Mathematics · May / June 2022 · Paper 21 · Question 5
YMA01
As printed in the paper
The question as text
5. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. Solve, for −180° < θ 180° , the equation 3 tan (θ + 43°) = 2 cos (θ + 43°) (6)
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
(
)
(
)
(
)
3
tan
θ
+
43
°=
2
cos
θ
+
43
°
3
=
2c
os
θ
+
4
3
°
(
)
c s
o
θ
+
4
3
°
(
)
(
)
3sin θ
+
43
°=
2cos
2
θ
+
43
°
(
)
(
)
(
)
3sin θ
+
43
°=
2 1 sin
−
2
θ
+
43
°
2sin
2
(
θ
+
43
°+
)
3sin
(
θ
+
43
°−
)
2
=
0
(
)
(
)
2sin
(
θ
+
43
°−
)
1
sin
(
θ
+
43
°+
)
2
=
0
sin
(
θ
+
43
°=
)
...
1
(
)
sin
θ+
43
°=
2
"
"
1
°
θ=
arcsin
−
43
2
θ= −
13 , 107
°
°
Notes:
sin..
2 ...
M1: Uses
and multiplies through to form an equation of the form
A
sin...
B
cos
tan..
=
=
cos..
Condone poor notation e.g.:
s
i
n
(
)
(
)
(
)
(
)
3
ta
n
θ
+
43
°=
2cos
θ
+
43
°
3
θ
+
43
°
=
2
co
s
θ
+
43
°
cos
(
)
(
)
3sin θ
+
43
°=
2cos
2
θ
+
43
°
(with or without brackets)
M1: Applies Pythagorean identity to obtain a 3 term quadratic equation in sin.
Allow use of
cos ...
2
= ±±
1 sin ...
2
M1: Solves a 3 term quadratic in
sin
(
43
° by any valid means.
)
θ+
This may be implied by at least one correct root for their quadratic.
(
)
sin
θ+
43
°=
x
43
Allow if they have
or another variable or e.g. sin α where
α
=
θ
+
°
(
)
(
)
A1: Correct value of
sin
θ+
43
°. If
sin
θ+
43
°=
x
is used, it must be clear they mean
(
)
sin
θ+
43
°but this may be implied if they have e.g. sin α = ½ where
α
=
θ
+
43
°
1
If
is left as a final answer it is A0.
x =
2
(
)
M1: Correct method for solving
sin
θ+
43
°=
k k
,|
| 1
<, look for use of inverse sine followed by
1 their )
−
subtraction of 43 from
sin
(
k
. Implied by one correct solution for their k
Do not allow mixing of degrees and radians for this mark.
A1: Correct solutions and no others in the range.
Make a free account to save your answers and see your weak topics and ask the AI why a wrong answer is wrong.