Mathematics · Oct / Nov 2020 · Paper 31 · Question 5
YMA01
As printed in the paper
Show the answer and mark scheme
5(a)[4]
sin3
x
≡
sin(2
x
+
x
)
≡
sin 2 cos
x
x
+
cos2 sin
x
x
(
)
2
≡
2sin
x
cos cos
x
x
+
1
−
2sin
x
sin
x
(
) (
)
2
2
≡
2sin
x
1 sin
−
x
+
1
−
2sin
x
sin
x
3
≡
3sin
x
−
4sin
x
*
π
π
∫
∫
3
1
3
3
5(b)[4]
sin
3
x x
d
sin
x
sin3 d
x x
=
−
4
4
0
0
π
3
1
3
= −
cos
x
+
cos3
x
4
12
0
5
=
24
(a)
Uses sin3
x
sin(2
x
x
)
sin 2 cos
x
x
cos2 sin
x
x
M1
=
+
=
±
±
M1
Uses the correct identity for sin 2
x
=
2sin cos
x
x
and any correct identity for cos2x
ddM1 Dependent upon both previous M's. It is for using
cos
2
x
= −
1 sin
2
x
to get an equation in only sin x
A1*
Fully correct solution with correct notation within their proof. Examples of incorrect notation include use of
sin x instead of
2
sin x or use of sin instead of sin x and so on. Penalise in the A mark only for such.
2
sin
2
1 cos
2
Note: The ddM mark and final A mark may be score by substituting
x
= −
x
into the right hand side of the
equation to reach an identical expression to an expanded left hand side (“working from both sides”), with the A
mark then awarded for correct work leading to identical expressions and an minimal conclusion given (e.g. //)
Note
You may see use of De Moivre’s Theorem from an FP3 candidate. This can score full credit if carried out
correctly. If there are errors or you are unsure then send to review.
If attempted in reverse:
1
(
)
3sin
x
−
4sin
3
x
=
3sin
x
−
2sin
2
x
sin
x
−
2sin (1 cos
x
−
2
x
)
=
sin
x
−
2sin
x
1 cos2
−
x
M1
2
1
Uses
sin
3
x
=
sin
x
×
sin
2
x
with either
sin
2
x
=
(1 cos2 )
−
x
OR
s
in
2
x
= −
1
c
os
2
x
and 2sin cos
x
2
M1
Uses both steps above to get to an equation with sin2x and cos2x
ddM1 Gathers terms to reach
=
cos2 sin
x
x
+
sin 2 cos
x
x
Completes the proof uses cos2 sin
sin 2 cos
sin3
A1
x
x
+
x
x
=
x
with no errors seen.
(b)
∫
∫
M1
Attempts to use part (a) to simplify. Accept
sin
3
x x
d
=
A
sin
x
+
B
sin3 d
x x
∫
sin
sin3 d
cos
cos3
dM1
A
x
+
B
x x
→
a
x
+
b
x
3
1
A1
−
cos
x
+
cos3
x
oe (not a multiple of this (unless recovered))
4
12
5
A1
CSO
=
24
π
π
π
5 (b)
∫
∫
(
)
∫
3
3
3
2
2
sin
3
x x
d
=
sin
x
1 cos
−
x
d
x
=
sin
x
−
sin cos
x
x
d
x
Alt 1
0
0
0
π
3
3
cos
x
= −
cos
x
−−
3
0
5
=
24
π
1
(b)
d
u
∫
∫
3
2
2
u
=
cos
x ⇒
=
−
si
n
x
⇒
sin
3
x x
d
=
u
−
1d
u
Alt 2
d
x
0
1
1
3
2
u
=
−
u
3
1
5
=
24
Notes: First three marks as follows (final A is as main scheme).
Alt 1
sin
3
x
=
sin sin
x
2
x
s
in
2
x
= −
1
c
os
2
x
M1:
Splits as
and applies
to get the integrand into and integrable form.
sin cos
x
2
x
K
c s
o
3
x
dM1
for sin
x →±
cos x
and
→
3
cos
x
A1
−
cos
x
−−
oe (not a multiple of this (unless recovered))
3
Alt 2
, finds d
u
M1
Sets
u
=
cos
x
x = ±
si x
n
and makes a full substitution using both of these to get an integral in terms of
d
u only. (Limits not needed for this mark).
dM1
for
a
u
2
−
b
→
A
u
3
−
b
u
1
3
2
u
A1
For reaching
−
u
including correct limits or for undoing the substitution and reaching
3
1
x
1
x
1
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