Mathematics · Oct / Nov 2020 · Paper 21 · Question 7
YMA01
As printed in the paper
Show the answer and mark scheme
7(i)[3]
1
sin
θ
1
tan
θ
+
=
+
sin
tan
θ
cos
θ
θ
sin
θ
cos
θ
Uses
tan
θ
≡
on both terms
or
cos
θ
1
sin
θ
cos
θ
tan
θ
+
=
+
tan
θ
cos
θ
sin
θ
sin
cos
sin
2
cos
2
sin
2
cos
2
θ
θ
θ
+
θ
θ
θ
+
=
+
or
cos
θ
sin
θ
sin
θ
cos
θ
sin
θ
cos
θ
sin
θ
cos
θ
sin
θ
1
cos
θ
Uses
tan
θ
≡
and
≡
and attempts common denominator of sinθcosθ
cos
θ
tan
θ
sin
θ
with a 2 term numerator one of which is correct. Or attempts 2 separate fractions with a
denominator of sinθcosθ one of which is correct. Depends on the first mark.
1
Correct proof with no notation errors or
=
*
sin
θ
cos
θ
missing variables but allow “≡” instead
1
of “=”. If there are any spurious “= 0”’s
or
alongside the proof score A0.
cos sin
θ
θ
Alternative 1 for (i)
1
tan
2
1
tan
2
1
θ
+
θ
tan
θ
+
=
or
+
Attempts common denominator of tanθ
tan
θ
tan
θ
tan
θ
tan
θ
sec
2
θ
1
cos
θ
=
=
×
tan
θ
cos
2
θ
sin
θ
Applies appropriate and correct
Or
identities to obtain in terms of sinθ and
sin
2
θ
cosθ only and eliminates “double
+
1
1
cos
θ
cos
2
θ
decker” fractions if necessary
=
=
×
sin
θ
cos
2
θ
sin
θ
cos
θ
1
=
*
Correct proof with no notation errors or
sin
cos
θ
θ
missing variables but allow “≡” instead
1
of “=”. If there are any spurious “= 0”’s
or
alongside the proof score A0.
cos sin
θ
θ
Alternative 2 for (i)
1
1
sin
2
1
θ
tan
θ
+
=
⇒
+
cos
θ
=
tan
θ
sin
θ
cos
θ
cos
θ
cos
θ
sin
θ
Uses
tan
θ
≡
and multiplies through by sinθ or cosθ
cos
θ
sin
2
cos
2
1
⇒
θ
+
θ
=
sin
θ
Uses
tan
θ
≡
and multiplies through by sinθ and cosθ
cos
θ
Fully correct work reaching a correct
identity with a conclusion. If there are
sin
2
θ
+
cos
2
θ
= is true hence proved
1
any spurious “= 0”’s alongside the proof
score A0.
7(ii)[4]
1
1
(
)
(
)
(
)
( )
3cos
2
2
x +
10
° =
1
⇒
cos
2
2
x
+
10
° =
⇒
cos 2
x
+
10
° = ±
3
3
Divides by 3 and takes square root of both sides. The “±” is not required.
−
1
( )
2
x
+
10
° =
cos
1
"
±
"
−
1
cos
1
"
( )
"
10
3
±
±
°
3
Applies
x
=
−
1
2
cos
1
"
( )
"
10
±
−
°
3
You may need to check their values if
⇒
x
=
no working is shown.
2
For reference 2
10
54.735..., 125.264...
x +
° =
° or
°
Awrt one of these
x =
22.4
x =
57.6
x =
22.4
° and
x =
57.6
°
Awrt both with no extras in range
If mixing degrees and radians allow the method marks.
Alternative 1 for part (b)
(
)
(
(
)
)
3cos
2
2
x +
10
° =
1
⇒
3 1 sin
−
2
2
x
+
10
°
= ⇒
1
2
2
(
)
(
)
( )
⇒
sin
2
2
x
+
10
° =
⇒
sin 2
x
+
10
° = ±
3
3
Uses a correct identity, rearranges and takes square root of both sides.
The “±” is not required.
−
2
( )
−
2
2
x
+
10
° =
sin
1
"
±
"
( )
sin
1
"
±
"
±
10
°
3
3
Applies
x
=
−
2
2
( )
sin
1
"
±
"
−
10
°
3
You may need to check their values if
⇒
x
=
no working is shown.
2
x =
22.4
° or
x =
57.6
°
Awrt one of these
x =
22.4
° and
x =
57.6
°
Awrt both with no extras in range
Alternative 2 for part (b)
(
)
1
+
cos 4
x
+
20
1
(
)
(
)
3cos
2
2
x
+
10
° =
3
⇒
cos 4
x
+
20
= −
2
3
cos 4
(
x +
20
)
Uses a correct identity, rearranges to make
the subject
1
−
−
1
2
x
+
10
° =
cos
1
"
−
"
cos
1
"
−
"
−
20
°
3
3
Applies
⇒
x
=
−
1
4
cos
1
"
−
"
−
20
°
3
You may need to check their values if
⇒
x
=
no working is shown.
4
For reference 4
x +
20
109.47..., 250.52...
° =
x =
22.4
° or
x =
57.6
°
Awrt one of these
° and
°
x =
22.4
x =
57.6
Awrt both with no extras in range
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