IB Math HL Y2
Mr. Jauk
1.
(a)
Review Trig IB Questions MS
Either finding depths graphically, using sin
t
  1 or solving
6
h(t) = 0 for t
(M1)
h (t)max = 12 (m), h (t)min = 4 (m)
(b)
Attempting to solve 8 + 4 sin
t
 8 algebraically or graphically
6
t [0, 6]  [12, 18]  {24}
A1A1
N3
(M1)
A1A1
N3
[6]
Not as well done as expected with most successful candidates using a
graphical approach. Some candidates confused t and h and subsequently
stated the values of t for which the water depth was either at a maximum and
a minimum. Some candidates simply gave the maximum and minimum
coordinates without stating the maximum and minimum depths.
In part (b), a large number of candidates left out t = 24 from their final
answer. A number of candidates experienced difficulties solving the
inequality via algebraic means. A number of candidates specified incorrect
intervals or only one correct interval.
2.
cos θ 
5 2  7 2  6 2 25  49  36 38

  θ  0.997
2  5 7
70
70
 2 = 1.99...
M1
A1
IB Math HL Y2
Mr. Jauk
cos  
Review Trig IB Questions MS
7 2  6 2  5 2 49  36  25 60

  0.775
2 7 6
84
84
 2 = 1.55...
A1
Required area =
1 2
1
5 1.99  sin 1.99   6 2 1.55  sin 1.55
2
2
M1A1
= 23.4 cm2
A1
N0
[6]
On the whole candidates found this question difficult, with relatively few
gaining the correct answer. A significant number did not actually start the
question and those that did often made assumptions about the triangles PRQ
and PQS being right-angled.
(a)
cos (A + B) = cos A cos B – sin A sin B,
cos (A – B) = cos A cos B + sin A sin B
(M1)(A1)
Hence cos (A + B) + cos (A – B) = 2cos A cos B
(AG)
3.
(b)
(c)
(i)
T1 (x) = cos (arccos x)
=x
(M1)
(A1)
(ii)
T2 (x) = cos (2 arccos x)
= 2 cos (arccos x) – 1
= 2x2 – 1
(A1)
(A2)
(AG)
(i)
Tn+1(x) + Tn–1(x) = cos [(n + 1)arccos x] + cos [(n – 1)arccos x]
Using part (a) with A = n arccos x, B = arccos x
Tn+1(x) + Tn–1(x) = 2 cos (n arccos x) cos (arccos x)
= 2x cos (arccos x)
= 2xTn(x)
(A1)
(M1)
(A1)
(A1)
(AG)
(ii)
Let Pn be the statement: Tn (x) is a polynomial of degree, n 
T1(x) = x, a polynomial of degree one.
So P1 is true.
T2(x) = 2x2 – 1, is a polynomial of degree two.
So P2 is true.
Assume that Pk is true.
From part (c)(i), Tk+1(x) = 2xTk(x) – Tk–1 (x)
Assume Pk–1 is true as well.
Tk(x) has degree k
 2xTk (x) has degree (k + 1)
and as Tk+1(x) has degree (k – 1)
 Tk+1(x) has degree (k + 1)
+
(A1)
(A1)
(M1)
(M1)
(M1)
(A1)
(A1)
2
5
IB Math HL Y2
Mr. Jauk
Review Trig IB Questions MS
By the principle of mathematical induction, Pn is true for
all positive integers n.
Notes: These arguments may be in a different order.
There is a maximum of 6 marks in part (ii) for
candidates who do not consider a two stage process.
(R1)
12
[19]
Most candidates solved part (a) correctly and many were successful in part (b). In
part (b), those who ‘proved’ that cos(2 arccos x) = 2x2 – 1 by plotting the graphs of
both functions and showing that they were coincident gained no credit. In part (c),
few candidates realised that in order to demonstrate the truth of the proposition for n
= k + 1 it was necessary to assume true = for both n = k and k = –1 and also to show
it true for n = 1 and n = 2.
4.
PR = h tan 55, QR = h tan 50 where RS = hM1A1A1
Use the cosine rule in triangle PQR.
(M1)
2
2
2
2
2
20 = h tan 55 + h tan 50  2h tan 55 h tan 50 cos 45
A1
400
(A1)
h2 
tan 2 55  tan 2 50  2 tan 55 tan 50 cos 45
= 379.9...
(A1)
h = 19.5 (m)
A1
[8]
Correct answers to this question were rarely seen. Although many candidates
expressed RS correctly in two different ways, they failed to go on to use the
cosine rule.
5.
C
104 m
A2
A
30°
D
x
30°
A1
65 m
B
IB Math HL Y2
Mr. Jauk
(a)
(b)
(c)
Review Trig IB Questions MS
Using the cosine rule (a2 = b2 + c2 – 2bc cos A)
Substituting correctly
BC2 = 652 + 1042 – 2(65)(104) cos 60°
= 4225 + 10816 – 6760 = 8281
 BC = 91 m
Finding the area using = 1 bc sin A
2
Substituting correctly, area = 1 (65)(104) sin 60°
2
= 1690 3 (Accept p = 1690)
(i)
Smaller area A1 =  1  (65)(x) sin 30°
2
65x
=
4
Larger area A2 =  1  (104)(x) sin 30°
2
= 26x
(ii)
Using A1 + A2 = A
A1
A1
A1
A1
N2 3
(M1)A1
AG
N0
M1
A1
N1
(M1)
65x + 26x = 1690 3
4
A1
Simplifying
169x = 1690 3
4
A1
Solving x 
4  1690 3
169
using sin rule in ΔADB and ΔACD
65
 BD  sin 30
Substituting correctly BD 
sin 30 sin AD̂B
65 sin AD̂B
DC  104  DC  sin 30
and
sin 30 sin AD̂C 104 sin AD̂C
Since AD̂B + AD̂C = 180°
It follows that sin AD̂B = sin AD̂C
N2 3
(M1)
Substituting
 x = 40 3 (Accept q = 40)
(d)
(M1)
A1
N1 8
(M1)
A1
A1
R1
R1
BD  DC  BD  65
65 104
DC 104
A1
 BD  5
DC 8
AG
N0 6
[20]
IB Math HL Y2
Mr. Jauk
Review Trig IB Questions MS
6.
METHOD 1
b
a–b
1
a
1
a – b= 12  12  2(1)(1) cos 
=
(M1)
2(1  cos  )
(A1)
1
4 sin 2 
2
1
= 2sin .
2
=
(A1) (C3)
METHOD 2
O
–12
a
b
M
A
a–b
B
1
.
2
1
Therefore, a – b= 2 sin .
2
In OAM, AM = OA sin
AB=a–b
(M1)(A1)
(A1) (C3)
[3]
This question turned out to be one of the most difficult for the candidates. Most were
unable to even start the question. Those who did often went into incredibly
complicated and inconclusive procedures.