Mathematics · May / June 2024 · Paper 31 · Question 7
YMA01
As printed in the paper
The question as text
7. 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
7(a)[4]
x
x
(
)
2 sin cos45
x
°+
cos sin 45
x
°=
cos cos60
x
°+
sin sin60
x
°
1
3
sin
x
+
cos
x
=
cos
x
+
sin
x
2
2
(
)
cos
=
3
−
2 sin
x
x
x
1
3
+
2
tan
=
=
= −−
2
3
*
3
2
−
1
−
States or uses
x +
45
°= θ o.e.
2
7(b)[4]
Proceeds from e.g.
(
)
tan 2
θ
−
45
°= −−
2
3
2
θ
−
45
°=
105 ,285
°
°
Correct order of operations to find one angle
θ
=
75 ,165
°
°
(a)
Condone working in mixed variables
Attempts to use the compound angle identities to produce an equation in sin x and cos x
M1:
(
)
Look for
2
×
sin cos45
°±
cos sin 45
°=
cos cos60
°±
sin sin60
° o.e.
x
x
x
x
1
3
May be implied by a first line of e.g.
sin
±
cos
=
cos
±
sin
o.e.
x
x
x
x
2
2
Condone missing brackets.
A1: Correct equation in sin x and cos x.
(
)
E.g.
2 sin cos45
°+
cos sin 45
°=
cos cos60
°+
sin sin60
°o.e. (may be implied by further
x
x
x
x
cos
(
60
)
cos60
work). May also see use of equivalent angles e.g.
−
°=
°
M1:
For an attempt to solve the problem. Look for an attempt (condoning slips) at two of the three elements
required to complete the proof, namely
use of correct exact trigonometric values for sin 45 ,cos45 ,cos60 ,sin60°
°
°
°
collection of like terms in sinx
and cos
x or tan x
sin
x
use oftan
=
x
cos
x
A1*: Proceeds to the given answer with all previous marks scored.
There should be no errors in the manipulation and no bracket omissions, other than a missing trailing
bracket.
e.g.
2(sin cos45
°+
cos sin 45
°=
cos cos60
°+
sin sin60
°
x
x
x
x
When proceeding from
sin
=
cos
(where A or B may be 1) to the given answer it is acceptable
A
x
B
x
to proceed from 1
−+
2
3
cos
=
sin
to the given answer as evidence of the use of
x
x
2
2
Condone poor notation to be recovered provided the final answer line is written correctly. Condone
working in mixed variables provided the final given answer is all in terms of x.
Allow surd work to be done via a calculator.
x
+
45
°
B1:
States or uses
x +
45
°= θ o.e. e.g.
2
θ
=
2
This is implied for sight of the equation
tan 2
(
45
)
2
3
θ−
°= −−
(
)
M1:
Proceeds from
tan 2
θ
±
α
°= −−
2
3
2
θ
±
α
°=
105 or 285
°
°
where
α≠
0
The attempt must either achieve an angle of 105 or 285
°
°or equivalent expression (Allow in radians
2
75
(
180n
(3sf) which are 1.83, 4.97), or allow a general solution of e.g.
θ
±
α
°= −
°+
Maybe implied by further work which is not 75 or 165
°
°
dM1: Correct order of operations to solve their 2
θ
±
α
°=
...
This is dependent on the previous method mark.
(
)
Note that
tan 2
θ−
45
°= −−
2
3
75
° does not imply this mark. We must see either e.g.
2
θ−
45
°=
105
°θ=
...
or some intermediate stage before seeing 75°
A1:
Both angles
with no others given within the range
θ
=
75 ,165
°
°
(
)
Note that
tan 2
θ
−
45
°= −−
2
3
2
θ
−
45
°=
105
°
θ=
75 ,165
°
° is acceptable for full marks
Alt (b) via use of cos(2
θ
−
105 )
°=
cos2 cos105
θ
°+
sin 2 sin105
θ
°
cos105
°
2 sin 2
θ=
cos(2
θ−
105 )
°
tan 2
θ
=
θ
=
75 ,165
°
°
2
sin105
−
°
Note the order of the marks needs to match up to the main scheme so 0110 is possible.
cos105
°
3
B1:
For achieving
o.e. so allow
tan 2
θ= −
tan 2
θ
=
=
awrt
−
0.58
3
2
−
sin105
°
3 tan
2
6tan
3
0
Or via double angle identities
θ−
θ−
=
o.e.
M1:
Attempts to use the compound angle identities (allowing sign slips when using them) to reach a form
tan 2
θ=
k
where k is a constant not 2
−−
3
(or expression in trigonometric terms such as cos105°
π
as seen above). Allow 2
θ= −
30
° o.e. (allow in radians
−
) to imply this mark. Do not be concerned
6
by the mechanics of their rearrangement.
Alternatively, via double angle identities reaches a 3TQ in tanθ
dM1: Correct order of operations from tan 2
θ=
proceeding to
θ=
...
e.g. 2
θ
= −
30
°
k
tan
−
1
k
must be in degrees) can score this mark. You may need to check this if
θ
=
2
Alternatively, correctly solves their
3 tan
2
θ−
6tan
θ−
3
=
0
proceeding to
...
θ=
A1:
Both angles
θ
=
75 ,165
°
°
with no others given within the range
0.02
x
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