Mark scheme Pure Mathematics Year 2 Unit Test 6: Trigonometry
Pearson
Progression Step
Q Scheme Marks AOs
and Progress
descriptor
1 M1 2.2a 6th
Use Pythagoras’ theorem to show that the length of OB =2 3
Solve problems
or OD =2 3 or states BD =4 3
involving arc
length and sector
Makes an attempt to find Ð DAB or Ð DCB . M1 2.2a
area in context.
2
cos DAO
For example, 4 is seen.
2 2 A1 1.1b
DAB DCB
Correctly states that 3 or 3
Makes an attempt to find the area of the sector with a radius of M1 2.2a
2
4 and a subtended angle of 3
1 2 2
A 4
For example, 2 3 is shown.
16 A1 1.1b
Correctly states that the area of the sector is 3
Recognises the need to subtract the sector area from the area of M1 3.2a
the rhombus in an attempt to find the shaded area.
16
8 3
For example, 3 is seen.
Recognises that to find the total shaded area this number will M1 3.2a
16
2 8 3
need to be multiplied by 2. For example, 3
Using clear algebra, correctly manipulates the expression and A1 1.1b
2
gives a clear final answer of 3
16 24 3
(8 marks)
Notes
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Mark scheme Pure Mathematics Year 2 Unit Test 6: Trigonometry
Pearson
Progression Step
Q Scheme Marks AOs
and Progress
descriptor
2a æ 9q 2 ö M1 2.1 6th
2cos3q » 2 ç1- ÷ =2 - 9q 2
è 2 ø Understand
Shows that small-angle
approximations
2cos3q - 1 » 1- 9q 2 = 1- 3q 1+ 3q M1 1.1b
Shows that ( )( ) for sin, cos and
tan (angle in
radians).
Shows 1+ sin q + tan 2q =1+ q + 2q =1+ 3q M1 2.1
1 sin tan 2 1 3 1 A1 1.1b
Recognises that
2cos3 1 1 3 1 3 1 3
(4)
2b 1 A1 1.1b 7th
1
When θ is small, 1 3 Use small-angle
approximations to
solve problems.
(1)
(5 marks)
Notes
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Mark scheme Pure Mathematics Year 2 Unit Test 6: Trigonometry
Pearson
Progression Step
Q Scheme Marks AOs
and Progress
descriptor
3a Writes tanx and secx in terms of sinx and cosx. For example, M1 2.1 5th
sin x 1 Understand the
tan x sec x cos x cos x functions sec,
cosec and cot.
1 sin x 1 sin x
1
sin x 1 1 M1 1.1b
Manipulates the expression to find cos x 1 sin x
1 A1 1.1b
- =- sec x
Simplifies to find cos x
(3)
3b B1 2.2a 6th
States that - sec x = 2 or sec x =- 2
Use the functions
æ 1 ö M1 1.1b sec, cosec and cot
1
cos x =- x =cos - 1 ç - to solve simple
Writes that 2 or è 2÷ø trigonometric
problems.
3p 5p A1 1.1b
x= ,
Finds 4 4
(3)
(6 marks)
Notes
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Mark scheme Pure Mathematics Year 2 Unit Test 6: Trigonometry
Pearson
Progression Step
Q Scheme Marks AOs
and Progress
descriptor
4 BD M1 3.1 6th
sin
States that 1 and concludes that BD =sin q Prove
sec2x = 1 + tan2x
AD M1 3.1 and
cos cosec2x = 1 + cot2x.
States that 1 and concludes that AD =cosq
States that Ð DBC =q M1 2.2a
DC sin 2 M1 3.1
tan DC
States that sin and concludes that cos oe.
sin M1 3.1
cos
States that BC and concludes that BC =tan q oe.
Recognises the need to use Pythagoras’ theorem. For example, M1 2.2a
2 2 2
AB + BC =AC
Makes substitutions and begins to manipulate the equation: M1 1.1b
2
cos sin 2
1 tan
2
1 cos
2
cos 2 sin 2
1 tan
2
cos
Uses a clear algebraic progression to arrive at the final answer: A1 1.1b
2
1
1 tan 2
cos
1+ tan 2 q =sec 2 q
(8 marks)
Notes
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Mark scheme Pure Mathematics Year 2 Unit Test 6: Trigonometry
Pearson
Progression Step
Q Scheme Marks AOs
and Progress
descriptor
5 Uses the double-angle formulae to write: M1 2.2a 6th
6sin cos 60 6cos sin 60 8 3 cos Use the
double-angle
1 3 M1 1.1b formulae for sin,
cos 60 sin 60 cos and tan.
Uses the fact that 2 and 2 to write:
3sinq + 3 3cosq =8 3 cosq
5 3 M1 1.1b
tan q =
Simplifies this expression to 3
Correctly solves to find q =70.9°, 250.9° A1 1.1b
(4 marks)
Notes
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Mark scheme Pure Mathematics Year 2 Unit Test 6: Trigonometry
Pearson
Progression Step
Q Scheme Marks AOs
and Progress
descriptor
6a Writes sin 3 cos3 sin 3 cos3 sin 3 cos3 M1 1.1b 7th
2
Use addition
sin 2 3 2sin 3 cos3 cos 2 3
formulae and/or
double-angle
A2 2.2a
Uses sin 3q + cos 3q º 1 and 2sin3q cos3q º sin6q to write:
2 2
formulae to solve
2 equations.
( sin3q + cos3q ) º 1+ sin 6q
Award one mark for each correct use of a trigonometric
identity.
(3)
6b States that: B1 2.2a 7th
2 2 Use addition
1 sin 6 formulae and/or
2
double-angle
Simplifies this to write: M1 1.1b formulae to solve
equations.
2
sin 6
2
3 9 11 M1 1.1b
6 , , ,
Correctly finds 4 4 4 4
Additional answers might be seen, but not necessary in order to
award the mark.
p 3p A1 1.1b
q= ,
States 24 24
9p 11p
q¹ ,
Note that 24 24 . For these values 3θ lies in the third
quadrant, therefore sin 3 and cos3 are both negative and
cannot be equal to a positive surd.
(4)
(7 marks)
Notes
6b
Award all 4 marks if correct final answer is seen, even if some of the 6θ angles are missing in the preceding
step.
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Mark scheme Pure Mathematics Year 2 Unit Test 6: Trigonometry
Pearson
Progression Step
Q Scheme Marks AOs
and Progress
descriptor
7a States: M1 1.1b 6th
Rcos(q + a ) º Rcosq cos a - Rsin q sin a Understand how
to use identities
Or: to rewrite
5cosq - 8sin q º Rcosq cos a - Rsin q sin a acosx + bsinx.
Deduces that: M1 1.1b
5 =Rcos a 8 =Rsin a
A1 1.1b
States that R = 89
Use of sin q + cos q =1 might be seen, but is not necessary to
2 2
award the mark.
Finds that a =1.0122 A1 1.1b
8
tan
5 might be seen, but is not necessary to award the mark.
(4)
7b Uses the maths from part a to deduce that A1 3.4 7th
Tmax =1100 + 89 =1109.43°C Solve problems
involving
Recognises that the maximum temperature occurs when M1 3.4 acosx + bsinx.
x
cos 1.0122 1
3
x M1 1.1b
=2p - 1.0122
Solves this equation to find 3
Finds x = 15.81 hours A1 1.1b
(4)
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Mark scheme Pure Mathematics Year 2 Unit Test 6: Trigonometry
7c æx ö M1 3.4 8th
1097 =1100 + 89 cos ç +1.0122 ÷
Deduces that è3 ø Use trigonometric
functions and
Begins to solve the equation. For example, M1 1.1b identities to solve
problems in a
x 3
cos 1.0122 range of
3 89 is seen. unfamiliar
contexts.
x M1 1.1b
+1.0122 =1.8944, 2p - 1.8944, 2p +1.8944
States that 3
Further values may be seen, but are not necessary in order to
award the mark.
Finds that x = 2.65 hours, 10.13 hours, 21.50 hours A1 1.1b
(4)
(12 marks)
Notes
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