IB Math – SL: Trig Practice Problems
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Circular Functions and Trig - Practice Problems (to 07)
1.
In the triangle PQR, PR = 5 cm, QR = 4 cm and PQ = 6 cm.
Calculate
the size of PQ̂R ;
the area of triangle PQR.
(a)
(b)
(Total 6 marks)
2.
The following diagram shows a triangle ABC, where AĈB is 90, AB = 3, AC = 2 and BÂC is .
5
.
3
4 5
Show that sin 2 =
.
9
Show that sin  =
(a)
(b)
Find the exact value of cos 2.
(c)
3.
(Total 6 marks)
The following diagram shows a sector of a circle of radius r cm, and angle  at the centre. The perimeter of the
sector is 20 cm.
20 2r
.
r
(a)
Show that  =
(b)
2
The area of the sector is 25 cm . Find the value of r.
(Total 6 marks)
4.
The following diagram shows the triangle AOP, where OP = 2 cm, AP = 4 cm and AO = 3 cm.
A
diagram not to
scale
O
(a)
P
Calculate AÔP , giving your answer in radians.
(3)
The following diagram shows two circles which intersect at the points A and B. The smaller circle C1 has centre
O and radius 3 cm, the larger circle C2 has centre P and radius 4 cm, and OP = 2 cm. The point D lies on the
circumference of C1 and E on the circumference of C2.Triangle AOP is the same as triangle AOP in the diagram
above.
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IB Math – SL: Trig Practice Problems
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A
C2
C1
D
O
E
P
diagram not to
scale
B
(b)
Find AÔB , giving your answer in radians.
(2)
(c)
Given that AP̂B is 1.63 radians, calculate the area of
(i)
sector PAEB;
(ii)
sector OADB.
(5)
(d)
2
The area of the quadrilateral AOBP is 5.81 cm .
(i)
Find the area of AOBE.
(ii)
Hence find the area of the shaded region AEBD.
(4)
(Total 14 marks)
5.
The following diagram shows a pentagon ABCDE, with AB = 9.2 cm, BC = 3.2 cm, BD = 7.1 cm, AÊD =110,
AD̂E = 52 and AB̂D = 60.
(a)
Find AD.
(b)
Find DE.
(4)
(4)
2
(c)
The area of triangle BCD is 5.68 cm . Find DB̂C .
(d)
Find AC.
(e)
Find the area of quadrilateral ABCD.
(4)
(4)
(5)
(Total 21 marks)
6.
 x
2
The diagram below shows the graph of f (x) = 1 + tan   for −360  x  360.
(a)
On the same diagram, draw the asymptotes.
(2)
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IB Math – SL: Trig Practice Problems
(b)
Write down
(i)
the period of the function;
(ii)
the value of f (90).
(c)
Solve f (x) = 0 for −360  x  360.
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(2)
(2)
(Total 6 marks)
7.
(a)
2
Consider the equation 4x + kx + 1 = 0. For what values of k does this equation have two equal roots?
Let f be the function f ( ) = 2 cos 2 + 4 cos  + 3, for −360    360.
2
(b)
Show that this function may be written as f ( ) = 4 cos  + 4 cos  + 1.
8.
(c)
Consider the equation f ( ) = 0, for −360    360.
(i)
How many distinct values of cos  satisfy this equation?
(ii)
Find all values of  which satisfy this equation.
(d)
Given that f ( ) = c is satisfied by only three values of , find the value of c.
(3)
(1)
(5)
(2)
(Total 11 marks)
A Ferris wheel with centre O and a radius of 15 metres is represented in the diagram below. Initially seat A is at
ground level. The next seat is B, where AÔB =
(a)
Find the length of the arc AB.
(b)
Find the area of the sector AOB.
π
.
6
(2)
(2)
(c)
The wheel turns clockwise through an angle of
2π
. Find the height of A above the ground.
3
(3)
The height, h metres, of seat C above the ground after t minutes, can be modelled by the function
h (t) = 15 − 15 cos  2t  π  .

(d)
4
π
.
4
(i)
Find the height of seat C when t =
(ii)
(iii)
Find the initial height of seat C.
Find the time at which seat C first reaches its highest point.
(8)
(e)
Find h′ (t).
(f)
For 0  t  ,
(i)
sketch the graph of h′;
(ii)
find the time at which the height is changing most rapidly.
(2)
(5)
(Total 22 marks)
9.
2
Let f (x) = a (x − 4) + 8.
(a)
Write down the coordinates of the vertex of the curve of f.
(b)
Given that f (7) = −10, find the value of a.
(c)
Hence find the y-intercept of the curve of f.
(Total 6 marks)
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IB Math – SL: Trig Practice Problems
10.
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The following diagram shows a circle with radius r and centre O. The points A, B and C are on the circle and
AÔC =.
The area of sector OABC is
Find the value of r and of .
4
2
 and the length of arc ABC is .
3
3
(Total 6 marks)
11.
Let ƒ (x) = a sin b (x − c). Part of the graph of ƒ is given below.
Given that a, b and c are positive, find the value of a, of b and of c.
(Total 6 marks)
12.
The points P(−2, 4), Q (3, 1) and R (1, 6) are shown in the diagram below.
(a)
(b)
Find the vector PQ .
Find a vector equation for the line through R parallel to the line (PQ).
(Total 6 marks)
13.
The diagram below shows a circle of radius r and centre O. The angle AÔB = .
2
The length of the arc AB is 24 cm. The area of the sector OAB is 180 cm .
Find the value of r and of .
(Total 6 marks)
14.
The diagram below shows a quadrilateral ABCD. AB = 4, AD = 8, CD =12, B Ĉ D = 25, BÂD =.
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IB Math – SL: Trig Practice Problems
(a)
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Use the cosine rule to show that BD = 4 5  4 cos  .
(2)
Let  = 40.
(b)
(i)
Find the value of sin CB̂D .
(ii)
Find the two possible values for the size of CB̂D .
(iii)
Given that CB̂D is an acute angle, find the perimeter of ABCD.
(12)
(c)
Find the area of triangle ABD.
(2)
(Total 16 marks)
15.
2
Let y = –16x + 160x –256. Given that y has a maximum value, find
(i)
the value of x giving the maximum value of y;
(ii)
this maximum value of y.
The triangle XYZ has XZ = 6, YZ = x, XY = z as shown below. The perimeter of triangle XYZ is 16.
(a)
(4)
(b)
(i)
Express z in terms of x.
(ii)
2
Using the cosine rule, express z in terms of x and cos Z.
(iii)
Hence, show that cos Z =
5 x 16
.
3x
(7)
Let the area of triangle XYZ be A.
2
2 2
(c)
Show that A = 9x sin Z.
(2)
2
2
(d)
Hence, show that A = –16x + 160x – 256.
(e)
(i)
(ii)
(4)
Hence, write down the maximum area for triangle XYZ.
What type of triangle is the triangle with maximum area?
(3)
(Total 20 marks)
16.
The function f is defined by f : x  30 sin 3x cos 3x, 0  x 
(a)
(b)
17.
π
.
3
Write down an expression for f (x) in the form a sin 6x, where a is an integer.
Solve f (x) = 0, giving your answers in terms of .
(Total 6 marks)
The following diagram shows two semi-circles. The larger one has centre O and radius 4 cm. The smaller one has
centre P, radius 3 cm, and passes through O. The line (OP) meets the larger semi-circle at S. The semi-circles
intersect at Q.
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IB Math – SL: Trig Practice Problems
(a)
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(i)
Explain why OPQ is an isosceles triangle.
(ii)
Use the cosine rule to show that cos OP̂Q =
(iii)
Hence show that sin OP̂Q =
(iv)
Find the area of the triangle OPQ.
1
.
9
80
.
9
(7)
(b)
Consider the smaller semi-circle, with centre P.
(i)
(ii)
Write down the size of OP̂Q.
Calculate the area of the sector OPQ.
(3)
(c)
Consider the larger semi-circle, with centre O. Calculate the area of the sector QOS.
(d)
Hence calculate the area of the shaded region.
(3)
(4)
(Total 17 marks)
18.
The graph of a function of the form y = p cos qx is given in the diagram below.
y
40
30
20
10
/2

x
–10
–20
–30
–40
(a)
(b)
19.
Write down the value of p.
Calculate the value of q.
(Total 6 marks)
A farmer owns a triangular field ABC. One side of the triangle, [AC], is 104 m, a second side, [AB], is 65 m and
the angle between these two sides is 60°.
(a)
Use the cosine rule to calculate the length of the third side of the field.
(3)
(b)
Given that sin 60° =
3
, find the area of the field in the form p 3 where p is an integer.
2
(3)
Let D be a point on [BC] such that [AD] bisects the 60° angle. The farmer divides the field into two parts A1 and
A2 by constructing a straight fence [AD] of length x metres, as shown on the diagram below.
C
104 m
A2
A
30°
D
x
30°
A1
65 m
B
(c)
(i)
65x
Show that the area of Al is given by
.
4
(ii)
Find a similar expression for the area of A2.
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IB Math – SL: Trig Practice Problems
(iii)
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Hence, find the value of x in the form q 3 , where q is an integer.
(7)
(d)
(i)
(ii)
Explain why sin AD̂C  sin AD̂B .
Use the result of part (i) and the sine rule to show that
BD 5
 .
DC 8
20.
(5)
(Total 18 marks)
2
The following diagram shows a circle of centre O, and radius r. The shaded sector OACB has an area of 27 cm .
Angle AÔB = θ = 1.5 radians.
A
C
B
r
O
(a)
Find the radius.
(b)
Calculate the length of the minor arc ACB.
Working:
Answers:
(a) ..................................................................
(b) ..................................................................
(Total 6 marks)
21.
Consider y = sin  x 

  .
9
(a)
The graph of y intersects the x-axis at point A. Find the x-coordinate of A, where 0  x  π.
(b)
Solve the equation sin  x 

  = – 1 , for 0  x  2.
9
2
Working:
Answers:
(a) ..................................................................
(b) ..................................................................
22.
(Total 6 marks)
The diagram shows a triangular region formed by a hedge [AB], a part of a river bank [AC] and a fence [BC]. The
hedge is 17 m long and BÂC is 29°. The end of the fence, point C, can be positioned anywhere along the river
bank.
(a)
Given that point C is 15 m from A, find the length of the fence [BC].
15 m
A
river bank
C
29°
17 m
B
(3)
(b)
The farmer has another, longer fence. It is possible for him to enclose two different triangular regions with
this fence. He places the fence so that AB̂C is 85°.
(i)
Find the distance from A to C.
(ii)
Find the area of the region ABC with the fence in this position.
(5)
(c)
To form the second region, he moves the fencing so that point C is closer to point A.
Find the new distance from A to C.
(d)
Find the minimum length of fence [BC] needed to enclose a triangular region ABC.
(4)
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IB Math – SL: Trig Practice Problems
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(2)
(Total 14 marks)
23.
Let f (x) =
(a)
1 sin 2x + cos x for 0  x  2.
2
Find f (x).
(i)
2
One way of writing f (x) is –2 sin x – sin x + 1.
2
(ii)
Factorize 2 sin x + sin x – 1.
(iii) Hence or otherwise, solve f (x) = 0.
(6)
The graph of y = f (x) is shown below.
y
A
b
0 a
2 x
B
There is a maximum point at A and a minimum point at B.
(b)
Write down the x-coordinate of point A.
(1)
(c)
The region bounded by the graph, the x-axis and the lines x = a and x = b is shaded in the diagram above.
(i)
Write down an expression that represents the area of this shaded region.
(ii)
Calculate the area of this shaded region.
(5)
(Total 12 marks)
24.
2
In triangle PQR, PQ is 10 cm, QR is 8 cm and angle PQR is acute. The area of the triangle is 20 cm . Find the
size of angle PQ̂R.
(Total 6 marks)
25.
Let f (x) = 6 sin x , and g (x) = 6e
maximum value at B (0.5, b).
–x
– 3 , for 0  x  2. The graph of f is shown on the diagram below. There is a
y
B
0
(a)
(b)
(c)
1
2
x
Write down the value of b.
On the same diagram, sketch the graph of g.
Solve f (x) = g (x) , 0.5  x  1.5.
(Total 6 marks)
26.
Consider the equation 3 cos 2x + sin x = 1
(a)
(b)
(c)
27.
2
Write this equation in the form f (x) = 0 , where f (x) = p sin x + q sin x + r , and p , q , r 
Factorize f (x).
Write down the number of solutions of f (x) = 0, for 0  x  2.
.
(Total 6 marks)
The diagram below shows two circles which have the same centre O and radii 16 cm and 10 cm respectively. The
two arcs AB and CD have the same sector angle  = 1.5 radians.
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IB Math – SL: Trig Practice Problems
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A
B
D
C
O
28.
Find the area of the shaded region.
Let f (x) = sin (2x + 1), 0  x  π.
(a)
Sketch the curve of y = f (x) on the grid below.
(Total 6 marks)
y
2
1.5
1
0.5
0
0.5
1
1.5
2
2.5
3
3.5 x
–0.5
–1
–1.5
–2
(b)
29.
Find the x-coordinates of the maximum and minimum points of f (x), giving your answers correct to one
decimal place.
(Total 6 marks)
2
In a triangle ABC, AB = 4 cm, AC = 3 cm and the area of the triangle is 4.5 cm .
Find the two possible values of the angle BÂC .
Working:
Answer:
…………………………………………..
(Total 6 marks)
30.
2
Solve the equation 2 cos x = sin 2x for 0  x  π, giving your answers in terms of π.
Working:
Answer:
…………………………………………..
(Total 6 marks)
31.
The depth y metres of water in a harbour is given by the equation
t
 2
y = 10 + 4 sin   ,
where t is the number of hours after midnight.
(a)
Calculate the depth of the water
(i)
when t = 2;
(ii)
at 2100.
(3)
The sketch below shows the depth y, of water, at time t, during one day (24 hours).
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IB Math – SL: Trig Practice Problems
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y
15
14
13
12
11
10
9
depth (metres)
8
7
6
5
4
3
2
1
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 t
time (hours)
(b)
(i)
(ii)
Write down the maximum depth of water in the harbour.
Calculate the value of t when the water is first at its maximum depth during the day.
(3)
The harbour gates are closed when the depth of the water is less than seven metres. An alarm rings when the gates
are opened or closed.
(c)
(i)
How many times does the alarm sound during the day?
(ii)
Find the value of t when the alarm sounds first.
(iii) Use the graph to find the length of time during the day when the harbour gates are closed. Give
your answer in hours, to the nearest hour.
(7)
(Total 13 marks)
32.
The following diagram shows a triangle ABC, where BC = 5 cm, B̂ = 60°, Ĉ = 40°.
A
B
40°
60°
5 cm
C
(a)
Calculate AB.
(b)
Find the area of the triangle.
Working:
Answers:
(a) …………………………………………..
(b) …………………………………………..
33.
(Total 6 marks)
The diagram below shows a circle of radius 5 cm with centre O. Points A and B are on the circle, and AÔB is
0.8 radians. The point N is on [OB] such that [AN] is perpendicular to [OB].
A
5 cm
O
0.8
N
B
Find the area of the shaded region.
Working:
Answer:
…………………………………………........
34.
Let f (x) = 1 + 3 cos (2x) for 0  x  π, and x is in radians.
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(Total 6 marks)
Page 10 of 23
IB Math – SL: Trig Practice Problems
(a)
(i)
(ii)
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Find f (x).
Find the values for x for which f (x) = 0, giving your answers in terms of .
(6)
The function g (x) is defined as g (x) = f (2x) – 1, 0  x 
(b)
35.
π
.
2
(i)
The graph of f may be transformed to the graph of g by a stretch in the x-direction with scale factor
1
followed by another transformation. Describe fully this other transformation.
2
(ii)
Find the solution to the equation g (x) = f (x)
(4)
(Total 10 marks)
Part of the graph of y = p + q cos x is shown below. The graph passes through the points (0, 3) and (, –1).
y
3
2
1
0

2
x
–1
Find the value of
(a)
p;
(b)
q.
Working:
Answers:
(a) ..................................................................
(b) ..................................................................
36.
37.
Find all solutions of the equation cos 3x = cos (0.5x), for 0  x  .
Working:
Answer:
..................................................................
(Total 6 marks)
(Total 6 marks)
The diagram below shows a triangle and two arcs of circles.
The triangle ABC is a right-angled isosceles triangle, with AB = AC = 2. The point P is the midpoint of [BC].
The arc BDC is part of a circle with centre A.
The arc BEC is part of a circle with centre P.
E
B
D
2
A
(a)
(b)
P
C
2
Calculate the area of the segment BDCP.
Calculate the area of the shaded region BECD.
(Total 6 marks)
38.
7
10
The diagram shows a parallelogram OPQR in which OP =   , OQ =  .
 3
1
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IB Math – SL: Trig Practice Problems
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y
P
Q
O
x
R
(a)
Find the vector OR .
(3)
(b)
Use the scalar product of two vectors to show that cos OP̂Q = –
15
.
754
(4)
(c)
(i)
Explain why cos PQ̂R = –cos OP̂Q.
(ii)
Hence show that sin PQ̂R =
23
.
754
(iii)
39.
Calculate the area of the parallelogram OPQR, giving your answer as an integer.
(7)
(Total 14 marks)
The points P, Q, R are three markers on level ground, joined by straight paths PQ, QR, PR as shown in the
diagram. QR = 9 km, PQ̂R = 35°, PR̂Q = 25°.
P
Diagram not to scale
Q
(a)
35°
25°
9 km
R
Find the length PR.
(3)
–1
(b)
Tom sets out to walk from Q to P at a steady speed of 8 km h . At the same time, Alan sets out to jog
–1
from R to P at a steady speed of a km h . They reach P at the same time. Calculate the value of a.
(c)
The point S is on [PQ], such that RS = 2QS, as shown in the diagram.
(7)
P
S
Q
R
Find the length QS.
(6)
(Total 16 marks)
40.
Consider the function f (x) = cos x + sin x.
(a)
π
) = 0.
4
(i)
Show that f (–
(ii)
Find in terms of , the smallest positive value of x which satisfies f (x) = 0.
(3)
x
The diagram shows the graph of y = e (cos x + sin x), – 2  x  3. The graph has a maximum turning point at C(a,
b) and a point of inflexion at D.
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IB Math – SL: Trig Practice Problems
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y
6
C(a, b)
4
D
2
–2
–1
1
2
3
x
dy
.
dx
(b)
Find
(c)
Find the exact value of a and of b.
(3)
(4)
π
2e 4
(d)
Show that at D, y =
.
(e)
Find the area of the shaded region.
(5)
(2)
(Total 17 marks)
41.
The graph of the function f (x) = 3x – 4 intersects the x-axis at A and the y-axis at B.
(a)
Find the coordinates of
(i)
A;
(ii)
B.
(b)
Let O denote the origin. Find the area of triangle OAB.
Working:
Answers:
(a) (i) ...........................................................
(ii) ...........................................................
(b) ..................................................................
(Total 6 marks)
2
Factorize the expression 3 sin x – 11 sin x + 6.
2
(b)
Consider the equation 3 sin x – 11 sin x + 6 = 0.
(i)
Find the two values of sin x which satisfy this equation,
(ii)
Solve the equation, for 0°  x  180°.
Working:
Answers:
(a) ..................................................................
(b) (i) ...........................................................
(ii) ...........................................................
42.
(a)
43.
(Total 6 marks)
The diagram below shows a circle, centre O, with a radius 12 cm. The chord AB subtends at an angle of 75° at the
centre. The tangents to the circle at A and at B meet at P.
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A
12 cm
P diagram not to
scale
O 75º
B
(a)
Using the cosine rule, show that the length of AB is 12 21 – cos 75 .
(b)
Find the length of BP.
(c)
Hence find
(i)
the area of triangle OBP;
(ii)
the area of triangle ABP.
(d)
Find the area of sector OAB.
(e)
Find the area of the shaded region.
(2)
(3)
(4)
(2)
(2)
(Total 13 marks)
44.
Note: Radians are used throughout this question.
A mass is suspended from the ceiling on a spring. It is pulled down to point P and then released. It oscillates up
and down.
diagram not to
scale
P
Its distance, s cm, from the ceiling, is modelled by the function s = 48 + 10 cos 2πt where t is the time in seconds
from release.
(a)
(i)
What is the distance of the point P from the ceiling?
(ii)
How long is it until the mass is next at P?
(5)
(b)
(i)
ds
Find
.
dt
(ii)
Where is the mass when the velocity is zero?
(7)
A second mass is suspended on another spring. Its distance r cm from the ceiling is modelled by the function r =
60 + 15 cos 4t. The two masses are released at the same instant.
(c)
Find the value of t when they are first at the same distance below the ceiling.
(2)
(d)
In the first three seconds, how many times are the two masses at the same height?
(2)
(Total 16 marks)
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IB Math – SL: Trig Practice Problems
45.
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The following diagram shows a circle of centre O, and radius 15 cm. The arc ACB subtends an angle of 2 radians
at the centre O.
C
A
B
15
cm
Diagram not to scale
2 rad
O
AÔB = 2 radians
OA = 15 cm
Find
(a)
the length of the arc ACB;
(b)
the area of the shaded region.
Working:
Answers:
(a) ..................................................................
(b) ..................................................................
46.
47.
(Total 6 marks)
–1
Two boats A and B start moving from the same point P. Boat A moves in a straight line at 20 km h and boat B
–1
moves in a straight line at 32 km h . The angle between their paths is 70°.
Find the distance between the boats after 2.5 hours.
(Total 6 marks)
Let f (x) = sin 2x and g (x) = sin (0.5x).
(a)
Write down
(i)
the minimum value of the function f ;
(ii)
the period of the function g.
(b)
Consider the equation f (x) = g (x).
Find the number of solutions to this equation, for 0  x 
3π
.
2
Working:
Answers:
(a) (i) ..........................................................
(ii) ..........................................................
(b) .................................................................
(Total 6 marks)
48.
Consider the following statements
x
A:
log10 (10 ) > 0.
B:
–0.5  cos (0.5x)  0.5.
π
π
 arctan x  .
2
2
C:
–
(a)
Determine which statements are true for all real numbers x. Write your answers (yes or no) in the table
below.
Statement
(a) Is the statement true for all
(b) If not true, example
real numbers x? (Yes/No)
A
B
C
(b)
If a statement is not true for all x, complete the last column by giving an example of one value of x for
which the statement is false.
Working:
(Total 6 marks)
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IB Math – SL: Trig Practice Problems
49.
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The diagram shows a triangle ABC in which AC = 7
2
, BC = 6, AB̂C = 45°.
2
A
2
7 2
Diagram
not to scale
B
(a)
Use the fact that sin 45° =
45°
C
6
6
2
to show that sin BÂC = .
7
2
(2)
The point D is on (AB), between A and B, such that sin BD̂C =
(b)
(i)
(ii)
(iii)
6
.
7
Write down the value of BD̂C + BÂC .
Calculate the angle BCD.
Find the length of [BD].
(6)
(c)
Show that
Area of BDC
BD
=
.
Area of BAC
BA
(2)
(Total 10 marks)
50.
In triangle ABC, AC = 5, BC = 7, Â = 48°, as shown in the diagram.
C
7
5
A
diagram not to scale
48°
B
Find B̂, giving your answer correct to the nearest degree.
Working:
Answer:
......................................................................
(Total 6 marks)
51.
Given that sin x =
1
, where x is an acute angle, find the exact value of
3
(a)
cos x;
(b)
cos 2x.
Working:
Answers:
(a) ..................................................................
(b) ..................................................................
(Total 6 marks)
52.
2
Consider the trigonometric equation 2 sin x = 1 + cos x.
2
Write this equation in the form f (x) = 0, where f (x) = a cos x + b cos x + c,
and a, b, c  .
(b)
Factorize f (x).
(c)
Solve f (x) = 0 for 0°  x  360°.
Working:
Answers:
(a) ..................................................................
(a)
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IB Math – SL: Trig Practice Problems
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(b) ..................................................................
(c) ..................................................................
(Total 6 marks)
53.
The following diagram shows a triangle with sides 5 cm, 7 cm, 8 cm.
5
7
Diagram not to scale
8
Find
(a)
the size of the smallest angle, in degrees;
(b)
the area of the triangle.
Working:
Answers:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
54.
2
2
(a)
Write the expression 3 sin x + 4 cos x in the form a cos x + b cos x + c.
(b)
Hence or otherwise, solve the equation
2
3 sin x + 4 cos x – 4 = 0,
0  x  90.
Working:
Answers:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
55.
In the following diagram, O is the centre of the circle and (AT) is the tangent to the circle at T.
T
O
A
Diagram not to scale
If OA = 12 cm, and the circle has a radius of 6 cm, find the area of the shaded region.
Working:
Answer:
.......................................................................
(Total 4 marks)
56.
In the diagram below, the points O(0, 0) and A(8, 6) are fixed. The angle OP̂A
varies as the point P(x, 10) moves along the horizontal line y = 10.
y
P(x, 10)
y=10
A(8, 6)
O(0, 0)
x
Diagram to scale
(a)
(i)
Show that AP 
x – 16x  80.
2
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IB Math – SL: Trig Practice Problems
(ii)
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Write down a similar expression for OP in terms of x.
(2)
(b)
Hence, show that
cos OP̂A 
x 2 – 8 x  40
 {( x 2 – 16x  80)(x 2  100)}
,
(3)
(c)
Find, in degrees, the angle OP̂A when x = 8.
(2)
(d)
Find the positive value of x such that OP̂A  60 .
(4)
Let the function f be defined by
f ( x)  cos OP̂A 
(e)
57.
x 2 – 8 x  40
 {( x 2 – 16x  80)( x 2  100)}
, 0  x  15.
Consider the equation f (x) = 1.
(i)
Explain, in terms of the position of the points O, A, and P, why this
equation has a solution.
(ii)
Find the exact solution to the equation.
(5)
(Total 16 marks)
The diagram below shows a sector AOB of a circle of radius 15 cm and centre O. The angle  at the centre of the
circle is 2 radians.
Diagram not to scale
B
A
O
(a)
Calculate the area of the sector AOB.
(b)
Calculate the area of the shaded region.
Working:
Answers:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
58.
The diagrams below show two triangles both satisfying the conditions
AB = 20 cm, AC = 17 cm, AB̂C = 50°.
Diagrams not
to scale
Triangle 1
Triangle 2
A
A
B
C
B
C
(a)
Calculate the size of AĈB in Triangle 2.
(b)
Calculate the area of Triangle 1.
Working:
Answers:
(a) ..................................................................
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IB Math – SL: Trig Practice Problems
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(b) ..................................................................
(Total 4 marks)
59.
The depth, y metres, of sea water in a bay t hours after midnight may be represented by the function
 2 
y  a  b cos 
t  , where a, b and k are constants.
 k 
60.
The water is at a maximum depth of 14.3 m at midnight and noon, and is at a minimum depth of 10.3 m at 06:00
and at 18:00.
Write down the value of
(a)
a;
(b)
b;
(c)
k.
Working:
Answers:
(a) ..................................................................
(b) ..................................................................
(c) ..................................................................
(Total 4 marks)
Town A is 48 km from town B and 32 km from town C as shown in the diagram.
C
32km
A
48km
B
Given that town B is 56 km from town C, find the size of angle CÂB to the nearest degree.
(Total 4 marks)
61.
62.
(a)
2
Express 2 cos x + sin x in terms of sin x only.
2
(b)
Solve the equation 2 cos x + sin x = 2 for x in the interval 0  x  , giving your answers exactly.
(Total 4 marks)
Note: Radians are used throughout this question.
(a)
Draw the graph of y =  + x cos x, 0  x  5, on millimetre square graph paper, using a scale of 2 cm per
unit. Make clear
(i)
the integer values of x and y on each axis;
(ii)
the approximate positions of the x-intercepts and the turning points.
(5)
(b)
Without the use of a calculator, show that  is a solution of the equation
 + x cos x = 0.
(3)
(c)
Find another solution of the equation  + x cos x = 0 for 0  x  5, giving your answer to six significant
figures.
(2)
(d)
Let R be the region enclosed by the graph and the axes for 0  x  . Shade R on your diagram, and write
down an integral which represents the area of R .
(2)
(e)
Evaluate the integral in part (d) to an accuracy of six significant figures. (If you consider it necessary, you
can make use of the result
d
( x sin x  cos x )  x cos x .)
dx
(3)
(Total 15 marks)
63.
A formula for the depth d metres of water in a harbour at a time t hours after midnight is
 
d  P  Q cos  t , 0  t  24,
6 
where P and Q are positive constants. In the following graph the point (6, 8.2) is a minimum point and the point
(12, 14.6) is a maximum point.
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IB Math – SL: Trig Practice Problems
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d
15
(12, 14.6)
10.
(6, 8.2)
5
0
6
12
18
(a)
Find the value of
(i)
Q;
(ii)
P.
(b)
Find the first time in the 24-hour period when the depth of the water is 10 metres.
(c)
(i)
(ii)
24
t
(3)
(3)
Use the symmetry of the graph to find the next time when the depth of the water is 10 metres.
Hence find the time intervals in the 24-hour period during which the water is less than 10 metres
deep.
64.
(4)
Solve the equation 3 cos x = 5 sin x, for x in the interval 0°  x  360°, giving your answers to the nearest degree.
(Total 4 marks)
65.
If A is an obtuse angle in a triangle and sin A =
66.
(a)
(b)
(c)
5
, calculate the exact value of sin 2A.
13
(Total 4 marks)
Sketch the graph of y =  sin x – x, –3  x  3, on millimetre square paper, using a scale of 2 cm per unit
on each axis.
Label and number both axes and indicate clearly the approximate positions of the
x-intercepts and the local maximum and minimum points.
(5)
Find the solution of the equation
 sin x – x = 0,
x > 0.
(1)
Find the indefinite integral
 ( sin x  x)dx
and hence, or otherwise, calculate the area of the region enclosed by the graph, the
x-axis and the line x = 1.
(4)
(Total 10 marks)
67.
Given that sin θ =
(a)
(b)
68.
1 , cos θ = – 3 and 0° ≤ θ ≤ 360°,
2
2
find the value of θ;
write down the exact value of tan θ.
(Total 4 marks)
The diagram shows a vertical pole PQ, which is supported by two wires fixed to the horizontal ground at A and B.
P
36 B
30
Q
70
A
BQ = 40 m
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IB Math – SL: Trig Practice Problems
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PB̂Q = 36°
BÂQ = 70°
AB̂Q = 30°
Find
(a)
(b)
the height of the pole, PQ;
the distance between A and B.
(Total 4 marks)
69.
The diagram shows a circle of radius 5 cm.
1 radian
Find the perimeter of the shaded region.
(Total 4 marks)
70.
f (x) = 4 sin  3x 

  .
2
For what values of k will the equation f (x) = k have no solutions?
71.
(Total 4 marks)
In this question you should note that radians are used throughout.
2
(a)
(i)
Sketch the graph of y = x cos x, for 0  x  2 making clear the approximate positions of the
positive x-intercept, the maximum point and the end-points.
(ii)
Write down the approximate coordinates of the positive x-intercept, the maximum point and the
end-points.
(7)
(b)
Find the exact value of the positive x-intercept for 0  x  2.
(2)
Let R be the region in the first quadrant enclosed by the graph and the x-axis.
(c)
(i)
Shade R on your diagram.
(ii)
Write down an integral which represents the area of R.
(3)
(d)
Evaluate the integral in part (c)(ii), either by using a graphic display calculator, or by using the following
information.
d 2
2
(x sin x + 2x cos x – 2 sin x) = x cos x.
dx
(3)
(Total 15 marks)
72.
In this part of the question, radians are used throughout.
The function f is given by
2
f (x) = (sin x) cos x.
The following diagram shows part of the graph of y = f (x).
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IB Math – SL: Trig Practice Problems
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y
A
C
x
B
O
The point A is a maximum point, the point B lies on the x-axis, and the point C is a point of inflexion.
(a)
Give the period of f.
(1)
(b)
From consideration of the graph of y = f (x), find to an accuracy of one significant figure the range of f.
(c)
(i)
Find f (x).
(ii)
Hence show that at the point A, cos x =
(iii)
Find the exact maximum value.
(1)
1.
3
(9)
(d)
Find the exact value of the x-coordinate at the point B.
(1)
(e)
 f (x) dx.
(i)
Find
(ii)
Find the area of the shaded region in the diagram.
(4)
(f)
73.
74.
3
Given that f (x) = 9(cos x) – 7 cos x, find the x-coordinate at the point C.
(4)
(Total 20 marks)
A triangle has sides of length 4, 5, 7 units. Find, to the nearest tenth of a degree, the size of the largest angle.
(Total 4 marks)
O is the centre of the circle which has a radius of 5.4 cm.
O
A
B
2
The area of the shaded sector OAB is 21.6 cm . Find the length of the minor arc AB.
(Total 4 marks)
75.
6
The circle shown has centre O and radius 6. OA is the vector   , OB is the vector
0
 5 
 .
vector 
 11 
  6
  and OC is the
 0 
y
C
B
O
A
x
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IB Math – SL: Trig Practice Problems
(a)
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Verify that A, B and C lie on the circle.
(3)
(b)
Find the vector AC .
(c)
ˆC .
Using an appropriate scalar product, or otherwise, find the cosine of angle OA
(2)
(3)
(d)
Find the area of triangle ABC, giving your answer in the form a 11 , where a 
.
(4)
(Total 12 marks)
76.
2
2
Solve the equation 3 sin x = cos x, for 0°  x  180°.
77.
(Total 4 marks)
The diagrams show a circular sector of radius 10 cm and angle θ radians which is formed into a cone of slant
height 10 cm. The vertical height h of the cone is equal to the radius r of its base. Find the angle θ radians.
10cm
10cm
h
r
(Total 4 marks)
78.
The diagram shows the graph of the function f given by

f (x) = A sin  x  + B,
2 
for 0  x  5, where A and B are constants, and x is measured in radians.
y
(1,3)
(5, 3)
2
(0, 1)
x
0
1
2
3
4
5
(3, –1)
The graph includes the points (1, 3) and (5, 3), which are maximum points of the graph.
(a)
Write down the values of f (1) and f (5).
(2)
(b)
Show that the period of f is 4.
(2)
The point (3, –1) is a minimum point of the graph.
(c)
Show that A = 2, and find the value of B.
(5)
(d)

Show that f (x) =  cos  x  .
2 
The line y = k – x is a tangent line to the graph for 0  x  5.
(e)
Find
(i)
the point where this tangent meets the curve;
(ii)
the value of k.
(f)
Solve the equation f (x) = 2 for 0  x  5.
(4)
(6)
(5)
(Total 24 marks)
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IB Math – SL: Trig Practice Problems: MarkScheme
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Circular Functions and Trig - Practice Problems (to 07) MarkScheme
1.
(a)
Evidence of using the cosine rule
(M1)
p r q
, q 2  p 2  r 2  2 pr cos PQ̂R
2 pr
2
eg cos PQ̂R =
2
2
Correct substitution
eg
A1
4 6 5
, 5 2  4 2  6 2  2  4  6 cos Q
2 4 6
2
2
2
cos PQ̂R =
27
 0.5625
48
(A1)
PQ̂R = 55.8 (0.973 radians)
(b)
Area =
A1N2
1
pr sin PQ̂R
2
For substituting correctly
1
 4  6 sin 55.8
2
2
= 9.92 (cm )
A1
A1N1
[6]
2.
Note:
(a)
Throughout this question, do not accept methods which involve
finding  .
Evidence of correct approach
A1
BC
, BC  32  2 2  5
AB
5
sin  =
3
eg sin  =
(b)
AG
Evidence of using sin 2 = 2 sin  cos 
(M1)
 5  2 

 3   3 


4 5
=
9
= 2
(c)
N0
Evidence of using an appropriate formula for cos 2 
4 5
4
5  80 
 , 2   1 , 1  2  , 1  
9 9
9
9  81 
1
cos 2 = 
9
A1
AGN0
M1
eg
A2
N2
[6]
3.
(a)
For using perimeter = r + r + arc length
20 = 2r + r
20 2r
r
1  20  2r 
2
Finding A = r 2 
 10r  r
2  r 
(M1)
A1

(b)

For setting up equation in r
Correct simplified equation, or sketch
2
2
eg 10r – r = 25, r – 10r + 25 = 0
r = 5 cm
AG

N0
(A1)
M1
(A1)
A1
N2
[6]
4. Notes: Candidates may have differing answers due to using approximate answers from previous parts or using
answers from the GDC. Some leeway is provided to accommodate this.
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IB Math – SL: Trig Practice Problems: MarkScheme
(a)
METHOD 1
Evidence of using the cosine rule
(M1)
a b c
, a 2  b 2  c 2  2bc cos A
2ab
2
eg cos C =
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2
2
Correct substitution
eg cos AÔP =
32  2 2  4 2 2 2
, 4  3  2 2  2  3 2 cos AÔP
2  3 2
A1
cos AÔP = 0.25
 26 
AÔP = 1.82  
 (radians)
 45 
METHOD 2
Area of AOBP = 5.81 (from part (d))
Area of triangle AOP = 2.905
(M1)
2.9050 = 0.5  2  3  sin AÔP
A1
AÔP = 1.32 or 1.82
 26 
AÔP = 1.82  
 (radians)
 45 
(b)
(c)
A1
(= 2  3.64)
AÔB = 2(  1.82)
 38 
= 2.64  
 (radians)
 45 
(A1)
(i)
(M1)
Appropriate method of finding area
eg area =
1 2
 4 1.63
2
2
= 13.0 (cm )
(accept the exact value 13.04)
(ii)
(i)
(ii)
N2
N2
A1N2
1 2
θr
2
Area of sector PAEB =
(d)
A1
A1
A1N2
1 2
Area of sector OADB =  3  2.64
2
A1
2
= 11.9 (cm )
Area AOBE = Area PAEB  Area AOBP (= 13.0  5.81)
= 7.19 (accept 7.23 from the exact answer for PAEB)
Area shaded = Area OADB  Area AOBE (= 11.9  7.19)
= 4.71 (accept answers between 4.63 and 4.72)
A1N1
M1
A1N1
M1
A1N1
[14]
5.
(a)
(b)
Evidence of choosing cosine rule
2
2
2
eg a = b + c  2bc cos A
Correct substitution
2
2
2
eg (AD) = 7.1 + 9.2  2(7.1) (9.2) cos 60
2
(AD) = 69.73
AD = 8.35 (cm)
180  162 = 18
Evidence of choosing sine rule
Correct substitution
eg
(c)
(M1)
A1
(A1)
A1N2
(A1)
(M1)
A1
DE
8.35
=
sin 18 sin 110
DE = 2.75 (cm)
Setting up equation
A1
(M1)
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N2
Page 2 of 34
IB Math – SL: Trig Practice Problems: MarkScheme
eg
1
1
ab sin C = 5.68,
bh = 5.68
2
2
Correct substitution
A1
1
1
eg 5.68 = (3.2) (7.1) sin DB̂C ,  3.2  h = 5.68, (h = 3.55)
2
2
sin DB̂C = 0.5
DB̂C 30 and/or 150
(d)
(e)
Alei - Desert Academy
Finding A B̂ C (60 + D B̂ C)
Using appropriate formula
2
2
2
2
2
2
eg (AC) = (AB) + (BC) , (AC) = (AB) + (BC)  2 (AB)
(BC) cos ABC
Correct substitution (allow FT on their seen AB̂C )
2
2
2
eg (AC) = 9.2 + 3.2
AC = 9.74 (cm)
For finding area of triangle ABD
Correct substitution Area =
1
 9.2  7.1 sin 60
2
= 28.28...
Area of ABCD = 28.28... + 5.68
2
= 34.0 (cm )
(A1)
A1
N2
(A1)
(M1)
A1
A1
(M1)
N3
A1
A1
(M1)
A1N3
[21]
6.
(a)
y
10
5
–360°
–180°
0
360° x
180°
–5
–10
(b)
(c)
Correct asymptotes
(i)
Period = 360
(ii)
f (90) = 2
270, 90
Notes:
(accept 2)
A1A1
A1
A1
A1A1
N2
N1
N1
N1N1
Penalize 1 mark for any additional values.
Penalize 1 mark for correct answers given
 3 

,  , or 4.71, 1.57  .
2
 2

in radians 
[6]
7.
(a)
METHOD 1
Using the discriminant  = 0
2
k =441
k = 4, k =  4
METHOD 2
Factorizing
2
(2x  1)
k = 4, k =  4
(M1)
A1A1
N3
(M1)
A1A1
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N3
Page 3 of 34
IB Math – SL: Trig Practice Problems: MarkScheme
(b)
(c)
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2
Evidence of using cos 2 = 2 cos   1
2
eg 2(2 cos   1) + 4 cos  + 3
2
f () = 4 cos  + 4 cos  + 1
(i)
1
(ii)
METHOD 1
Attempting to solve for cos 
M1
AG
A1
N0
N1
M1
1
cos  = 
2
(A1)
 = 240, 120,  240, 120 (correct four values only)
METHOD 2
2
Sketch of y = 4 cos  + 4 cos  + 1
A2N3
M1
y
9
–360
(d)
–180
Indicating 4 zeros
 = 240, 120, 240, 120 (correct four values only)
Using sketch
c=9
180
x
360
(A1)
A2N3
(M1)
A1
[11]
(a)
Note: Accept exact answers given in terms of .
Evidence of using l = r
arc AB = 7.85 (m)
(b)
1
Evidence of using A  r 2 θ
2
(M1)
2
Area of sector AOB = 58.9 (m )
METHOD 1
A1
8.
(c)
N2
(M1)
A1
N2
N2
π6
2 π
3
angle =

30
6
(A1)

6

height = 15 + 15 sin
6
attempt to find 15 sin
M1
= 22.5 (m)
METHOD 2
A1
π
N2
3
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Page 4 of 34
IB Math – SL: Trig Practice Problems: MarkScheme
angle =
Alei - Desert Academy

60
3
(A1)

3

height = 15 + 15 cos
3
attempt to find 15 cos
(d)
(i)
M1
= 22.5 (m)
A1

  
h  1515cos   
4
2 4
(M1)
= 25.6 (m)
N2
A1N2




4
(ii)
h(0) = 15  15 cos  0 
(iii)
= 4.39(m)
METHOD 1
Highest point when h = 30
(M1)
A1N2
R1


30 = 15  15 cos  2t  
4



cos  2t   = 1
4

3 

t = 1.18  accept

8 

M1
(A1)
A1N2
METHOD 2
h
30
2π
t
Sketch of graph of h
Correct maximum indicated
t = 1.18
METHOD 3
Evidence of setting h(t) = 0


sin  2t 
M2
(A1)
A1N2
M1

0
4
(A1)
Justification of maximum
eg reasoning from diagram, first derivative test, second
derivative test


t = 1.18  accept


(e)
h(t) = 30 sin  2t 
(f)
(i)
3 

8 
R1
A1N2

 (may be seen in part (d))
4
A1A1
N2
h(t)
30
π
2
π
t
–30
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Page 5 of 34
IB Math – SL: Trig Practice Problems: MarkScheme
Alei - Desert Academy
A1A1A1 N3
Notes: Award A1 for range 30 to 30, A1 for two zeros. Award A1 for approximate
correct sinusoidal shape.
(ii)
METHOD 1
Maximum on graph of h
(M1)
t = 0.393
A1N2
METHOD 2
Minimum on graph of h
(M1)
t = 1.96
A1N2
METHOD 3
Solving h(t) = 0
(M1)
One or both correct answers
A1
t = 0.393, t = 1.96
N2
[22]
9.
(a)
(b)
(c)
Vertex is (4, 8)
A1A1
N2
2
Substituting 10 = a(7  4) + 8
a = 2
For y-intercept, x = 0
y = 24
M1
A1N1
(A1)
A1
N2
[6]
10.
METHOD 1
Evidence of correctly substituting into A =
1 2
r θ
2
Evidence of correctly substituting into l = r
For attempting to eliminate one variable …
leading to a correct equation in one variable
r=4
=

6
(= 0.524, 30)
METHOD 2
Setting up and equating ratios
4
2


3 3
r 2 2r
A1
A1
(M1)
A1
A1A1
N3
(M1)
A1A1
Solving gives r = 4
2  1 2 4 
  or r    
3  2
3 

 0.524, 30
=
6

 0.524, 30
r=4
=
6
r =
A1
A1
A1
N3
[6]
11.
  3 
a = 4, b = 2, c =
 or etc 
2  2

A2A2A2 N6
[6]
12.
(a)

 5 
PQ =  
  3
(b)
Using r = a + tb
 x  1  5 
      t  
 y   6    3
A1A1
N2
A2A1A1 N4
[6]
13.
METHOD 1
Evidence of correctly substituting into l = r
A1
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Page 6 of 34
IB Math – SL: Trig Practice Problems: MarkScheme
Evidence of correctly substituting into A =
1 2
r 
2
For attempting to solve these equations
eliminating one variable correctly
r = 15
 = 1.6 (= 91.7)
METHOD 2
Setting up and equating ratios
N3
(M1)
A1A1
Solving gives r = 15
 = 1.6
r = 15
A1
(M1)
A1
A1A1
24 180

2r r 2
r = 24
Alei - Desert Academy
A1
 1 2

 or r θ 180
 2

A1
(= 91.7)
 = 1.6 (= 91.7)
A1
N3
[6]
14.
(a)
For correct substitution into cosine rule
BD =
4 2  8 2  2  4  8 cos θ
For factorizing 16, BD =
165  4 cos θ 
= 4 5  4 cos θ
(b)
(i)
(ii)
BD = 5.5653 ...
(c)
Area =
A1
AGN0
(A1)
sin CB̂D sin 25

12
5.5653
sin CB̂D = 0.911
(accept 0.910, subject to AP)
CB̂D = 65.7
Or CB̂D = 180  their acute angle
= 114
(iii)
A1
M1A1
A1N2
A1
(M1)
A1N2
BD̂C = 89.3
BC
5.5653
BC
12
(or cosine rule)

or

sin 89.3 sin 25 sin 89.3 sin 65.7
(A1)
BC = 13.2
(accept 13.17…)
Perimeter = 4 + 8 + 12 + 13.2
= 37.2
A1
1
 4  8  sin 40
2
N1
M1A1
A1N2
A1
= 10.3
A1
N1
[16]
15.
(a)
METHOD 1
Note:
There are many valid algebraic approaches
to this problem (eg completing the square,
using x 
b
) . Use the following mark
2a
allocation as a guide.
(i)
dy
0
Using
dx
32x + 160 = 0
x=5
2
(ii)
ymax = 16(5 ) + 160(5)  256
ymax = 144
METHOD 2
(M1)
A1
A1N2
A1N1
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Page 7 of 34
IB Math – SL: Trig Practice Problems: MarkScheme
(i)
(b)
(ii)
(i)
(ii)
(iii)
Alei - Desert Academy
Sketch of the correct parabola (may be seen in part (ii))
x=5
ymax = 144
z = 10  x
(accept x + z = 10)
2
2
2
z = x + 6 2  x  6  cos Z
Substituting for z into the expression in part (ii)
2
2
Expanding 100  20x + x = x + 36  12x cos Z
Simplifying 12x cos Z = 20x  64
M1
A2N2
A1
A1
Isolating cos Z =
A1
cos Z =
20x  64
12x
5 x 16
3x
A2
(M1)
N1
N1
N2
A1
A1
AGN0
Note:
(c)
Expanding, simplifying and isolating may
be done in any order, with the final A1
being awarded for an expression that
clearly leads to the required answer.
Evidence of using the formula for area of a triangle
1


 A   6  x  sin Z 
2


1


A  3x sin Z  A 2   3 6 x 2  sin 2 Z 
4


(d)
M1
A1
2
2 2
A = 9x sin Z
2
2
Using sin Z = 1  cos Z
AG
(A1)
5 x 16
Substituting
for cos Z
3x
2
 25x 2 160x  256 
 5 x 16 

for expanding 
 to

9x 2
 3x 


(e)
N0
A1
A1
for simplifying to an expression that clearly leads to the required answer
2
2
2
eg A = 9x  (25x  160x + 256)
2
2
A = 16x + 160x  256
2
(i)
144 (is maximum value of A , from part (a))
Amax = 12
(ii)
Isosceles
A1
Evidence of choosing the double angle formula
f (x) = 15 sin (6x)
Evidence of substituting for f (x)
eg 15 sin 6x = 0, sin 3x = 0 and cos 3x = 0
6x = 0, , 2
(M1)
A1
(M1)
AG
A1
A1N1
A1
N1
[20]
16.
(a)
(b)
x = 0,
 
,
6 3
N2
A1A1A1 N4
[6]
17.
(a)
(i)
OP = PQ (= 3cm)
So  OPQ is isosceles
(ii)
Using cos rule correctly eg cos OP̂Q =
cos OP̂Q =
9  9 16  2 
 
18
 18 
R1
AGN0
32  32  4 2
2  3 3
(M1)
A1
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Page 8 of 34
IB Math – SL: Trig Practice Problems: MarkScheme
cos OP̂Q =
(iii)
Alei - Desert Academy
1
9
AGN0
2
2
Evidence of using sin A + cos A = 1
M1
1 
80 
sin OP̂Q = 1 

81 
81 
80
sin OP̂Q =
9
(iv)
A1
AGN0
Evidence of using area triangle OPQ =
1
 OP  PQ sin P
2
(i)
1
80 9
3 3
,  0.9938
2
9 2
80
Area triangle OPQ =
2
OP̂Q = 1.4594...
(ii)
OP̂Q = 1.46
Evidence of using formula for area of a sector
M1
eg
(b)
eg Area sector OPQ =

20
  4.47
(d)
A1N1
(M1)
1 2
 3 1.4594
2
= 6.57
(c)
A1N1
 1.4594
 0.841
2
1 2
Area sector QOS =  4  0.841
2
QÔP =
A1N2
(A1)
A1
= 6.73
Area of small semi-circle is 4.5 (= 14.137...)
Evidence of correct approach
eg Area = area of semi-circle  area sector OPQ  area sector QOS +
area triangle POQ
Correct expression
eg 4.5  6.5675...  6.7285... + 4.472..., 4.5  (6.7285... + 2.095...),
4.5 (6.5675... + 2.256...)
Area of the shaded region = 5.31
A1N2
A1
M1
p = 30
METHOD 1
A22
Period =
2
q
(M2)
=

2
(A1)
A1
A1
N1
[17]
18.
(a)
(b)
q=4
METHOD 2
A1
Horizontal stretch of scale factor =
scale factor =
q=4
1
q
1
4
4
(M2)
(A1)
A1
4
[6]
19.
(a)
2
2
using the cosine rule (A2) = b + c –2bc cos Â
(M1)
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Page 9 of 34
IB Math – SL: Trig Practice Problems: MarkScheme
2
2
2
substituting correctly BC = 65 +104 –2 (65) (104) cos 60°
= 4225 + 10816 – 6760 = 8281
 BC = 91 m
(b)
(c)
1
bc sin Â
2
substituting correctly, area = 1 (65) (104) sin 60°
2
= 1690 3 (Accept p = 1690)
1
(i)
A1 =   (65) (x) sin 30°
2
65x
=
4
1
(ii)
A2 =   (104) (x) sin 30°
2
finding the area, using
= 26x
(iii)
(d)
65x
starting A1 + A2 = A or substituting
+ 26x = 1690 3
4
169x
simplifying
= 1690 3
4
4  1690 3
x=
169
 x = 40 3 (Accept q = 40)
(i)
Recognizing that supplementary angles have equal sines
(ii)
eg AD̂C = 180 – AD̂B  sin AD̂C = sin AD̂B
using sin rule in ΔADB and ΔACD
65  BD  sin 30
BD 
sin 30 sin AD̂B
65 sin AD̂B
sin 30
and DC  104  DC 
sin 30 sin AD̂B 104 sin AD̂C
substituting correctly
Alei - Desert Academy
A1
A13
(M1)
A1
A1
3
A1
AG
1
M1
A1
2
(M1)
A1
A1
A1
4
R1
(M1)
A1
M1
since sin AD̂B = sin AD̂C
BD  DC  BD  65
65 104 DC 104
BD  5

DC 8
A1
AG
5
[18]
20.
(a)
(b)
1
A  r 2
2
1
27  (1.5) r 2
2
2
r  36
r  6 cm
Arc length  r  1.5 6
Arc length = 9 cm
Note: Penalize a total of (1 mark) for missing units.
21.
(a)
when y  0 (may be implied by a sketch)
x
8π
or 2.79
9
(M1)(A1)
(A1)
(A1)
(M1)
(A1)
(C4)
(C2)
[6]
(A1)
(A1)
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(C2)
Page 10 of 34
IB Math – SL: Trig Practice Problems: MarkScheme
(b)
Alei - Desert Academy
METHOD 1
Sketch of appropriate graph(s)
Indicating correct points
(M1)
(A1)
(A1)(A1) (C2)(C2)
x  3.32 or x  5.41
METHOD 2
π
1

sin  x    
9
2

π 11π
π 7π
, x 

9 6
9
6
7π π
11π π
x
 , x

6 9
6 9
19π
31π
x
, x
( x  3.32, x  5.41)
18
18
x
22.
(a)

for using cosine rule a  b  c  2ab cos C
2
2
2
(A1)(A1)
(A1)(A1) (C2)(C2)
[6]

(M1)
BC2  152  172  2 15 17  cos 29
(A1)
BC  8.24 m
(A1)
(N0)
Notes: Either the first or the second line may be implied, but not both.
Award no marks if 8.24 is obtained by assuming a right (angled) triangle
(BC = 17 sin 29).
3
(i)
C
A
29°
17
85°
B
ACB  180  (29  85)  66
for using sine rule (may be implied)
(ii)
(c)
(M1)
AC
17

sin85 sin 66
17sin85
AC 
sin 66
AC  (18.5380 )  18.5 m
1
Area  17 18.538... sin 29
2
 76.4 m2 (Accept 76.2 m2 )
(A1)
(A1)
(N2)
(A1)
(A1)
AĈB from previous triangle  66
ˆ  180  66  114 (or 29  85)
Therefore alternative ACB
ˆ  180  (29  114)  37
ABC
(N1)
5
(A1)
C
114°
A
29°
17
37°
B
AC
17

sin 37 sin114
AC  (11.19906 )  11.2 m
(M1)(A1)
(A1)
(N1)
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Page 11 of 34
IB Math – SL: Trig Practice Problems: MarkScheme
Alei - Desert Academy
(d)
A
C
29°
17
B
Minimum length for BC when AĈB = 90°or diagram
showing right triangle
CB
sin 29 
17
CB  17sin 29
CB  (8.2417 )  8.24 m
(M1)
(A1)
(N1)
2
[14]
23.
(a)
1
f ( x)   2cos 2 x  sin x
2
 cos2 x  sin x
(i)
(A1)(A1)
(N2)
Note: Award (A1)(A1) for 2sin x  sin x  1 only if work shown, using
product rule on sin x cos x  cos x .
2
2sin x  sin x 1  (2sin x 1)(sin x  1) or
2
(ii)
2(sin x  0.5) (sin x  1)
2sin x  1 or sin x  1
1
sin x 
2
π
5π
3π
x   (0.524) x 
 (2.62) x 
 (4.71)
6
6
2
(iii)
(A1)
(A1)(A1)(A1)
(N1)
(N1)
(N1)(N1) 6
(b)
x
(c)
(i)
π
  0.524 
6
(A1)
(N1) 1
EITHER
curve crosses axis when x 
π
Area   π2 f ( x)dx 
6

5π
6
π
2
π
(may be implied)
2
f ( x)dx
(A1)
(M1)(A1) (N3)
OR
5
6

6

Area =
(ii)
f ( x) dx
Area  0.875  0.875
 1.75
(M1)(A2) (N3)
(M1)
(A1)
(N2) 5
[12]
24.
1
Using area of a triangle =
ab sin C
2
1
20 (10)(8)sin Q
2
(M1)
(A1)(A1)(A1)
Note: Accept any letter for Q
sin Q = 0.5
PQ̂R = 30 or
(A1)

or 0.524
6
(A1)
(C6)
(A1)
(C1)
[6]
25.
(a)
b=6
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Page 12 of 34
IB Math – SL: Trig Practice Problems: MarkScheme
(b)
y
B
1
(c)
Alei - Desert Academy
2
x
(accept (1.05, −0.896) ) (correct answer only, no additional
solutions)
x = 1.05
(A3)
(C3)
(A2)
(C2)
[6]
26.
(a)
(b)
(c)
2
3(1  2 sin x) + sin x = 1
2
6 sin x  sin x  2 = 0 (p = 6, q = 1, r = 2)
(3 sin x  2)(2 sin x + 1)
4 solutions
(A1)
(A1)
(C2)
(A1)(A1) (C2)
(A2)
(C2)
[6]
27.
Area of large sector
1 2
1 2
r θ = 16 × 1.5
2
2
(M1)
= 192
(A1)
1 2
1
2
Area of small sector r θ =
× 10 × 1.5
2
2
(M1)
= 75
Shaded area = large area – small area = 192 – 75
= 117
(A1)
(M1)
(A1)
(C6)
[6]
28.
(a)
y
2
1.5
1
0.5
0
0.5
1
1.5
2
2.5
3.5 x
3
–0.5
–1
–1.5
–2
(b)
(A1)(A1) (C2)
Note: Award (A1) for the graph crossing the y-axis between 0.5 and 1, and
(A1) for an approximate sine curve crossing the x-axis twice. Do not
penalize for x >3.14.
π 1
 
 4 2
(Maximum) x = 0.285… 
x = 0.3 (1 dp)
 3π 1 
(Minimum) x = 1.856… 
 
 4 2
x = 1.9 (1 dp)
(A1)
(A1)
(C2)
(A1)
(A1)
(C2)
[6]
29.
Area of a triangle =
1
× 3 × 4 sin A
2
(A1)
1
× 3 × 4 sin A = 4.5
2
(A1)
sin A = 0.75
(A1)
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Page 13 of 34
IB Math – SL: Trig Practice Problems: MarkScheme
A = 48.6° and A = 131° (or 0.848, 2.29 radians)
Note: Award (C4) for 48.6° only, (C5) for 131° only.
Alei - Desert Academy
(A1)(A2) (C6)
[6]
30.
METHOD 1
2
2 cos x = 2 sin x cos x
2
2 cos x – 2 sin x cos x = 0
2 cos x(cos x – sin x) = 0
cos x = 0, (cos x – sin x) = 0
x=
(M1)
(M1)
(A1)(A1)
π
π
,x=
2
4
(A1)(A1) (C6)
METHOD 2
Graphical solutions
EITHER
2
for both graphs y = 2 cos x, y = sin 2 x,
OR
2
for the graph of y = 2 cos x – sin 2 x.
THEN
Points representing the solutions clearly indicated
1.57, 0.785
x=
π
π
,x=
2
4
(M2)
(M2)
(A1)
(A1)
(A1)(A1) (C6)
Notes: If no working shown, award (C4) for one correct answer.
Award (C2)(C2) for each correct decimal answer 1.57, 0.785.
Award (C2)(C2) for each correct degree answer 90°, 45°. Penalize a total
of [1 mark] for any additional answers.
[6]
31.
(a)
(b)
(i)
(ii)
(i)
10 + 4 sin 1 = 13.4
(A1)
At 2100, t = 21
(A1)
10 + 4 sin 10.5 = 6.48
(A1)
Note: Award (A0)(A1) if candidates use t = 2100 leading to y = 12.6. No
other ft allowed.
14 metres
(A1)
(ii)
14 = 10 + 4 sin    sin   = 1
(i)
4
(ii)
10 + 4 sin   = 7
t
2
t
2
 t = π (3.14) (correct answer only)
(c)
(iii)
t
2
t
 sin   = –0.75
2
 t = 7.98
depth < 7 from 8 –11 = 3 hours
from 2030 – 2330 = 3 hours
therefore, total = 6 hours
(N2) 3
(M1)
(A1)
(A1)
(N2) 3
(M1)
(A1)
(A1)
(M1)
(M1)
(A1)
(N3)
(N3) 7
[13]
32.
(a)
(b)
Angle A  80
AB
5

sin 40 sin80
AB  3.26 cm
1
1
Area  ac sin B  (5)(3.26)sin 60
2
2
 7.07 (accept 7.06) cm 2
(A1)
(M1)
(A1)
(C3)
(M1)(A1)
(A1)
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(C3)
Page 14 of 34
IB Math – SL: Trig Practice Problems: MarkScheme
Alei - Desert Academy
Note: Penalize once in this question for absence of units.
[6]
33.
METHOD 1
1 2
(5) (0.8)
2
 10
ON  5cos0.8   3.483...
Area sector OAB 
AN  5sin 0.8
(M1)
(A1)
(A1)
  3.586.....
(A1)
1
ON  AN
2
 6.249... (cm2 )
 10  6.249..
 3.75 (cm2 )
Area of  AON 
Shaded area
(A1)
(A1)
(C6)
METHOD 2
A
O
N
B
F
Area sector ABF 
1 2
(5) (1.6)
2
 20
(M1)
(A1)
1 2
(5) sin1.6
2
 12.5
Twice the shaded area  20 12.5 ( 7.5)
1
Shaded area  (7.5)
2
 3.75 (cm2 )
Area OAF 
34.
(a)
(i)
(ii)
(M1)
(A1)
(M1)
(A1)
(C6)
[6]
f ( x)  6sin 2 x
(A1)(A1)
EITHER
f ( x)  12sin x cos x  0
 sin x  0 or cos x  0
(M1)
OR
sin 2 x  0 ,
for 0  2 x  2
(M1)
THEN
π
x  0, , π
2
(b)
(i)
(ii)
translation
in the y-direction of –1
1.11
(1.10 from TRACE is subject to AP)
(A1)(A1)(A1)
(N4)
6
(A1)
(A1)
(A2)
4
[10]
35.
3 = p + q cos 0
3=p+q
–1 = p + q cos 
–1 = p – q
(a)
p=1
(b)
q=2
(M1)
(A1)
(M1)
(A1)
(A1)
(A1)
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(C3)
(C3)
Page 15 of 34
IB Math – SL: Trig Practice Problems: MarkScheme
Alei - Desert Academy
[6]
36.
Method 1
y
1.80
0
2.51
x
0
1.80 [3 sf]
2.51 [3 sf]
Method 2
3x = ±0.5x + 2 (etc.)
 3.5x = 0, 2, 4 or 2.5x = 0, 2, 4
7x = 0, 4, (8) or 5x = 0, 4, (8)
(C2)
(G2)
(G2)
(C2)
(C2)
(M1)
(A1)
(A1)
4π
4π
or x = 0,
7
5
4π 4π
x = 0,
,
7 5
x = 0,
(A1)(A1)(A1)
(C2)(C2)(C2)
[6]
37.
(a)
1
2
area of sector ΑΒDC =
π(2) = π
4
(A1)
area of segment BDCP = π – area of ABC
=π–2
(b)
(M1)
(A1)
2
BP =
(C3)
(A1)
area of semicircle of radius BP =
1
2
π( 2 ) = π
2
(A1)
area of shaded region = π – (π – 2) = 2
(A1)
(C3)
[6]
38.
(a)
OR  PQ
=q–p
10   7 
  –  
 1   3
 3 

= 
 – 2
=
(b)
cos OP̂Q 
PO 
(A1)(A1)
(A1)
PO  PQ
(A1)
PO  PQ
– 72  – 32 =
58 ,
PO  PQ = –21 + 6 = –15
– 15
– 15

cos OP̂Q 
58 13
754
(c)
(i)
Since
3
OP̂Q + PQ̂R = 180°
PQ  32  – 2 = 13
2
(A1)(A1)
(A1)
(AG)
4
(R1)
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Page 16 of 34
IB Math – SL: Trig Practice Problems: MarkScheme
cos
(ii)
Alei - Desert Academy
15 

PQ̂R = –cos OP̂Q  

754 

 15 
sin PQ̂R = 1 – 

 754 
(AG)
2
(M1)
529
754
23
754
=
=
(A1)
(AG)
OR
15
754
cos  =
754
15
x
P
(M1)
2
therefore x = 754 – 225 = 529  x = 23
 sin  =
23
754
(A1)
(AG)
Note: Award (A1)(A0) for the following solution.
15
  = 56.89°
754
cos  =
 sin  = 0.8376
23
= 0.8376  sin  =
754
(iii)
23
754
Area of OPQR = 2 (area of triangle PQR)
1
=2×
PQ  QR  sin PQ̂R
2
1
23
13 58
=2×
2
754
= 23 sq units.
OR
Area of OPQR = 2 (area of triangle OPQ)
=2
1
 7  1 – 3  10
2
= 23 sq units.
Notes: Other valid methods can be used.
Award final (A1) for the integer answer.
(M1)
(A1)
(A1)
(A1)
(M1)
(A1)(A1)
(A1)
7
[14]
39.
(a)
PR
9
Sine rule

sin35 sin 120
9 sin 35
PR =
sin 120
= 5.96 km
(b)
(M1)(A1)
(A1)
3
EITHER
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Page 17 of 34
IB Math – SL: Trig Practice Problems: MarkScheme
Alei - Desert Academy
Sine rule to find PQ
PQ =
9 sin 25
sin 120
= 4.39 km
(M1)(A1)
(A1)
OR
2
2
2
Cosine rule: PQ = 5.96 + 9 – (2)(5.96)(9) cos 25
= 19.29
PQ = 4.39 km
4.39
Time for Tom =
8
5.96
Time for Alan =
a
4.39 5.96
Then
=
8
a
(c)
a = 10.9
2
2
RS = 4QS
2
2
4QS = QS + 81 – 18 × QS × cos 35
2
2
3QS + 14.74QS – 81 = 0 (or 3x + 14.74x – 81 = 0)
 QS = –8.20 or QS = 3.29
therefore QS = 3.29
OR
QS
2QS

sinSR̂Q sin35
1
 sin SR̂Q  sin 35
2
SR̂Q = 16.7°
Therefore,
(M1)(A1)
(A1)
(A1)
(A1)
(M1)
(A1)
7
(A1)
(M1)(A1)
(A1)
(G1)
(A1)
(M1)
(A1)
(A1)
QŜR = 180 – (35 + 16.7)
= 128.3°
9
QS  SR 



sin 128.3 sin16.7  sin 35 
9 sin 16.7 
9 sin 35 
QS =


sin 128.3  2 sin 128.3 
= 3.29
(A1)
(M1)
(A1)
6
[16]
40.
(a)
(i)
(ii)
1
 π 1
 π
, sin  –   –
–  
2
2
 4
 4
 π
 π
therefore cos  –   sin  –  = 0
 4
 4
cos
cos x + sin x = 0  1 + tan x = 0
 tan x = –l
3π
x=
4
(A1)
(AG)
(M1)
(A1)
Note: Award (A0) for 2.36.
OR
x=
3π
4
(G2)
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3
Page 18 of 34
IB Math – SL: Trig Practice Problems: MarkScheme
(b)
Alei - Desert Academy
x
y = e (cos x + sin x)
dy
x
x
= e (cos x + sin x) + e (–sin x + cos x)
dx
(M1)(A1)(A1)
3
x
= 2e cos x
(c)
dy
x
= 0 for a turning point  2e cos x = 0
dx
 cos x = 0
(M1)
(A1)
π
π
x=
a=
2
2
π
π
π
π
2
y = e (cos
+ sin ) = e 2
2
2
(A1)
π
b=e2
(A1)
4
Note: Award (M1)(A1)(A0)(A0) for a = 1.57, b = 4.81.
(d)
d2 y
=0
dx 2
At D,
(M1)
x
x
2e cos x – 2e sin x = 0
x
2e (cos x – sin x) = 0
 cos x – sin x = 0
x=
π
4
π
(e)
(A1)
(A1)
 y = e 4 (cos
=
(A1)
2e
π
π
+ sin )
4
4
π
4
Required area =
(A1)
(AG)

3
4
0
e x (cos x + sin x)dx
5
(M1)
= 7.46 sq units
(G1)
OR
Αrea = 7.46 sq units
(G2)
2
Note: Award (M1)(G0) for the answer 9.81 obtained if the calculator is in
degree mode.
[17]
41.
(a)
(b)
4 
, 0
3 
(i)
A is 
(ii)
B is (0, –4)
(A1)(A1) (C2)
Note: In each of parts (i) and (ii), award C1 if A and B are interchanged,
C1 if intercepts given instead of coordinates.
1
4
×4×
2
3
8
= (= 2.67)
3
Area =
(A1)(A1) (C2)
(M1)
(A1)
(C2)
[6]
42.
(a)
(b)
(3 sin x – 2)(sin x – 3)
(i)
(A1)(A1) (C2)
2
Note: Award A1 if 3x – 11x + 6 correctly factorized to give
(3x – 2)(x – 3) (or equivalent with another letter).
(3 sin x – 2)(sin x – 3) = 0
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Page 19 of 34
IB Math – SL: Trig Practice Problems: MarkScheme
sin x =
2
3
sin x = 3
Alei - Desert Academy
(A1)(A1) (C2)
(ii)
x = 41.8°, 138°
(A1)(A1) (C2)
Notes: Penalize [1 mark] for any extra answers and [1 mark] for answers in radians. ie Award A1 A0 for
41.8°, 138° and any extra answers. Award A1 A0 for 0.730, 2.41. Award A0 A0 for 0.730, 2.41 and any extra
answers.
[6]
43.
Note: Do not penalize missing units in this question.
(a)
2
AB
12 × cos 75°
2
= 12 (2 – 2 cos 75°)
2
= 12 × 2(1 cos 75°)
2
2
= 12 + 12 – 2 × 12 ×
(A1)
AB = 12 2(1  cos 75)
(AG)
(A1)
2
Note: The second (A1) is for transforming the initial expression to any
simplified expression from which the given result can be clearly seen.
(b)
(c)
PÔB = 37.5°
BP = 12 tan 37.5°
= 9.21 cm
OR
(A1)
(M1)
(A1)
BP̂A = 105°
BÂP = 37.5°
AB
BP

sin 105 sin 37.5
AB sin 37.5
BP =
= 9.21(cm)
sin 105
(A1)
(i)
Area ∆OBP =
1
 1

 12  9.21  or  12  12 tan 37.5 
2
2


2
2
= 55.3 (cm ) (accept 55.2 cm )
(ii)
1
2
Area ∆ABP = (9.21) sin105°
2
(M1)
(A1)
(M1)
(A1)
(M1)
2
2
= 41.0 (cm ) (accept 40.9 cm )
(A1)
1
π  75

 122  75 
 π  122 
 or
2
180  360

(M1)
(d)
Area of sector =
(e)
2
2
= 94.2 (cm ) (accept 30π or 94.3 (cm ))
Shaded area = 2 × area ∆OPB – area sector
2
2
2
= 16.4 (cm ) (accept 16.2 cm , 16.3 cm )
3
4
(A1)
(M1)
2
(A1)
2
[13]
44.
(a)
(i)
(ii)
(b)
(i)
Note: Do not penalize missing units in this question.
At release(P), t = 0
s = 48 + 10 cos 0
= 58 cm below ceiling
58 = 48 +10 cos 2πt
cos 2πt = 1
t = 1sec
OR
t = 1sec
ds
= –20π sin 2πt
dt
(M1)
(A1)
(M1)
(A1)
(A1)
(G3)
5
(A1)(A1)
Note: Award (A1) for –20π, and (A1) for sin 2t.
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Page 20 of 34
IB Math – SL: Trig Practice Problems: MarkScheme
(ii)
v=
Alei - Desert Academy
ds
= –20π sin 2πt = 0
dt
(M1)
sin 2 πt = 0
t = 0,
(A1)
s = 48 + 10 cos 0 or s = 48 +10 cos π
(M1)
= 58 cm (at P)
= 38 cm (20 cm above P)
(A1)(A1) 7
Note: Accept these answers without working for full marks. May be
deduced from recognizing that amplitude is 10.
48 +10 cos 2πt = 60 + 15 cos 4πt
(M1)
t = 0.162 secs (A1)
OR
t = 0.162 secs
(G2)
2
12 times
(G2)
2
Note: If either of the correct answers to parts (c) and (d) are missing and suitable graphs have been
sketched, award (G2) for sketch of suitable graph(s); (A1) for t = 0.162; (A1) for 12.
(c)
(d)
45.
1
... (at least 2 values)
2
[16]
l = r
(a)
(b)
or ACB = 2 × OA
= 30 cm
(M1)
(A1)
AÔB (obtuse) = 2 – 2
1
2 1
2
Area =  r = (2 – 2)(15)
2
2
(A1)
(C2)
(M1)(A1)
2
= 482 cm (3 sf)
(A1)
(C4)
[6]
46.
B
d
A
50
80
70°
P
(M1)(A2)
OR
2.5 × 20 = 50
2.5 × 32 = 80
2
2
2
d = 50 + 80 – 2 × 50 × 80 × cos 70°
d = 78.5 km
(M1)(A1)
(A1)
(M1)(A1)
(A1)
(C6)
[6]
47.
(a)
(i)
(ii)
–1
4 (accept 720°)
(b)
(A1)
(A2)
(C1)
(C2)
(G1)
(A2)
(C3)
y

32
number of solutions: 4
[6]
48.
Statement
A
B
C
(a) Is the statement true for all
real numbers x? (Yes/No)
No
No
Yes
(b) If not true, example
x = –l (log10 0.1 = –1)
x = 0 (cos 0 = 1)
N/A
(a) (A3) (C3)
(b) (A3) (C3)
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Page 21 of 34
IB Math – SL: Trig Practice Problems: MarkScheme
Alei - Desert Academy
Notes: (a) Award (A1) for each correct answer. (b) Award (A) marks for statements A and B only if NO in column
(a). Award (A2) for a correct counter example to statement A, (A1) for a correct counter example to statement B
(ignore other incorrect examples). Special Case for statement C: Award (A1) if candidates write NO, and give a
valid reason (eg arctan 1 =
5π
).
4
[6]
49.
(a)
7 2
6
 2
sin A sin 45
2
2

sin A = 6 
2 7 2
6
=
7
(M1)
(A1)
(AG)
2
(b)
A
D
B
(i)
(ii)
(A1)
=> A = 59.0° or 121° (3 sf)
(A1)(A1)
BĈD = 180° – (121° + 45°)
= 14.0° (3 sf)
7 2
BD
 2
sin 14 sin 45
=>BD = 1.69
(c)
C
BD̂C + BÂC = 180°
6
sin A =
7
=>
(iii)
h
1
 BD  h
Area BDC 2

Area BAC 1
 BA  h
2
BD
=
BA
(A1)
(M1)
(A1)
6
(M1)(A1)
(AG)
2
OR
1
BD  6 sin 45
Area ΔBCD 2

Area ΔBAC 1
BA  6 sin 45
2
BD
=
BA
50.
Using sine rule:
 sin B =
sin B sin 48

5
7
5
sin 48° = 0.5308…
7
 B = arcsin (0.5308) = 32.06°
(M1)(A1)
(AG)
2
[10]
(M1)(A1)
(M1)
(M1)(A1)
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IB Math – SL: Trig Practice Problems: MarkScheme
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= 32° (nearest degree)
(A1)
Note: Award a maximum of [5 marks] if candidates give the answer in
radians (0.560).
(C6)
[6]
51.
(a)
x is an acute angle => cos x is positive.
2
2
2
cos x + sin x = 1 => cos x = 1 – sin x
=> cos x =
=
(b)
1
1–  
 3
(M1)
(M1)
2
(A1)
8
2 2
(=
)
9
3
1
2
cos 2x = 1 – 2 sin x = 1 – 2  
(A1)
2
(M1)
3
=
(C4)
7
9
(A1)
(C2)
Notes: (a) Award (M1)(M0)(A1)(A0) for
cos  sin –1  1   = 0.943.


 3 

(b) Award (M1)(A0) for cos  2 sin –1  1   = 0.778.


 3 

[6]
52.
(a)
2
2
2
2 sin x = 2(1 – cos x) = 2 – 2 cos x = l + cos x
2
=> 2 cos x + cos x – l = 0
(M1)
(A1)
(C2)
(A1)
(C1)
2
Note: Award the first (M1) for replacing sin x by
2
1 – cos x.
(b)
2
2 cos x + cos x – 1 = (2 cos x – 1)(cos x +1)
(c)
1
cos x =
or cos x = –l
2
=> x = 60°, 180° or 300°
(A1)(A1)(A1)
(C3)
Note: Award (A1)(A1)(A0) if the correct answers are given in radians (ie

5
, ,
, or 1.05, 3.14, 5.24)
3
3
[6]
53.
(a)
The smallest angle is opposite the smallest side.
82  7 2  52
28 7
88 11

=
= 0.7857
112 14
(b)
cos θ =
(M1)
Therefore, θ = 38.2°
(A1)
Area =
1
× 8 × 7 × sin 38.2°
2
2
= 17.3 cm
(C2)
(M1)
(A1)
(C2)
[4]
54.
(a)
(b)
2
2
3 sin x + 4 cos x = 3(1 – cos x) + 4cos x
2
= 3 – 3 cos + 4 cos x
2
2
3 sin x + 4 cos x – 4 = 0 3 – 3 cos x + 4 cos x – 4 = 0
(A1)
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(C1)
Page 23 of 34
IB Math – SL: Trig Practice Problems: MarkScheme
2
 3 cos x – 4 cos x + 1 = 0
(3 cos x – 1)(cos x – 1) = 0
cos x =
Alei - Desert Academy
(A1)
1
or cos x = 1
3
x = 70.5° or x = 0°
(A1)(A1) (C3)
Note: Award (C1) for each correct radian answer, ie x = 1.23 or x = 0.
[4]
55.
OT̂A = 90°
(A1)
12  6
2
AT =
2
= 6 3
TÔA = 60° =
π
3
(A1)
Area = area of triangle – area of sector
=
1
1
π
×6× 6 3 –
×6×6×
2
2
3
(M1)
2
= 12.3 cm (or 18 3 – 6)
(A1)
(C4)
OR
TÔA = 60°
(A1)
1
× 6 × 12 × sin 60
2
1
π
Area of sector =
×6×6×
2
3
Area of  =
(A1)
(A1)
2
Shaded area = 18 3 – 6 = 12.3 cm (3 sf)
(A1)
(C4)
[4]
56.
(a)
(b)
(i)
AP =
( x  8)  (10  6)  x  16x  80
(M1) (AG)
(ii)
OP =
( x  0) 2  (10  0) 2  x 2  100
(A1)
=
2
2
2 x 2  16x  80
2 x 2  16x  80 x 2  100
x 2  8 x  40
{( x 2  16x  80)(x 2  100)}
For x = 8, cos OP̂A = 0.780869
arccos 0.780869 = 38.7° (3 sf)
OR
tan OP̂A 
8
10
OP̂A = arctan (0.8) = 38.7° (3 sf)
OP̂A = 60°  cos OP̂A = 0.5
x 2  8 x  40
0.5 =
2
2
2 x 2  16x  80 x 2  100
cos OP̂A 
(d)
2
AP  OP  OA
2AP  OP
2
( x  16x  80)  ( x 2  100)  (8 2  6 2 )
2
cos OP̂A 
=
(c)
2
{( x 2  16x  80)( x 2  100)}
(M1)
(M1)
(M1)
(AG)
3
(M1)
(A1)
(M1)
(A1)
2
(M1)
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IB Math – SL: Trig Practice Problems: MarkScheme
2
2x – 16x + 80 –
{( x 2  16x  80)(x 2  100)} = 0
x = 5.63
(e)
(i)
(ii)
Alei - Desert Academy
(M1)
(G2)
f (x) = 1 when cos OP̂A = 1
(R1)
hence, when OP̂A = 0.
This occurs when the points O, A, P are collinear.
(R1)
(R1)
The line (OA) has equation y =
When y = 10, x =
3x
4
40
(= 13 13 )
3
4
(M1)
(A1)
OR
x=
40
(= 13 13 )
3
(G2)
5
Note: Award (G1) for 13.3.
[16]
57.
(a)
(b)
1 2
1
2
Area = r   (15 )(2)
2
2
(M1)
2
= 225 (cm )
(A1)
Area ∆OAB =
1 2
15 sin 2 = 102.3
2
2
Area = 225 – 102.3 = 122.7 (cm )
= 123 (3 sf)
(C2)
(A1)
(A1)
(C2)
[4]
58.
(a)
(b)
sin (AĈB) sin 50

20
17
20sin 50
 sin (AĈB) 
= 0.901
17
AĈB > 90°  AĈB = 180° – 64.3° = 115.7°
AĈB = 116 (3 sf)
In Triangle 1, AĈB = 64.3°
 BÂC = 180° – (64.3° + 50°)
(M1)
(A1)
= 65.7°
(A1)
1
2
Area = (20)(17) sin 65.7° = 155 (cm ) (3 sf)
2
(A1)
(C2)
(C2)
[4]
59.
METHOD 1
The value of cosine varies between –1 and +1. Therefore:
t = 0  a + b = 14.3
t = 6  a – b = 10.3
 2a = 24.6  a = 12.3
 2b = 4.0  b = 2
Period = 12 hours 
2π(12)
= 2π
k
 k = 12
(A1)
(A1)
(C1)
(C1)
(M1)
(A1)
(C2)
METHOD 2
y
14.3
10.3
6
12
18
24
t (h)
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Page 25 of 34
IB Math – SL: Trig Practice Problems: MarkScheme
From consideration of graph:
Midpoint = a = 12.3
Amplitude = b = 2
2π
= 12
2π
k
Period =
 k = 12
Alei - Desert Academy
(A1)
(A1)
(C1)
(C1)
(M1)
(A1)
(C2)
[4]
60.
C
32km
A
B
2
48km
cos CÂB 
48  32  562
2(48)(32)
2
(M1)(A1)
CÂB = arccos(0.0625)
(A1)
(A1)
 86°
[4]
61.
(a)
(b)
2
2
2 cos x + sin x = 2(1 – sin x) + sin x
2
= 2 – 2 sin x + sin x
2
2 cos x + sin x = 2
2
 2 – 2 sin x + sin x = 2
2
sin x – 2 sin x = 0
sin x(1 – 2 sin x) = 0
sin x = 0 or sin x =
1
2
(A1)
(M1)
sin x = 0  x = 0 or  (0° or 180°)
Note: Award (A1) for both answers.
(A1)
1
5π
π
x=
or
(30° or 150°)
2
6
6
(A1)
sin x =
Note: Award (A1) for both answers.
[4]
62.
(a)
y
4
x<1
{0.5<
3.5<y<4
(A1)
MAXIMUM
POINT
3
2
{
integers (A1)
on axis
1
–1
(b)
(c)
(d)
5
1
2
3
4
LEFT
(A1)
3.5<x<4 (A1)
INTERCEPT 3<x<3.5
3.2<x<3.6 MINIMUM
(A1)
–0.2<y <0 POINT
x
RIGHT
INTERCEPT
{
 is a solution if and only if  +  cos  = 0.
Now  +  cos  =  + (–1)
=0
By using appropriate calculator functions x = 3.696 722 9...
 x = 3.69672 (6sf)
See graph:
5
(M1)
(A1)
(A1)
(M1)
(A1)
(A1)
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3
2
Page 26 of 34
IB Math – SL: Trig Practice Problems: MarkScheme
Alei - Desert Academy
π
 (π  x cos x)dx
(A1)
2
(A3)
3
0
π
(e)
EITHER
 (π  x cos x)dx = 7.86960 (6 sf)
0
Note: This answer assumes appropriate use of a calculator eg
 fnInt(Y1 , X , 0, π)  7.869604401
with Y1  π  x cos x
‘fnInt’: 
π
OR
 (π  x cos x)dx  [πx  x sin x  cos x]
0
π
0
= ( – 0) + ( sin  – 0 × sin 0) + (cos  – cos 0)
2
=  + 0 + –2 = 7.86960 (6 sf)
(A1)
(A1)
3
[15]
63.
(a)
(i)
Q=
1
(14.6 – 8.2)
2
(M1)
= 3.2
(ii)
P=
(A1)
1
(14.6 + 8.2)
2
(M0)
= 11.4
(b)
(c)
(A1)
π 
10 = 11.4 + 3.2 cos  t 
6 
7
π 
so
= cos  t 
16
6 
7 π
therefore arccos 
 t
 16  6
π
which gives 2.0236... = t or t = 3.8648. t = 3.86(3 sf)
6
(i)
(ii)
By symmetry, next time is 12 – 3.86... = 8.135... t = 8.14 (3 sf)
From above, first interval is 3.86 < t < 8.14
This will happen again, 12 hours later, so
15.9 < t < 20.1
3
(M1)
(A1)
(A1)
3
(A1)
(A1)
(M1)
(A1)
4
[10]
64.
3 cos x = 5 sin x

sin x 3

cos x 5
(M1)
 tan x = 0.6
x = 31° or x = 211° (to the nearest degree)
Note: Deduct [1 mark] if there are more than two answers.
(A1)
(A1)(A1) (C2)(C2)
[4]
65.
5
12
sin A =
 cos A = 
13
13
But A is obtuse  cos A = –
(A1)
12
13
(A1)
sin 2A = 2 sin A cos A
(M1)
5  12 
=2×
  
13  13 
120
=–
169
66.
(a)
(A1)
y =  sin x – x
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(C4)
[4]
Page 27 of 34
IB Math – SL: Trig Practice Problems: MarkScheme
Alei - Desert Academy
y
3
2
(1.25, 1.73)
1
(2.3, 0)
–3
(–2.3, 0)
–2
–1
1
2
3
x
–1
(–1.25, –1.73)
–2
–3
(b)
(c)
(A5)
5
Notes: Award (A1) for appropriate scales marked on the axes.
Award (A1) for the x-intercepts at (2.3, 0).
Award (A1) for the maximum and minimum points at (1.25, 1.73).
Award (A1) for the end points at (3, 2.55).
Award (A1) for a smooth curve.
Allow some flexibility, especially in the middle three marks here.
(A1)
x = 2.31

x2
( π sin x  x)dx   π cos x 
C
2
1
(A1)(A1)
Note: Do not penalize for the absence of C.
1
Required area =
 (π sin x  x)dx
(M1)
0
= 0.944
area = 0.944
OR
(G1)
(G2)
4
[10]
67.
(a)
30º
Acute angle 30°
(M1)
Note: Award the (M1) for 30° and/or quadrant diagram/graph seen.
2nd quadrant since sine positive and cosine negative
  = 150°
(A1)
(b)
tan 150° = –tan 30° or tan 150° =
1
2
(C2)
(M1)
3

2
tan 150° = –
1
(A1)
(C2)
3
[4]
68.
(a)
PQ
= tan 36°
40
 PQ  29.1 m (3 sf)
(A1)
(C1)
(b)
B
40m
Q
30
70
A
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IB Math – SL: Trig Practice Problems: MarkScheme
AQ̂B = 80°
AB
40

sin 80 sin 70
Alei - Desert Academy
(A1)
(M1)
Note: Award (M1) for correctly substituting.
 AB = 41 9. m (3 sf)
(A1)
(C3)
[4]
69.
Perimeter = 5(2π – 1) + 10
(M1)(A1)(A1)
Note: Award (M1) for working in radians; (A1) for 2π – 1; (A1) for +10.
= (10π + 5) cm (= 36.4, to 3 sf)
(A1)
(C4)
[4]
70.


From sketch of graph y = 4 sin  3x 
or by observing sin   1.
k > 4, k < –4
π

2
(M2)
(A1)(A1) (C2)(C2)
4
3
2
1
–2
–
0

0
2
–1
–2
–3
–4
[4]
71.
(a)(i) & (c)(i)
y
1
(1.1, 0.55)
R
(1.51, 0)
0
1
2
x
–1
(2, –1.66)
–2
(A3)
Notes: The sketch does not need to be on graph paper. It should have the correct shape, and the points (0, 0), (1.1,
0.55), (1.57, 0) and (2, –1.66) should be indicated in some way.
Award (A1) for the correct shape.
Award (A2) for 3 or 4 correctly indicated points, (A1) for 1 or 2 points.
(ii)
Approximate positions are
positive x-intercept (1.57, 0)
(A1)
maximum point (1.1, 0.55)
(A1)
end points (0, 0) and (2, –1.66)
(A1)(A1) 7
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IB Math – SL: Trig Practice Problems: MarkScheme
(b)
2
x cos x = 0
x ≠ 0 ⇒ cos x = 0
(M1)
π
x=
2
(c)

(ii)
(d)
(A1)
Note: Award (A2) if answer correct.
see graph
(i)

2
0
Alei - Desert Academy
x 2 cos x dx
2
(A1)
(A2)
3
Note: Award (A1) for limits, (A1) for rest of integral correct (do not
penalize missing dx).
Integral = 0.467
(G3)
OR

Integral = x sin x  2 x cos x  2 sin x
π
2

π/2
0

π
(1)  2 (0)  2(1) – [0 + 0 – 0]
2
 4

π
=
– 2 (exact) or 0.467 (3 sf)
2
(M1)
2
= 
(M1)
(A1)
3
(A1)
(A1)
1
1
[15]
72.
(a)
(b)
From graph, period = 2π
Range = {y –0.4 < y < 0.4}
(c)
(i)
f (x) =
(ii)
2
3
= cos x (2 sin x cos x) – sin x (sin x) or –3 sin x + 2 sin x
(M1)(A1)(A1)
Note: Award (M1) for using the product rule and (A1) for each part.
f (x) = 0
(M1)
2
 sin x{2 cos x – sin x} = 0 or sin x{3 cos x – 1} = 0
(A1)
2
 3 cos x – 1 = 0
d
2
{cos x (sin x) }
dx
1
 3
 cos x = ±  
(A1)
At A, f (x) > 0, hence cos x =

 1  
f (x) =  1 –
 3  

2 1
2
= 

3
3 9
(iii)
(d)
x=
(e)
(i)
(ii)
(f)
 1  
 
 3  
2
1
 
 3





3
π
2
1
dx  sin 3 x  c
3
3


π/2
1
π

(cosx)(sin x) 2 dx   sin   (sin 0) 3 
Area =
0
3
2



1
=
3
 (cos x)(sin x)
2

At C f (x) = 0
3
 9 cos x – 7 cos x = 0
(R1)(AG)
(M1)
(A1)
9
(A1)
1
(M1)(A1)
(M1)
(A1)
4
(M1)
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IB Math – SL: Trig Practice Problems: MarkScheme
Alei - Desert Academy
2
 cos x(9 cos x – 7) = 0
(M1)
π
7
x=
(reject) or x = arccos
= 0.491 (3 sf)
2
3
(A1)(A1) 4
[20]
73.
42  52  7 2
2 45
cos  =
=–
Note: Award (M1) for identifying the largest angle.
(M1)
1
5
(A1)
  = 101.5°
OR Find other angles first
 = 44.4°
 = 34.0°
  = 101.6°
Note: Award (C3) if not given to the correct accuracy.
74.
(A1)
(M1)
(A1)(A1) (C4)
[4]
AB = r
1 2
2
r 
2
r
2
= 21.6 ×
5 .4
=
(M1)(A1)
(A1)
= 8 cm
(A1)
1
2
× (5.4)  = 21.6
2
4
=
(= 1.481 radians)
2 .7
OR
(M1)
AB = r
(A1)
4
= 5.4 ×
2 .7
(M1)
= 8 cm
(A1)
(C4)
[4]
75.
(a)
OA = 6

A is on the circle
(A1)
OB = 6

B is on the circle.
(A1)

C is on the circle.
(A1)
 5 

OC  
 11 
= 25  11
=6
(b)
AC  OC  OA
 5   6
   
= 
 11   0 
 1 

 11 
(c)
3
(M1)
= 
(A1)
AO  AC
cos OAˆ C 
AO AC
(M1)
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2
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IB Math – SL: Trig Practice Problems: MarkScheme
  6   1 
 .

0   11 

=
6 1  11
6
=
Alei - Desert Academy
(A1)
6 12
=
1

2 3
3
6
(A1)
ˆC 
OR cos OA
1
=
6 2  ( 12 ) 2  6 2
2  6  12
as before
(M1)(A1)
(A1)
12
OR using the triangle formed by AC and its horizontal and
vertical components:
(d)
AC  12
(A1)
1
cos OAˆ C 
12
(M1)(A1) 3
Note: The answer is 0.289 to 3 sf
A number of possible methods here
BC  OC  OB
 5    6
   
= 
 11   0 
(A1)
 11 

 11 
= 
(A1)
BC  =
132
1
ABC =  132  12
2
= 6 11
OR ABC has base AB = 12
(A1)
(A1)
11
(A1)
1
 12  11
2
(A1)
and height =
 area =
(A1)
= 6 11
ˆC 
OR Given cos BA
(A1)
3
6
33
1
33
sin BAˆ C 
 ABC   12  12 
6
2
6
= 6 11
(A1)(A1)(A1)
(A1)
4
[12]
76.
1
2
tan x =
(M1)
3
 tan x = 
1
(M1)
3
 x = 30° or x = 150°
(A1)(A1) (C2)(C2)
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IB Math – SL: Trig Practice Problems: MarkScheme
Alei - Desert Academy
[4]
77.
2
2
h = r so 2r = 100  r = 50
l = 10 = 2r
(M1)
(M1)
2π 50
10
2 π5 2
=
10
 =  2 = 4.44 (3sf)
=
(A1)
(A1)
(C4)
Note: Accept either answer.
[4]
78.
(a)
(b)
f (1) = 3
EITHER
f (5) = 3
distance between successive maxima = period
=5–1
=4
OR
Period of sin kx =
so period =
2π
;
k
(M1)
2π
π
2
(A1)
=4
(c)
(A1)(A1) 2
(M1)
(A1)
(AG)
(AG)
π
 3π 
 + B = 3 and A sin   + B = –1
2
 2 
EITHER A sin 
(M1) (M1)
 A + B = 3, – A + B = –1
 A = 2, B = 1 (AG)(A1)
OR Amplitude = A
(A1)(A1)
(M1)
3  (1) 4

A=
2
2
(M1)
A=2
Midpoint value = B
B=
(AG)
(M1)
3  (1) 2

2
2
(M1)
B=1
(A1)
Note: As the values of A = 2 and B = 1 are likely to be quite obvious to a
bright student, do not insist on too detailed a proof.
(d)
2
π 
x + 1
2 
π
π 
f (x) =  2 cos  x  + 0
2
2 
5
f (x) = 2 sin 
(M1)(A2)
π
π
 , (A1) for 2 cos 
2
 
2
Note: Award (M1) for the chain rule, (A1) for 
π
2
=  cos 

x


x .

(A1)
4
Notes: Since the result is given, make sure that reasoning is valid. In
particular, the final (A1) is for simplifying the result of the chain rule
calculation. If the preceding steps are not valid, this final mark should not
be given. Beware of “fudged” results.
(e)
(i)
π 
x
2 
y = k – x is a tangent  – =  cos 
(M1)
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IB Math – SL: Trig Practice Problems: MarkScheme
π 
x
2 
 –1 = cos 

(ii)
(f)
π 
x + 1 = 2
2 
π  1
 sin  x  
2  2
π
π 5π 13π
 x  or
or
2
6
6
6
1 5 13
x = or or
3 3
3
(A1)
π
x =  or 3 or ...
2
 x = 2 or 6 ...
Since 0  x  5, we take x = 2, so the point is (2, 1)
Tangent line is: y = –(x – 2) + 1
y = (2 + 1) – x
k = 2 + 1
f (x) = 2  2 sin 
Alei - Desert Academy
(A1)
(A1)
(M1)
(A1)
6
(A1)
(A1)
(A1)(A1)(A1)
5
[24]
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