Mathematics · Oct / Nov 2025 · 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.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
5(i)[5]
4tan
θ
+
θ
=
5sin
0
θ
θ
sin
sin
States or uses
tan
θ
=
→
4
+
5sin
θ
=
0
θ
θ
cos
cos
(
)
sin
θ
4 5cos
+
θ
=
0
4
θ
= −
cos
5
4
°
cos
θ
=
−
θ
= awrt143 or awrt 217
°
5
4
cos
θ
=
−
θ
= awrt143 1 and awrt 216 9
.
°
.
°
5
sin
θ
=
0
θ
=
180 360
°
,
°
sin
x
cos
x
5
×
sin cos
x
x
5(ii)[4]
+
=
sin
2
x
+
cos
2
x
=
5sin
x
cos
x
sin
x
cos
x
5sin
x =
1
x =
awrt 0.201, 2.94
(i) Condone a lack of a degrees symbol throughout. Condone solutions that drop θ’s
in their working
sin
θ
sin
θ
M1: States
or attempts to use
in the given equation.
tan
θ
=
tan
θ
=
θ
θ
cos
cos
Condone slips in coefficients but writing 4tanθas 4sin
θ
θis M0.
4cos
dM1: Attempts all of the following
θ
sin
usestan
θ
=
condoning slips in the coefficients
θ
cos
multiplies by cosθand proceeds to a linear equation in cosθ via factorising orcancelling out the term in sinθ
achieves cosθ
where
k <
= k
1
4
A1: Achieves
cos
θ
=
−
and finds one correct value for θ to the nearest degree or in
5
4
radians to one dp. So, allow for example, awrt 2.5(0) following
cos
θ
=
−
5
4
A1: Achieves
θ
−
and finds both θ values in degrees correct to one dp.
cos
=
5
awrt 143 1 and awrt 216 9
.
°
.
°
5
B1: Correctly achieves sin
for their equation followed by
θ
=
°
°
θ
=
0
180 360
,
Condone 0 appearing as a solution
………………………………………………………………………………………………………………….
Alternative approaches exist and can be marked in a similar way:
If you see an approach that deserves credit and you don’t know how to score it,
please send to review.
Alt I, II and III are examples of alternative approaches.
Alt I Using sin
θ
=
tan cos
θ
θ
(
)
4tan
θ
+
5sin
θ
=
0
4tan
θ
+
5tan cos
θ
θ
=
0
tan
θ
4
+
5cos
θ
=
0
4
then solve
θ
−
and tan
cos
=
θ
=
0
5
M1:
Writingsinθ as cos tan
θ
θ
dM1:
Factorises out tanθand proceeds to cos
θ
= k
where
k <
1
Alt II Squaring approach
θ
+
θ
θ
=
θ
= −
θ
θ
2
θ
=
2
θ
2
θ
4sin
5sin cos
0
4sin
5sin cos
16sin
25sin
cos
3
This will produce
θ
±
and sin
sin
=
θ
=
0
5
M1:
As main method
dM1:
Squaring each term including the coefficients, factorising and cancelling
leading to sin
θ
where
k < Note that an approach
= k
1
a
+
b
=
0
a
2
+
would score
dM0. It must be
a
+
b
=
0
a
2
−
b
2
=
0
B1: Allow this to be scored even from an approach where dM1has not been awarded
Alt III using Pythagoras’ theorem
)
(
2
1
−
cos
θ
2
(
)
sin
θ
sin
θ
2
2
3
+
θ
=
+
θ
=
+
−
θ
=
θ
+
4
5sin
0
4
5sin
0
4
5 1
cos
0
5cos
4cos
θ
θ
θ
cos
cos
cos
Then solve by any means to find values for cosθ
4
cos
θ
= ±
1,
−
θ
=
180 ,360 ,awrt 143.1 and awrt 216.9
°
°
°
°
5
M1:
As main method. dM1: Correct method of producing a 4 term cubic in cosθ
followed by a valid attempt to solve to produce value(s) for cosθ
(ii) Condone solutions that drop x’s in parts of the working and notational errors e.g.
2
2
as long as the candidate proceeds in a manner equivalent to the
sin
x
sin
x
M1:
Attempts to multiply by sin cos
x
x , seen on the two terms on the lhs of the
equation and then uses
2
+
2
= . Condone slips when rearranging this
sin
x
cos
x
1
identity as long as the equation leads to sin
x
=
k k
,
<o.e
1
Alternatively uses a common denominator approach and simplifies sin
x
+
cos
x
1
x
sin cos
x
A1:
Proceeds to 5sin
x = or exact equivalent
1
dM1:
Correct attempt to solve 5sin
x = , implied by one correct solution to 2 sf in
1
radians or 1 dp in degrees within the given range. 0 to So
1
5sin
x
=
1
sin
x
=
x
=
0.2
only is dM0 but x = 0.20 or 2.9 is dM1.
5
Note that the degree answer is 11.5.
Also, solutions such as 0.064π (radians) are acceptable
Allow a miscopy on the 5. So, for example, if they had written sin
x
cos
x
+
=
cos
x
sin
x
would be for a solution of 3sin
x = .
1
A1:
x =
awrt 0.201, 2.94
and no extras within the given domain (condoning the
πwhich occurs from a solution of cos
appearance of
x =
. Award as soon as you
0)
2
see the two correct solutions, so long as there are no additional solutions inside the
range.
Also allow multiples of πso accept
x =
0.064 , 0.936
π
πwith the decimal aspect being
accurate to at least 3dp
Centre of circle is midpoint of (
)
and (
14, 6 = (
)
)
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