Mathematics · Oct / Nov 2023 · Paper 31 · Question 8
YMA01
As printed in the paper
Show the answer and mark scheme
8(a)[4]
2
−
2cos2
θ
Starting with the LHS:
2cosec 2 (1 cos2 )
2
θ
−
θ
=
sin 2
2
θ
2
−
2(1 2sin
−
2
θ
)
=
4sin
2
θ
cos
2
θ
=
sec
2
θ
= +
1 tan
2
θ
RHS
*
8(b)[4]
sec
2
x
−
3sec
x
−
4
=
0
sec
x
=
...
1
cos
x =
(ignore
−)
1
4
1
cos
x
=
x
=
...
4
x =
75.5 , 284.5
°
°
Notes
(a) The most common method and where to award marks is as follows:
1
M1:
Uses
cosec2
θ
=
(oe) at some stage in the proof to convert the cosec into sine.
sin 2
θ
M1
Attempts to use one of
cos2
θ
= ±±
1 2sin
2
θ
or sin 2
θ
=
2sin
θ
cos
θ
but condone
as an attempt at squaring.
sin 2
2
θ
=
2sin
2
θ
cos
2
θ
dM1: Depends on previous M. Attempts to use both
cos2
θ
= −
1 2sin
2
θ
oe and sin 2
θ
=
2sin
A1*:
Achieves the RHS with no mathematical errors seen and all stages of working shown. Must see
the
sec θ before the final answer. Condone minor notational errors (e.g. a missing θ) but not
2
consistent poor notation throughout. There will be other solutions but if they work from both
sides then they need a conclusion at the end.
For other methods or variations apply scheme as follows:
M1: Uses a correct identity for either cosec 2θ or cosec22θ or cot22θ to get the equation in terms of
sin and cosine only or to introduce cosec2θ if working in reverse. Condone the 2 becoming ½.
M1: Attempts to apply a double angle identity correct up to sign errors, e.g.
cos2
θ
= ±±
1 2sin
dM1: Depends on previous M. Use of at least two correct double angle identities and/or Pythagorean
identities to progress sufficiently towards the goal (e.g. reduces all terms to single argument
form in cosine or sine only, or identify an appropriate factor to cancel to such, or to get all
double arguments if working in reverse with the sin 2θ identified). There may be algebraic slips,
but all trig identities used up to this point must be correct (e.g. see stag reach in main scheme).
A1*: Fully correct proof with sufficient stages shown including use of
s
ec
2
θ
= +
1
t
an
2
θ
or
equivalent identity (e.g. other Pythagorean relation), and if necessary, a conclusion given.
1
2
For example
1
+
tan
2
θ
=
sec
2
θ
=
= +
2nd M1, use of a suitable double angle identity
co
s
2
θ
1 cos2
θ
(
)
(
)
(
)
2 1 cos2
−
θ
2 1 cos2
−
θ
2 1 cos2
−
θ
=
=
=
3rd dM1, correct initial double angles
(
)(
)
1 cos2
+
θ
1 cos2
−
θ
1 cos 2
−
2
θ
sin 2
2
θ
identity and Pythagorean identity used to identify the sin in the denominator, all identities correct
(
)
=
2
cosec
2
2θ
1 cos
−
2
θ
1st M1, use of correct identity for cosec22θ. (A1 if all correct)
(b)
Note: allow with x or θ and allow recovery of mixed variables in this part.
M1:
Uses part (a) to forms a 3 term quadratic in sec x or cos x, solves their quadratic and finds cos x
=… or sec x =… for at least one solution. For reference the quadratic in cos is
4cos
2
x +
3cos
x
−=
1
0
1
(ignore any reference to cos
x =−) Must have a correct cosine or implied by a
1
A1:
cos
x =
4
correct answer after reaching a secant if no incorrect working seen. May be implied if not seen
explicitly.
dM1: Correct method to find a value for x from cos x = k where |k| < 1 as a solution for their equation.
May be implied by one correct angle from sec x
=
k
or cos x
=
k
. Note radian answer awrt 1.3
implies dM1.
A1:
awrt 75.5° and awrt 284.5° Allow with or without 180° included, but there must be no other
angles in given interval. All previous marks must have been scored. Ignore answers outside the
domain.
(b)Alt
1
+
tan
x
=
4
+
3sec
x
9sec
x
=
tan
x
−
6tan
x
+
9
tan
x
−
15tan
x
=
0
2
2
4
2
4
2
tan
2
x
=
...(
≠
0)
tan
x = ±
(
)
1
5
tan
x
=
k ≠
0
x
=
...
x =
75.5 , 284.5
°
°
Notes
(b) Alt
M1: Makes sec x the subject, squares and solves the resulting quadratic in tan2 x leading to a value for
tan2 x. Note there may be variations on this approach, such as
1
+
tan
2
x
=
4
+
3
1
+
tan
used before squaring, or solving for “1+ tan2 x”. Score for a correct approach leading to a value
for tan2 x.
A1: Correct value for tan x. Need not give both values.
dM1: Solves from their tan x = k, k ≠ 0 to find a value for x
A1: Both correct values obtained and no invalid solutions in the range (ignore 180° as main scheme).
Must reject extra solutions.
Note: other methods or variations may be seen, which can be marked according to the same principles,
first M for correct approach to find a value for a trig ratio, A1 correct nontrivial value, dM1
solves for x.
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