Mathematics · Oct / Nov 2022 · Paper 21 · Question 5
YMA01
As printed in the paper
The question as text
5. 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
5(a)[3]
sin
θ
cos (3cos
tan )
2
cos (3cos
)
2
θ
θ
−
θ
=
θ
θ
−
=
cos
θ
or e.g.
cos (3cos
tan )
2
3cos
2
sin
2
θ
θ
−
θ
=
θ
−
θ
=
or e.g.
sin
θ
cos (3cos
θ
θ
−
tan )
θ
=
2
3cos
2
θ
−
cos
θ
=
2
cos
θ
3(1 sin
−
2
θ
)
−
sin
θ
=
2
3sin
2
θ
+
sin
θ
−=
1 0 *
5(b)[5]
) 1
−±
13
(
sin 2
x
=
(or awrt 0.43 and awrt −0.77 (or truncated −0.76))
6
2
x
=
sin (0.434...) or 2
−
1
x
=
sin ( 0.77...)
−
1
−
x
=
...
−
1.13,
−
0.438, 0.225, 1.35
(a)
sin
θ
M1: Uses
tan
θ
=
to write the equation in terms of sine and cosine only.
cos
θ
M1: Uses
sin
2
θ
+
cos
2
θ
= to obtain a quadratic equation in sine only.
1
A1*: Achieves
3sin
2
θ
+
sin
θ
−=
1
0 *
with no errors.
Condone one notational slip e.g.
3sinθinstead of
2
3sin θor e.g. 3 sin for 3sinθ in the working
2
but the printed answer must be correct but allow e.g.
0
=
sin
θ
+
3sin
2
θ
−
1
(b)
M1: Attempts to solve the quadratic
3sin
2
θ
+
sin
θ
−=
1
0
which may be in 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 roots must be correct.
A1: 1
−±
13
or a minimum of 1
−±
13
(or as decimals awrt 0.43 and awrt −0.77).
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