IB Math 11
Assignment: Trigonometry
NAME:_____________________________
NON-CALCULATOR:
1. The diagram shows two concentric circles with centre O.
The radius of the smaller circle is 8 cm and the radius of the larger circle is 10
cm. Points A, B and C are on the circumference of the larger circle such that
π
AÔB is
radians. (diagram not to scale)
3
(a) Find the length of the arc ACB.
(2)
(b) Find the area of the shaded region.
(4)
(Total 6 marks)
2. The diagram shows a circle of radius r and centre O. The angle AÔB = θ.
The length of the arc AB is 24 cm. The area of the sector OAB is 180 cm2.
Find the value of r and of θ.
(Total 6 marks)
3. A spring is suspended from the ceiling. It is pulled down and released, and then oscillates up and
down. Its length, l centimetres, is modelled by the function l = 33 + 5cos((720t)°), where t is time in
seconds after release.
(a)
Find the length of the spring after 1 second.
(2)
(b)
Find the minimum length of the spring.
(3)
(c)
Find the first time at which the length is 33 cm.
(3)
(d)
What is the period of the motion?
(2)
(Total 10 marks)
4. Part of the graph of y = p + q cos x is shown below. The graph passes through the points (0, 3) and (π, –1).
Find the value of
(a)
p;
(b)
q.
(Total 6 marks)
5. (a) Show that 4 – cos 2θ + 5 sin θ = 2 sin2 θ + 5 sin θ + 3.
(2)
(b) Hence, solve the equation 4 – cos 2θ + 5 sin θ = 0 for 0 ≤ θ ≤ 2π.
(5)
(Total 7 marks)
6. (a) Given that 2 sin2 θ + sinθ – 1 = 0, find the two values for sin θ.
(4)
(b) Given that 0° ≤ θ ≤ 360° and that one solution for θ is 30°, find the other two possible values for θ.
(2)
(Total 6 marks)
7. Solve the equation 3 sin2 x = cos2 x, for 0° ≤ x ≤ 180°.
(Total 4 marks)
CALCULATOR ALLOWED:
 x
1. The diagram below shows the graph of f (x) = 1 + tan   for −360° ≤ x ≤ 360°.
 2
(a)
On the same diagram, draw the asymptotes.
(2)
(b)
Write down
(i)
the period of the function;
(ii)
the value of f (90°).
(2)
(c)
Solve f (x) = 0 for −360° ≤ x ≤ 360°.
(2)
(Total 6 marks)
2. The following graph shows the depth of water, y metres, at a point P, during one day.
The time t is given in hours, from midnight to noon.
(a)
Use the graph to write down an estimate of the value of t when
(i)
the depth of water is minimum;
(ii)
the depth of water is maximum;
(iii)
the depth of the water is increasing most rapidly.
(3)
(b)
The depth of water can be modelled by the function y = A cos (B (t – 1)) + C.
(i)
Show that A = 8.
(ii)
Write down the value of C.
(iii)
Find the value of B.
(6)
(c)
A sailor knows that he cannot sail past P when the depth of the water is less than 12 m.
Calculate the values of t between which he cannot sail past P.
(2)
(Total 11 marks)
3. The function f is defined by f : x → 30 sin 3x cos 3x, 0 ≤ x ≤
π
.
3
(a) Write down an expression for f (x) in the form a sin 6x, where a is an integer.
(b) Solve f (x) = 0, giving your answers in terms of π.
(Total 6 marks)
Non-Calculator
1.
(a)
correct substitution in l = rθ
π 1
e.g. 10 × , × 2π × 10
3 6
20 π  10 π 
arc length =
=

6  3 
(b)
(A1)
A1 N2
area of large sector =
1
π  100 π 
× 10 2 ×  =

2
3
6 
(A1)
area of small sector =
1 2 π  64 π 
×8 × =

2
3
6 
(A1)
evidence of valid approach (seen anywhere)
1 π
e.g. subtracting areas of two sectors, × (10 2 − 8 2 )
2 3
36π


, etc. 
area shaded = 6π  accept
6


M1
A1 N3
[6]
2.
METHOD 1
Evidence of correctly substituting into l = rθ
Evidence of correctly substituting into A =
For attempting to solve these equations
eliminating one variable correctly
r = 15
θ = 1.6 (= 91.7°)
1 2
r θ
2
A1
A1
(M1)
A1
A1A1
N3
METHOD 2
Setting up and equating ratios
24 180
=
2πr πr 2
A1A1
Solving gives r = 15
rθ = 24
 1 2

 or r θ =180 
2


θ = 1.6 (= 91.7°)
r = 15
(M1)
θ = 1.6 (= 91.7°)
A1
A1
A1
N3
[6]
3.
(a)
When t = 1, l = 33 + 5 cos 720
l = 33 + 5 = 38
(M1)
A1 N2
(b)
Minimum when cos = –1
lmin = 33 – 5
= 28
(M1)
(M1)
A1 N3
(c)
33 = 33 + 5cos720t (0 = 5 cos 720t)
720t = 90
90  1 
t=
= 
720  8 
(d)
M1
A1
A1 N1
Evidence of dividing into 360
360  1 
period =
= 
720  2 
(M1)
A1 N2
[10]
4.
3 = p + q cos 0
3=p+q
–1 = p + q cos π
–1 = p – q
(M1)
(A1)
(M1)
(A1)
(a)
p=1
(A1) (C3)
(b)
q=2
(A1) (C3)
[6]
5.
(a)
attempt to substitute 1 – 2 sin2 θ for cos 2θ
correct substitution
e.g. 4 – (1 – 2 sin2 θ) + 5 sin θ
4 – cos 2θ + 5 sin θ = 2 sin2 θ + 5 sin θ + 3
(M1)
A1
(b)
(M1)
AG N0
evidence of appropriate approach to solve
e.g. factorizing, quadratic formula
correct working
e.g. (2 sin θ + 3)(sin θ + 1), (2x + 3)(x + 1) = 0, sin x =
A1
−5± 1
4

correct solution sin θ = –1  do not penalise for including sin θ = –

3

2
(A1)
θ=
3π
2
A2 N3
[7]
6.
(a)
Attempt to factorise
correct factors (2sin θ – 1) (sin θ + 1) = 0
1
sin θ = , sin θ = –1
2
(b)
(M1)
A1
A1A1 N2
other solutions are 150°, 270°
A1A1 N1N1
[6]
7. tan2 x =
1
3
(M
1)
⇒ tan x = ±
1
3
(M
1)
⇒ x = 30° or x = 150°
(A
1)(A1)
(C
2)(C2)
[4]
Calculator Allowed
1.
(a)
(i)
(b)
7
A1
N1
(ii)
1
A1
N1
(iii)
10
A1
N1
(i)
evidence of appropriate approach
M1
e.g. A =
18 − 2
2
A=8
AG
N0
(ii)
C = 10
A2
N2
(iii)
METHOD 1
period = 12
(A1)
evidence of using B × period = 2π (accept 360°)
(M1)
e.g. 12 =
B=
2π
B
π
(accept 0.524 or 30)
6
A1
N3
METHOD 2
evidence of substituting
e.g. 10 = 8 cos 3B + 10
(M1)
simplifying
(A1)
π

e.g. cos 3B = 0  3B = 
2

B=
(c)
π
(accept 0.524 or 30)
6
correct answers
e.g. t = 3.52, t = 10.5, between 03:31 and 10:29 (accept 10:30)
A1
N3
A1A1
N2
[11]
2.
(a)
y
10
5
–180°
–360°
0
180°
360° x
–5
–10
Correct asymptotes
(b)
(c)
(i)
Period = 360°
(ii)
f (90°) = 2
(accept 2π)
270°, −90°
A1A1
N2
A1
N1
A1
N1
A1A1 N1N1
Notes:
Penalize 1 mark for any additional values.
Penalize 1 mark for correct answers given
 3π π

in radians  , − , or 4.71, −1.57  .
2
2


[6]
3. (a) Evidence of choosing the double angle formula
f (x) = 15 sin (6x)
(b)
Evidence of substituting for f (x)
(M1)
A1
N2
(M1)
eg 15 sin 6x = 0, sin 3x = 0 and cos 3x = 0
6x = 0, π, 2π
x = 0,
π π
,
6 3
A1A1A1
N4
[6]