Mathematics · Oct / Nov 2022 · 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. Given that cos 2θ − sin 3θ ≠ 0

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Show the answer and mark scheme

9(a)[4]
=
(
)
cos2
θ
−
sin3
θ
c s2
o
θ
−
sin 2 c s
θ
o
θ
+
cos2 sin
θ
θ
cos
2
θ
=
(
)
1
−
2sin
2
θ
−
2sin
θ
cos
2
θ
−
sin
θ
1 2
−
sin
2
θ
(
)(
)
1 sin
−
2
θ
1 sin
−
θ
1 sin
+
θ
=
=
(
)(
)
1 3sin
−
θ
−
2sin
2
θ
+
4sin
3
θ
1 si
−
n
θ
1
−
2s
in
θ
−
4sin
2
θ
1 sin
+
θ
*
= −
1 2sin
θ
−
4sin
2
θ
1
+
si
n
θ
(
)
9(b)[5]
2
c
o
s
e
c
(
1 sin
)
sin
2
1
2si
n
4sin
2
=
θ
+
θ
θ
=
−
θ
−
θ
1
2
s n
i
4
sin
2
−
θ
−
θ
9sin
2
θ
+
5sin
θ
−
2
=
0
−±
5
25
−
4
××−
9
2
1 .
sin
θ
=
=
...
θ
=
sin
−
.
.
1
8
Two of
θ=
15.6 ,164.4 ,235.6 ,304.4
°
°
°
°
All of
°
θ=
15.6 ,164.4 ,235.6 ,304.4
°
°
°
Notes:
(a)
M1: Applies the compound angle formula tosin3θ. Accept
±
sin
θ
cos2
θ
±
sin 2 cos
θ
θ
Allow for students
who apply
s
in3
θ
=
3si
n
θ
−
4s n
i
3
θ
directly. May be seen anywhere (e.g. in separate working).
M1: Uses correct double angle formula for cos2θ and sin2θ (or assumed used correctly if correct sin3θ
formula was used) to achieve all terms in single angle. (May be scored after the third M.)
M1: Uses Pythagorean identity to identify the factor of 1 sinθ
−
in numerator. They need not have achieved
single angle arguments for all terms for this mark. (May be scored before the second M.) There must be an
intermediate step before the given answer with 1 sinθ
−
cancelled if the factor is not explicitly seen in the
numerator.
A1: Cancels the 1 sinθ
−
from numerator and denominator and simplifies to the given result with no
incorrect steps or consistently incorrect notation.
(b)
M1: Substitutes the result from (a) and cross multiplies to get an equation in sin θ only.
A1: Correct simplified quadratic in sin θ
M1: Solves the resulting quadratic by formula, completing square or calculator (allow by factorisation if their
quadratic factorises) and applies inverse sine to at least one root. May be implied by any correct arcsin of one
of their solutions, such as −55.6°. My be implied by a radians answer (e.g. 0.272, 2.869…)
A1: Any two correct solutions in the range, accept awrt. Answers in radians scores A0.
A1: Awrt all four solutions correct and no others in the range.
c
os
2
θ
cos
2
θ
9(a)
=
(
)
cos2
θ
−
sin3
θ
c s2
o
θ
−
sin 2 c s
θ
o
θ
+
cos2 sin
θ
θ
Example
1 sin
2
−
θ
=
Alt
(
)
cos2
(
1 sin
)
2sin
1 s
in
2
θ
−
θ
−
θ
−
θ
(
)(
)
1 sin
−
θ
1 sin
+
θ
=
(
)
(
)(
)
1 sin
−
θ
co
s
2
θ
−
2sin
θ
1 sin
+
θ
1
−
si
n
θ
(
)(
)
1 sin
−
θ
1 sin
+
θ
=
)(
)
(
1 s
−
in
θ
1
−
2sin
2
θ
−
2 in
s
θ
−
2s
i
n
2
θ
1 sin
+
θ
= −
*
1 2sin
θ
−
4sin
2
θ
Notes
The above approach can be marked via the main scheme but scores the 3rd M before the 2nd M.
Alternative approaches will hopefully fit the scheme similarly. Other approaches can be similarly applied as
each of the three method marks will generally be required. E.g. see below (allowing first M for students who
apply
s
in3
θ
=
3si
n
θ
−
4s n
i
3
θ
directly)
Note in Alt 2 below it is more likely the 1st M will be gained by aside work to expand sin3θ with substitution
later seen into the equation.
1
+
sin
θ
1 sin
−
θ
1 sin
−
2
θ
9(a)
×
=
1
−
2s
i
n
θ
−
4sin
2
θ
1 sin
−
θ
1 3 in
−
s
θ
−
2
sin
2
θ
+
4sin
3
θ
Alt2
cos
2
θ
=
In reverse
1 3sin
−
θ
−
2sin
2
θ
+
4
sin
3
θ
cos
2
θ
cos
2
θ
=
=
(
)
1 2sin
−
2
θ
−
3sin
θ
+
4sin
3
θ
cos2
3sin
4sin
3
θ
−
θ
−
θ
cos
2
θ
cos
2
θ
=
=
(
)
(
(
)
)
cos2
θ
−
sin
θ
3
−
4sin
2
θ
cos2
θ
−
sin
θ
1
+
2 1 sin
−
2
θ
−
2sin
2
θ
cos
2
θ
=
(
)
(
)
cos2
θ
−
sin
θ
1
+
2cos
2
θ
−
1 cos2
−
θ
cos
2
θ
=
(
)
cos2
θ
−
2sin
θ
cos
2
θ
+
sin
θ
cos2
θ
c
os
2
θ
=
(
)
cos2
θ
−
s n
i 2 cos
θ
θ
+
cos2 sin
θ
θ
cos
2
θ
*
=
co
s2
θ
−
s
i 3
n
θ
9(a)
cos
2
θ
1 sin
+
θ
= −
Alt3
cos2
θ
−
sin3
θ
1
2sin
θ
−
4sin
2
θ
(
)
(
)(
)
cos
2
θ
1
−
2sin
θ
−
4s n
i
2
θ
=
1 sin
+
θ
cos2
θ
−
sin3
θ
By cross
(
)
multiplying
(
)(
)
cos
2
θ
1
−
2s
in
θ
−
4sin
2
θ
=
1
+
s
in
θ
cos2
θ
−
sin 2 cos
θ
θ
−
cos2 s
θ
i
n
θ
(
)
cos
2
θ
1 2si
−
n
θ
−
4sin
2
θ
(
)
(
)
(
1 sin
)
1 2sin
2
2sin
cos
2
1 2sin
2
si
n
=
+
θ
−
θ
−
θ
θ
−
−
θ
θ
co
s
2
θ
−
2si
n
θ
co
2s
θ
−
4sin
2
θ
cos
2
θ
= −
1 2sin
2
θ
−
2sin
θ
cos
2
θ
−
sinθ
+
2sin θ
3
+
sinθ
−
2sin θ
3
−
2sin
2
θ
cos
2
θ
−
sin
2
θ
+
2sin
4
θ
(
)
cos
2
θ
−
2sin
2
θ
cos
2
θ
= −
1 3sin
2
θ
+
2sin
2
θ
1 cos
−
2
θ
cos
2
θ
−
2sin
2
θ
cos
2
θ
= −
1 3sin
2
θ
+
2sin
2
θ
−
2sin
2
θ
cos
2
θ
cos
2
1 sin
2
θ
= −
θ
which is true, hence given result is true.*
Notes:
(a)
M1: Applies the compound angle formula tosin3θ at some stage in the working. Accept
±
sin
θ
cos2
θ
±
sin 2 cos
θ
θ
Allow for students who apply
s
in3
3si
n
4s n
i
3
directly. May be seen
θ
=
θ
−
θ
anywhere (e.g. in separate working).
M1: Uses correct double angle formula for cos2θ and sin2θ (or sin3θ) to achieve all terms in single angle.
(May be before or after cross multiplying.)
M1: Cross multiplies and uses Pythagorean identity somewhere in the working in an attempt to achieve a true
statement by reaching the same expression on both sides, or reducing to a known true trigonometric identity
(as shown in scheme). Award at the stage the identity is applied to an appropriate term in an expression
where the angles have been reduced to just θ.
A1: Reaches a correct true statement with no errors seen and makes conclusion that the original result is true.
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