Mathematics · Oct / Nov 2023 · Paper 31 · Question 3
YMA01
As printed in the paper
Show the answer and mark scheme
3(a)[2]
(
)
cos2
A
cos
A
cos
A
−
sin
A
sin
A
cos
2
A
−
(1 cos
−
2
A
)
cos2A
2cos
2
A
− *
1
3(b)[4]
sin 6
x
(
)
3
−
2cos6
x
d
x
=
3
x
−
(
+
c
)
3
π
(
)
(
)
sin 6
x
sin 6
×
sin 6
×
1
2
2
π
π
π
π
−
8
3
x
−
=
3
−
8
−
3
−
12
=
π
+
3
8
3
12
3
8
6
π
12
Notes
(a)
M1:
Writes (
cos2
)
cos
cos
sin
sin
and attempts to use (not just stated separately) the
A
A
A
−
A
A
identity
±
sin
2
A
±
cos
2
A
= ± Note going directly to
1
cos2
A
cos
2
A
−
sin
2
A
is allowed M1 as
long as a use of the Pythagorean identity is clear. NB Watch out for answers which use double
angle identities in the proof (circular reasoning) and score M0 if no Pythagorean identity is used.
A1*: Achieves the given answer with no errors seen (and no circular reasoning). Must see
cos2
A
cos
A
cos
A
−
sin
A
sin
A
and clear attempt at substitution of
sin
2
A
+
cos
2
A
allow “LHS” for “cos 2A”. Going directly to
in the first step would
cos2
A
cos
2
A
−
sin
2
A
score A0. Correct notation must used throughout but condone a missing closing bracket.
Note :
cos2
A
=
cos
A
−
sin
A
=
2cos
A
− with no further evidence scores M0A0 as they are just
1
2
2
2
writing out known formulae.
(b)
M1:
Attempts to substitute in the given result in part (a) and integrates to the form ...
x
±
...sin6
be a full substitution in terms of x,
cos2
A →
k
sin 2
A
is M0 unless recovered, e.g by later
substitution may be implied by the limits substituted).
sin 6
x
A1:
3
x −
3
dM1: Shows evidence of substituting in both limits into an expression of the form ...
x
±
...sin6
subtracts either way round to achieve an exact (not necessarily simplified but trig must be
evaluated) answer.
1
1
2
A1:
Correct answer e.g as scheme or
π+
−
or exact equivalent (and isw after a correct
8
3
6
answer).
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