Mathematics · May / June 2023 · Paper 31 · Question 9
YMA01
As printed in the paper
The question as text
9. 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
9(a)[3]
sin
x
cos
x
sin
x
cos
x
2
1 2sin
−
x
2sin cos
x
x
=
+
sin
x
cos x
2
1
2sin
x
1
=
−sin
+
2sin
x
=
=
cosec
x
*
sin x
sin
x
x
2
2
9(b)[5]
E.g. Equation is
cosec
θ
=
6cot
θ
−
4
1 cot
+
θ
=
6cot
θ
−
4
2
cot
6cot
5
0
E.g.
θ
−
θ
+
=
1
E.g.
tan =
θ
,1
5
π
=0.197, 4
θ
π
4
π
cosec cot
x
x x
d
= −
cosec
x
4
9(c)[2]
π
π
6
6
=
2
−
2
Notes
(a) There are lots of ways of proving this statement. In general score as follows
B1
For applying at least one CORRECT double or compound angle identity during the proof or for
forming a CORRECT single fraction initially.
M1
For a correct overall strategy, e.g. applying double angle identities to reduce terms to single angle
arguments and cancelling down terms to eliminate cos x terms (score at the stage cos x terms could
be eliminated), or attempting a single fraction and applying relevant identities to achieve single
angle argument with common factor cos x in the numerator. Allow slips in signs, such as
cos2
x
= ±
1 2si
n
2
x
for the M but otherwise identities used must be correct.
A1*
Fully correct proof showing all necessary steps, though the left hand side may be implied (and
1
may follow initial lines of aside working). Must see the
x →
c
o
s
ec
x
during the proof . Do not
si
n
penalise minor notational slips such as missing an x in one term.
(b)
M1
Correctly applies the result of (a) and attempts to use relevant identities, allowing sign errors e.g.
2
2
±±
1 cot
θ
=
cosec
θ to produce an equation in cotθ or other single trig term only. An
alternative is
1
cos
θ
(
)
2
(
)
=
6
−
4
=
1
6sin
θ
cos
θ
−
4
sin
2
θ
1
+
4sin
2
θ
=
36s
in
2
θ
1 sin
−
2
θ
sin
2
θ
sin
θ
2
2
A1
Correct quadratic
cot
6cot
5
0
or
5tan
6tan
1
0
. In the alternative, a correct
θ
−
θ
+
=
θ
−
θ
+ =
quadratic in sin2θ or cos2θ e.g.
52sin θ
4
−
28sin
2
θ
+ =
1
0
. The “=0” may be implied by an
attempt to solve. May be implied by correct solutions following an unsimplified quadratic.
dM1 Attempts to solve quadratic to find at least one value for their trig term used. Usual rules, may use
calculator.
A1
One correct value for θ following from a correct value for the trig term they are working in －
must have solved a correct quadratic in the dM. Accept awrt 0.197 or 0.785. Degrees answer are
A0A0.
π only (must be
A1
Both values correct and no other values in the range. Accept awrt 0.197 and 4
exact but isw after correct value seen).
Note: Allow if a different variable used (such as x). For mixed variables allow the M’s but only allow
the first A (and final A’s) if recovered.
Note Answers without working score no marks.
(c)
M1
For using part (a) and achieving
±
k
cosec
x
oe for the integral (limits not required for this mark.
May arise from longer methods, but must achieve the correct form.
A1
2
−
2
Must have scored the M answer only with no integration shown is M0A0.
cos2
x
sin 2
x
cos2 cos
x
x
+
sin 2 sin
x
x
(a)
+
=
Correct single fraction
ALT I
sin
x
cos
x
sin cos
x
x
Single fraction with single
2
cos
x
1 2sin
−
x
+
2sin cos sin
x
x
x
arguments and common factor
=
cos x in numerator
sin cos
x
x
cos x
1
= sin cos
=
=
cosec
x
*
sin
x
x
x
cos2
x
sin 2
x
cos2 cos
x
x
+
sin 2 sin
x
x
(a)
+
Correct single fraction
ALT II
sin
x
cos
x
sin cos
x
x
(
)
Applies identity to reach single fraction
cos 2
x
−
x
with single arguments and common factor
sin cos
x
x
cos x in numerator
cos x
1
cosec
x
sin cos
*
x
x
sin
x
Note cos(
x
−
2 )
x
is equally correct for the M1.
cos2
x
sin 2
x
cos
2
x
−
sin
2
x
2sin cos
x
x
(a)
+
+
Correct identity
ALT III
sin
x
cos
x
sin
x
cos
x
Single fraction with single
3
2
2
cos
x
−
sin
x
cos
x
+
2sin
x
cos
x
arguments and common factor
cos x in numerator
sin cos
x
x
(cos
2
x
+
sin
2
x
)cos
x
1
cosec
x
*
sin cos
x
x
sin
x
(
)
(
)
4
y
3
y
−
3
−
3 2
y
2
+
6
2
y
2
6
d
x
+
x
=
=
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