Mathematics · May / June 2021 · Paper 21 · Question 8
YMA01
As printed in the paper
The question as text
8. In this question you must show all stages of your working. 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
8(i)[4]
(
)
(
)
(
)
3sin
θ
+
30
°
=
7cos
θ
+
30
°
⇒
tan
θ
+
30
° =
3
7
θ
+
30
° =
arctan
(
=
66.8 )
°
3
θ =
36.8
° or 216.8°(awrt)
(
)
8(ii)[6]
(a)
3
sin
3
x
=
5sin
x
−
7 in co
s
x
s
x
⇒
3s
i
n
x
1
−
co
s
2
x
=
5s n
i
x
−
7
s
in cos
x
x
(
)
⇒
sin
x
5
−
7cos
x
+
3cos
2
x
−
3
=
0
(
)
⇒
sin
x
3cos
2
x
−
7
cos
x
+
2
=
0
(b)
sin
x
=
0
⇒
x
=
0
(
)(
)
⇒
sin
x
3cos
x
−
1 cos
x
−
2
=
0
⇒
cos
x
=
...
1
( )
3co
s
x
=
1
⇒
x
=
arccos
= ±
1.23
3
Both of x = awrt −1.23, 1.23
Notes:
(i)
Note: answers only scores no marks.
B1: Divides through by
cos
(
30
° to get correct equation in tan.
)
θ +
M1: Applies arctan to find a value for
°. If using other approaches it is for correct work to
θ +
30
achieve one value for
30
° or just θ as appropriate. (Allow if “radians” are used.)
θ +
A1: One of awrt 36.8° or awrt 216.8°
A1: Both of awrt 36.8° and awrt 216.8° with no other solutions in the range.
(ii)(a)
Part (ii) may be marked as a whole.
sin
2
x
= −
1 cos
2
x
M1: Uses
in the equation and gathers terms on one side and factors out the sin x
(Allow the M if the sin x is cancelled) May be seen in (b)
A1: Achieves suitable form. Accept nonzero multiples of the quadratic in cos x. (Allow if sin x is
cancelled and put back in later). Result must be clearly stated but may be seen in (b).
(b)
B1: For x = 0 as one solution.
M1: Solves the quadratic in cos x, usual rules (may be implied by correct answers).
M1: Takes arccos of at least one value that has modulus less than 1 which is a root of their
quadratic. Allow if in degrees, but must reach a value.
A1: Both solutions from cosine equation correct and no others in the range.
Alternatives to (i)
7
3
8(i)
(
)
(
)
sin θ +
30
° =
or
cos θ +
30
° =
58
58
Alt 1
(3sin(
θ
+
30 ))
°
2
=
7(cos(
θ
+
30 ))
°
2
⇒
..sin
2
(
θ
+
30 )
°
=
..(1 sin
−
2
(
θ
+
30 ))
°
⇒
sin
(
θ
+
30 )
° =
..
⇒
θ
+
30
° =
...
Or
2
2
2
2
(3sin(
θ
+
30 ))
°
=
7(cos(
θ
+
30 ))
°
⇒
..(1 cos
−
(
θ
+
30 )
° =
..cos
(
θ
+
30 )
°
⇒
cos
(
θ
+
30 )
° =
..
⇒
θ
+
30
° =
...
36.8
° or 216.8°(awrt)
θ =
Alt:
7
3
B1: For reaching
sin θ +
(
30
° =
)
or
cos θ +
(
30
° =
)
58
58
M1: For a complete method of squaring both sides, applying
sin (...)
2
+
cos (...)
2
= to achieve an
1
equation in just sin or cos and then applying the inverse function to their solution to achieve one
value for
° or just θ as appropriate. . (Allow if “radians” are used.)
θ +
30
A1A1: As above, must have rejected any extra solutions from squaring.
8(i)
3
3
3
7 3
7
(
)
(
)
3sin
θ
+
30
°
=
7cos
θ
+
30
°
=
si
n
θ
+
c
o
s
θ
=
c s
o
θ
−
sin
θ
oe
2
2
2
2
Alt 2
(
)
(
)
3sin
θ
+
30
°
=
7c
os
θ
+
30
° ⇒
(
)
(
)
3 sin
θ
cos30
°±
cos sin30
θ
° =
7 cos sin30
θ
°±
sin
θ
sin30
°
sin
θ
tan
...
arctan(..)
...
⇒
=
θ
=
⇒
θ
=
=
c
os
θ
θ =
36.8
° or 216.8°(awrt)
Alt 2:
B1: Correct application of compound angle formula and trig ratios to reach
3
3 sin
3
7 3
7
oe
θ
+
cos
θ
=
cos
θ
−
s
in
θ
2
2
2
2
M1: Attempts compound angle formula (allow sign slips), rearranges to make tanθ the subject and
applies arctan to find at least one value for θ . (Allow if “radians” are used.)
A1A1: As main scheme.
500
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