Mathematics · Oct / Nov 2024 · Paper 21 · Question 9
YMA01
As printed in the paper
The question as text
9. In this question you must show detailed reasoning. 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]
) 2sin
θ
(
2tan
θ
=
3cos
θ
=
3cos
θ
cos
θ
2sin
θ
(
)
3cos
2sin
3cos
2
3 1 sin
2
θ=
θ
θ
=
θ
=
−
θ
cos
(
)
2sin
3 1 sin
2
3sin
2
2sin
3
0*
θ
=
−
θ
θ
+
θ
−
=
9(b)[4]
π
−±
1
10
sin 2
+
=
x
(May only see positive root)
3
3
NB decimal roots are: −1.387…, 0.7207…
π
2
+
=
sin (0.7207...)
−
1
=
...
x
x
3
−
0.121,
−
2.50, 0.645, 3.02
(a)
For full marks condone a complete proof entirely in x (or another variable) instead of θ
sin
θ
tan
M1:
Uses
θ
=
to write the equation in terms of sine and cosine only.
cos
θ
2sin
θ
Must be the correct identity so e.g.
2tan
θ
=
is M0
2cos
θ
May be implied by e.g.
2tan
θ
=
3cos
θ
2sin
θ
=
3cos
2
θ
M1:
Uses
cos
2
θ
= −
1 sin
2
θ
to obtain a quadratic equation in sine only.
Must be the correct identity so e.g.
3cos
2
θ
= −
1 3sin
2
θ
is M0
A1*: Correct work with all necessary steps shown leading to the given answer.
Condone e.g.
2sin
θ
+
3sin
2
θ
−
3
=
0
There should be no notational or bracketing errors and no mixed or missing
variables. E.g. we would consider
cos θ written as
2
cosθa notational error
2
sin
tan
θ
=
θ
as a missing variable
cos
Working backwards:
(
) (
)
3sin
2
θ
+
2sin
θ
−
3
=
0
3 1 cos
−
2
θ
+
2sin
θ
−
3
=
0
2sin
θ
−
3cos
2
θ
=
0
2sin
3cos
2
θ
=
θ
2tan
θ
=
3cos *
θ
Hence proven
M1:
Uses
sin
2
θ
= −
1 cos
2
θ
to obtain an equation in sine and cosine2 only.
sin
θ
M1:
Uses
tan
θ
=
which may be implied.
cos
θ
A1*: Reaches 2tan
θ
=
3cos
θ
with the same conditions as above but with a (minimal)
conclusion.
(b)
M1:
Attempts to solve the quadratic
3sin
2
+
2sin
−
3
=
0
to obtain a value for sin x
x
x
where x is any variable. Usual rules apply for solving a quadratic (via a calculator is
also acceptable and may imply this mark). If no working is shown then the root(s)
must be correct but condone premature rounding e.g. 0.72, −1.3
M1:
Attempts to find one angle within the range by finding the inverse sine of one of their
π and dividing by 2.
roots, subtracting
3
For this mark allow to work in degrees if done correctly. E.g. they would need to
π to 60° and then find the inverse sine of one of their roots in degrees,
change
3
subtract 60° and divide by 2.
May be implied by a correct value of x in degrees or radians.
NB the answers in degrees are: 143,
−
−
6.94, 36.9, 173
(3sf)
46.1...
Do not allow the mixing of degrees and radians for this mark e.g.
=
x
A1:
Any two of awrt 0.12,
−
−
2.5, 0.64 or 0.65, 3.0
(Must be in radians)
All four of awrt 0.121,
−
−
2.50, 0.645, 3.02
A1:
and no others in the range.
(Must be in radians)
Condone −2.5 for −2.50 but the others must be awrt as shown for this final mark.
Make a free account to save your answers and see your weak topics and ask the AI why a wrong answer is wrong.