CE471 MOCK MID SEMESTER
TEST
STUDENT NUMBER: .
Answer all the questions in the spaces provided. Continue on the back of the sheets if
necessary.
A formula book is provided.
Any approved calculator is allowed.
Marks for each question are shown in brackets on the right. The total number of
marks available is 55.
This test is worth 40% of your final mark for this module.
Question Mark
1 10
2 10
3 6
4 6
5 12
6 4
7 7
Total 55
Percentage
, 1. Find all the solutions (in radians) of
4 sin(𝜃) − 3 cos(𝜃) = 2
for 0 ≤ 𝜃 ≤ 2 𝜋. Give your answers rounded correctly to four decimal places.
(10)
Let
𝑅 sin(𝜃 − 𝜙) = 𝑅 sin(𝜃) cos(𝜙) − 𝑅 cos(𝜃) sin(𝜙) = 4 sin(𝜃) − 3 cos(𝜃)
(1 mark)
Hence 𝑅 cos(𝜙) = 4 and 𝑅 sin(𝜙) = 3.
(1 mark)
Squaring and adding gives √42 + 32 = √25 = 5.
(1 mark)
Taking the ratio gives
sin(𝜙) 3
= tan(𝜙) =
cos(𝜙) 4
(1 mark)
𝜙 = tan−1(0.75) = 0.6435
(1 mark)
Therefore we need to solve
5 sin(𝜃 − 0.6435) = 2 ⇒ sin(𝜃 − 0.6435) = 0.4
(1 mark)
For a calculator 𝜃 − 0.6425 = 0.4115 ⇒ 𝜃 = 1.0540.
(1 mark)
From the properties of the sine function we also have
𝜃 − 0.6425 = 𝜋 − 0.4115 ⇒ 𝜃 = 𝜋 − 0.4115 + 0.6425 = 3.3726.
(3 marks)
2 for method
1 for accuracy
TEST
STUDENT NUMBER: .
Answer all the questions in the spaces provided. Continue on the back of the sheets if
necessary.
A formula book is provided.
Any approved calculator is allowed.
Marks for each question are shown in brackets on the right. The total number of
marks available is 55.
This test is worth 40% of your final mark for this module.
Question Mark
1 10
2 10
3 6
4 6
5 12
6 4
7 7
Total 55
Percentage
, 1. Find all the solutions (in radians) of
4 sin(𝜃) − 3 cos(𝜃) = 2
for 0 ≤ 𝜃 ≤ 2 𝜋. Give your answers rounded correctly to four decimal places.
(10)
Let
𝑅 sin(𝜃 − 𝜙) = 𝑅 sin(𝜃) cos(𝜙) − 𝑅 cos(𝜃) sin(𝜙) = 4 sin(𝜃) − 3 cos(𝜃)
(1 mark)
Hence 𝑅 cos(𝜙) = 4 and 𝑅 sin(𝜙) = 3.
(1 mark)
Squaring and adding gives √42 + 32 = √25 = 5.
(1 mark)
Taking the ratio gives
sin(𝜙) 3
= tan(𝜙) =
cos(𝜙) 4
(1 mark)
𝜙 = tan−1(0.75) = 0.6435
(1 mark)
Therefore we need to solve
5 sin(𝜃 − 0.6435) = 2 ⇒ sin(𝜃 − 0.6435) = 0.4
(1 mark)
For a calculator 𝜃 − 0.6425 = 0.4115 ⇒ 𝜃 = 1.0540.
(1 mark)
From the properties of the sine function we also have
𝜃 − 0.6425 = 𝜋 − 0.4115 ⇒ 𝜃 = 𝜋 − 0.4115 + 0.6425 = 3.3726.
(3 marks)
2 for method
1 for accuracy
