of k in the interval [−180° ; 180°].
(4)
3 sin(x + 120°) = 0, so sin(x + 120°) = 0 (1)
x + 120° = 0° ; 180° ; 360° ; … (1)
x = −120° ; 60° ; 240° (1)
Only x = −120° and x = 60° lie in the interval (1)
3.4
Describe fully the transformation that maps y = 3 sin x onto k, and write down the period of y = 3 sin 2(x + 120°).
(3)
k is y = 3 sin x shifted 120° to the LEFT (2)
Doubling the value of the constant inside HALVES the period: 360° ÷ 2 = 180° (1)

Question 4

[10 MARKS]

The graph of h(x) = tan ½x is sketched below for x ∈ [−360° ; 360°]. The dashed lines are the asymptotes of h.

4.1
Write down the period of h, and the equations of the two asymptotes shown in the sketch.
(4)
Period of tan kx is 180° ÷ k, and here k = ½ (1)
∴ period = 360° (1)
tan is undefined when ½x = 90° + 180°n (1)
∴ x = −180° and x = 180° (1)
4.2
Determine the values of x ∈ [−360° ; 360°] for which h(x) = 1.
(3)
tan ½x = 1, so ½x = 45° + 180°n (1)
x = 90° + 360°n (1)
∴ x = −270° or x = 90° (1)
4.3
Use the sketch to write down the values of x, where −180° < x < 180°, for which h(x) < 0.
(3)
Between the two asymptotes the curve lies BELOW the x-axis to the left of the origin (1)
and h(0°) = 0, which is not less than zero (1)
∴ −180° < x < 0° (1)

TOTAL: 52 marks

This question paper consists of 4 questions.