Mathematics · May / June 2021 · Paper 31 · Question 2
YMA01
As printed in the paper
Show the answer and mark scheme
2(a)[3]
States or uses sin 2
x
=
2sin cos
x
x
(
)
1
−
1 2sin
−
2
x
1 cos2
−
x
2sin
2
x
1 tan
=
=
=
x
2sin 2
x
4sin cos
x
x
4sin cos
x
x
2
2(b)[6]
(
)
9 1 cos2
−
θ
1
2
2
=
2sec
θ
⇒×
9 ''
''tan
θ
=
2sec
θ
2sin 2
θ
2
Attempts
1
+
tan
2
θ
=
sec
2
θ
⇒
4tan
2
θ
−
9tan
θ
+
4
=
0
9
±
17
tan
(
1.64038...,0.60961...
)
⇒
θ
=
8
⇒
θ
=
awrt 31.4 , 58.6
°
°
(b)
9 1 cos2
(
−
θ
)
1
=
2sec
2
θ
⇒×
9 ''
''tan
θ
=
2sec
2
θ
Way 2
2sin 2
θ
2
9
9sin
θ
4
8
tan
θ
=
2sec
2
θ
⇒
=
⇒
9sin
θ
cos
θ
=
4
⇒
sin 2
θ
=
2
cos
cos
2
9
θ
θ
8
sin 2
2
62.7333...,117.266...
...
θ
=
⇒
θ
=
⇒
θ
=
9
⇒
θ
=
awrt 31.4 , 58.6
°
°
(a)
B1: States or uses sin 2x = 2sin x cos x
M1: Attempts to use 1 – cos 2x = ±2sin2x and 2sin 2x = Asin x cos x and to proceeds to k tan x
A1: Fully correct solution with correct bracketing if seen
(b)
M1: Uses part (a) to form an equation of the form Atan θ = Bsec2θ
dM1: Replaces sec2θ by ±1 ± tan2θ and proceeds to form a 3TQ equation in just tan θ
A1: Correct 3TQ. E.g. 9tan θ = 4 + 4 tan2θ
M1: Correct attempt to find at least one value for tanθ for their 3TQ in tan θ
Allow solutions from calculators and condone decimals as long as accuracy is to at least 2sf.
A1: One of either awrt 31.4° or awrt 58.6°
A1: Both awrt 31.4° and awrt 58.6°. Withhold this mark if there are extra solutions in range.
...........................................................................................................................
(b) Way 2
M1: Uses part (a) to form an equation of the form equation Atan θ = Bsec2θ
1
sin
θ
dM1: Uses
,
and sin 2
θ
=
...sin
θ
cos
θ
to produce
sec
2
θ
=
tan
θ
=
cos
2
θ
cos
θ
an equation of the form sin 2
θ
=
C
8
A1:
sin 2
θ
=
9
M1: Full method to find at least one value of θ from sin 2θ = C where −1 < C < 1.
E.g. 2θ = sin1C, θ = ½ sin1C. (NB radians gives 0.547…, 1.02… and implies the M mark)
A1, A1: As above
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