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Trig double angle identities [203 marks] 1. Solve , for . Markscheme correct application of (A1) eg correct equation without logs A1 eg recognizing double-angle identity (seen anywhere) A1 eg evaluating (A1) correct working A1 eg and , one correct final answer (do not accept additional values) A2 N0 [7 marks] log 2 (2 sin x) + log 2 (cos x) = −1 2π < x < 5π 2 log a + log b = log ab log 2 (2 sin x cos x), log 2 + log(sin x) + log(cos x) 2 sin x cos x =2 −1 , sin x cos x = , sin2x = 1 4 1 2 log(sin 2x), 2sin x cos x = sin 2x, sin2x = 1 2 sin −1 () = (30 ) 1 2 π 6 x = +2π, 2x = , , 750 , 870 , x = π 12 25π 6 29π 6 π 12 x = 5π 12 x = , 25π 12 29π 12 2a. Let , where is acute. Find . Markscheme evidence of valid approach (M1) eg right triangle, correct working (A1) eg missing side is 2, A1 N2 [3 marks] sin θ = 5 3 θ cos θ cos 2 θ = 1 − sin 2 θ 1− ( ) 2 5 3 cos θ = 2 3 2b. Find . Markscheme correct substitution into formula for (A1) eg A1 N2 [2 marks] cos2θ cos2θ () 2 −1, 1−2( ) 2 , () 2 ( ) 2 2 3 5 3 2 3 5 3 cos2θ =− 1 9 [7 marks] [3 marks] [2 marks]

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### Transcript of Trig double angle identities [203 marks] Trig double angle identities [203 marks]

1. Solve , for .

Markschemecorrect application of (A1)

eg

correct equation without logs A1

eg

recognizing double-angle identity (seen anywhere) A1

eg

evaluating (A1)

correct working A1

eg and , one correct final answer

(do not accept additional values) A2 N0

[7 marks]

log2(2 sin x) + log2(cosx) = −1 2π < x < 5π

2

loga + logb = logab

log2(2 sin xcosx), log2 + log(sin x) + log(cosx)

2 sin xcosx = 2−1, sin xcosx = , sin 2x =14

12

log(sin 2x), 2 sin xcosx = sin 2x, sin 2x = 12

sin−1( ) = (30∘)12

π

6

x = + 2π, 2x = , , 750∘, 870∘, x =π

1225π

629π

12x = 5π

12

x = , 25π

1229π

12

2a.

Let , where is acute.

Find .

Markschemeevidence of valid approach (M1)

egright triangle,

correct working (A1)

egmissing side is 2,

A1 N2

[3 marks]

sin θ =√53

θ

cosθ

cos2θ = 1 − sin2θ

√1 − ( )2√53

cosθ = 23

2b. Find .

Markschemecorrect substitution into formula for (A1)

eg

A1 N2

[2 marks]

cos2θ

cos2θ

2 × ( )2− 1, 1 − 2( )2

, ( )2− ( )2

23

√53

23

√53

cos2θ = − 19

[7 marks]

[3 marks]

[2 marks] 3a.

The following diagram shows a triangle ABC and a sector BDC of a circle with centre B and radius 6 cm. The points A , B and D are onthe same line.

is obtuse.

Find .

MarkschemeMETHOD 1

correct substitution into formula for area of triangle (A1)

eg

correct working (A1)

eg

(A1)

(A1)

A1 N3

METHOD 2

(using height of triangle ABC by drawing perpendicular segment from C to AD )

correct substitution into formula for area of triangle (A1)

eg

correct working (A1)

eg

height of triangle is 3 A1

(A1)

A1 N3

[5 marks]

AB = 2√3 cm, BC = 6 cm, area of triangle ABC = 3√3 cm2,ABC

ABC

(6)(2√3) sin B, 6√3 sin B, (6)(2√3) sin B = 3√312

12

6√3 sin B = 3√3, sin B = 3√3

(6)2√312

sin B = 12

(30∘)π

6

ABC = (150∘)5π

6

(2√3) (h) = 3√3, h√312

h√3 = 3√3

CBD = (30∘)π

6

ABC = (150∘)5π

6

3b. Find the exact area of the sector BDC.

Markschemerecognizing supplementary angle (M1)

eg

correct substitution into formula for area of sector (A1)

eg

A1 N2

[3 marks]

CBD = , sector = (180 − ABC)(62)π

612

× × 62, π(62)( )12

π

630360

area = 3π (cm2)

[5 marks]

[3 marks] 4a.

Given that, where is an obtuse angle,

find the value of

Markschemevalid approach (M1)

eg ,

correct working (A1)

eg

correct calculation (A1)

eg

A1 N3

[4 marks]

sin x = 34

x

cosx;

sin2x + cos2x = 1

42 − 32, cos2x = 1 − ( )234

, cos2x =√7

47

16

cosx = −√7

4

4b. find the value of

Markschemecorrect substitution (accept missing minus with cos) (A1)

eg

correct working A1

eg

A1 N2

[3 marks]

Total [7 marks]

cos2x.

1 − 2( )2, 2(− )2

− 1, ( )2− ( )23

4

√7

4

√7

434

2 ( ) − 1, 1 − , −716

1816

716

916

cos2x = − (= − )216

18

5a.

The following diagram shows a right-angled triangle,

, where

.

Show that.

ABC

sin A = 513

cosA = 1213

[4 marks]

[3 marks]

[2 marks] MarkschemeMETHOD 1

approach involving Pythagoras’ theorem (M1)

eg , labelling correct sides on triangle

finding third side is 12 (may be seen on diagram) A1

AG N0

METHOD 2

approach involving (M1)

eg

correct working A1

eg

AG N0

[2 marks]

52 + x2 = 132

cosA = 1213

sin2θ + cos2θ = 1

( )2+ cos2θ = 1, x2 + = 15

1325169

cos2θ = 144169

cosA = 1213

5b. Find.

Markschemecorrect substitution into

(A1)

eg

correct working (A1)

eg

A1 N2

[3 marks]

cos2A

cos2θ

1 − 2( )2, 2( )2

− 1, ( )2− ( )25

131213

1213

513

1 − , − 1, −50169

288169

144169

25169

cos2A = 119169

6a.

In triangle

,

and

. The area of the triangle is

.

Find the two possible values for.

ABC

AB = 6 cm

AC = 8 cm

16 cm2

A

[3 marks]

[4 marks] Markschemecorrect substitution into area formula (A1)

eg

correct working (A1)

eg

;

; A1A1 N3

(accept degrees ie;

)

[4 marks]

(6)(8)sin A = 16, sin A =12

1624

A = arcsin( )23

A = 0.729727656… ,2.41186499…(41.8103149∘,138.1896851∘)

A = 0.7302.41

41.8∘

138∘

6b. Given that is obtuse, find

.

Markschemeevidence of choosing cosine rule (M1)

eg

correct substitution into RHS (angle must be obtuse) (A1)

eg ,

A1 N2

[3 marks]

A

BC

BC2 = AB2 + AC2 − 2(AB)(AC) cosA, a2 + b2 − 2ab cosC

BC2 = 62 + 82 − 2(6)(8)cos2.41, 62 + 82 − 2(6)(8)cos138∘

BC = √171.55

BC = 13.09786

BC = 13.1 cm

[3 marks] 7a.

The diagram shows a circle of radius metres. The points ABCD lie on the circumference of the circle.

BC = m, CD =

, and .

Find AC.

Markschemeevidence of choosing cosine rule (M1)

eg ,

correct substitution A1

eg ,

AC (m) A1 N2

[3 marks]

8

BCD = 73∘

c2 = a2 + b2 − 2ab cosC

11.52 + 82 − 2 × 11.5 × 8 cos104196.25 − 184 cos104

= 15.5

7b. (i) Find .

(ii) Hence, find .

ACD

ACB

[3 marks]

[5 marks] Markscheme(i) METHOD 1

evidence of choosing sine rule (M1)

eg ,

correct substitution A1

eg

A1 N2

METHOD 2

evidence of choosing cosine rule (M1)

eg

correct substitution A1

e.g.

A1 N2

(ii) subtracting their from

(M1)

eg ,

A1 N2

[5 marks]

=sinAa

sinB

b

sinD

AC

=sinACD8

sin10415.516…

ACD = 30.0∘

c2 = a2 + b2 − 2ab cosC

82 = 11.52 + 15.516…2 − 2(11.5)(15.516…)cosC

ACD = 30.0∘

ACD73

73 − ACD70 − 30.017…

ACB = 43.0∘

7c. Find the area of triangle ADC.

Markschemecorrect substitution (A1)

eg area

area (m ) A1 N2

[2 marks]

= 44.6 2

7d. Hence or otherwise, find the total area of the shaded regions.

[2 marks]

[4 marks] Markschemeattempt to subtract (M1)

eg ,

area (A1)

correct working A1

eg ,

shaded area is (m ) A1 N3

[4 marks]

Total [6 marks]

circle − ABCDπr2 − ΔADC − ΔACB

ΔACB = (15.516…)(14)sin 42.9812

π(8)2 − 44.6336… − (15.516…)(14)sin 42.9812

64π − 44.6 − 74.1

82.4 2

8a.

The following diagram shows a triangle ABC.

The area of triangle ABC is cm , AB

cm , AC cm and

.

Find .

Markschemecorrect substitution into area formula (A1)

eg

setting their area expression equal to (M1)

eg

A1 N2

[3 marks]

80 2

= 18= x

BAC = 50∘

x

(18x)sin 5012

80

9x sin 50 = 80

x = 11.6

8b. Find BC.

[3 marks]

[3 marks] Markschemeevidence of choosing cosine rule (M1)

eg

correct substitution into right hand side (may be in terms of) (A1)

eg

BC A1 N2

[3 marks]

c2 = a2 + b2 + 2ab sin C

x

11.62 + 182 − 2(11.6)(18)cos50

= 13.8

9a. Let. Find an expression for

in terms of m.

MarkschemeNote: All answers must be given in terms of m. If a candidate makes an error that means there is no m in their answer, do not award thefinal A1FT mark.

METHOD 1

valid approach involving Pythagoras (M1)

e.g.

, labelled diagram

correct working (may be on diagram) (A1)

e.g. ,

A1 N2

[3 marks]

METHOD 2

valid approach involving tan identity (M1)

e.g.

correct working (A1)

e.g.

A1 N2

[3 marks]

sin 100∘ = m

cos100∘

sin2x + cos2x = 1

m2 + (cos100)2 = 1√1 − m2

cos100 = −√1 − m2

tan = sincos

cos100 = sin100tan100

cos100 = m

tan100

9b. Let . Find an expression for

in terms of m.sin 100∘ = m

tan100∘

[3 marks]

[1 mark] MarkschemeMETHOD 1

(accept

) A1 N1

[1 mark]

METHOD 2

A1 N1

[1 mark]

tan100 = − m

√1−m2

m

−√1−m2

tan100 = m

cos100

9c. Let. Find an expression for

in terms of m.

MarkschemeMETHOD 1

valid approach involving double angle formula (M1)

e.g.

(accept

) A1 N2

Note: If candidates find , award full FT in parts (b) and (c), even though the values may not have appropriate signs for the angles.

[2 marks]

METHOD 2

valid approach involving double angle formula (M1)

e.g. ,

A1 N2

[2 marks]

sin 100∘ = m

sin 200∘

sin 2θ = 2 sin θcosθ

sin 200 = −2m√1 − m2

2m (−√1 − m2)

cos100 = √1 − m2

sin 2θ = 2 sin θ cosθ

2m × m

tan100

sin 200 = (= 2m cos100)2m2

tan100

10a.

Let .

Show that can be expressed as

.

f(x) = (sin x + cos x)2

f(x)1 + sin 2x

[2 marks]

[2 marks] Markschemeattempt to expand (M1)

e.g. ; at least 3 terms

correct expansion A1

e.g.

AG N0

[2 marks]

(sin x + cosx)(sin x + cosx)

sin2x + 2 sin xcosx + cos2x

f(x) = 1 + sin 2x

10b. The graph of f is shown below for .

Let . On the same set of axes, sketch the graph of g for

.

Markscheme

A1A1 N2

Note: Award A1 for correct sinusoidal shape with period and range

, A1 for minimum in circle.

0 ≤ x ≤ 2π

g(x) = 1 + cosx

0 ≤ x ≤ 2π

[0, 2]

[2 marks] 10c. The graph of g can be obtained from the graph of f under a horizontal stretch of scale factor p followed by a translation by thevector

.

Write down the value of p and a possible value of k .

Markscheme ,

A1A1 N2

[2 marks]

( k

0)

p = 2k = −π

2

11a.

The following diagram shows a triangle ABC.

,

.

(i) Show that .

(ii) Find p .

Markscheme(i) evidence of valid approach (M1)

e.g. choosing cosine rule

correct substitution (A1)

e.g.

simplification A1

e.g.

AG N0

(ii)

A1 N1

Note: Award A0 for , i.e. not rejecting the negative value.

[4 marks]

BC = 6CAB = 0.7AB = 4pAC = 5pp > 0

p2(41 − 40 cos0.7) = 36

62 = (5p)2 + (4p)2 − 2 × (4p) × (5p)cos0.7

36 = 25p2 + 16p2 − 40p2cos0.7

p2(41 − 40 cos0.7) = 36

1.85995…

p = 1.86

p = ±1.86

[2 marks]

[4 marks] 11b.

Consider the circle with centre B that passes through the point C. The circle cuts the line CA at D, and is obtuse. Part of the circle is shown in the following diagram.

Write down the length of BD.

Markscheme A1 N1

[1 mark]

BD = 6

11c. Find .

Markschemeevidence of valid approach (M1)

e.g. choosing sine rule

correct substitution A1

e.g.

(A1)

A1 N3

[4 marks]

4p

sin0.76

π − 0.9253166… = 2.216275…

11d. (i) Show that radians, correct to 2 decimal places.

(ii) Hence, find the area of the shaded region.

CBD = 1.29

[1 mark]

[4 marks]

[6 marks] Markscheme(i) evidence of valid approach (M1)

e.g. recognize isosceles triangle, base angles equal

A1

AG N0

(ii) area of sector BCD (A1)

e.g.

area of triangle BCD (A1)

e.g.

evidence of subtraction M1

A1 N3

[6 marks]

π − 2(0.9253…)

CBD = 1.29

0.5 × (1.29) × (6)2

0.5 × (6)2sin 1.29

5.92496…

5.937459…

area = 5.94

12a.

The following diagram shows , where RQ = 9 cm,

and .

Find .

Markscheme A1 N1

[1 mark]

ΔPQRPRQ = 70∘

PQR = 45∘

RPQ

RPQ =65∘

12b. Find PR .

[1 mark]

[3 marks] Markschemeevidence of choosing sine rule (M1)

correct substitution A1

e.g.

7.021854078

A1 N2

[3 marks]

=PRsin45∘

9sin65∘

PR = 7.02

12c. Find the area of .

Markschemecorrect substitution (A1)

e.g.

A1 N2

[2 marks]

ΔPQR

area = × 9 × 7.02… × sin 70∘12

29.69273008

area = 29.7

13a.

Let , where

.

Find .

sin θ = 2√13

< θ < ππ2

cosθ

[2 marks]

[3 marks] MarkschemeMETHOD 1

evidence of choosing (M1)

correct working (A1)

e.g. ,

,

A1 N2

Note: If no working shown, award N1 for

.

METHOD 2

approach involving Pythagoras’ theorem (M1)

e.g. ,

finding third side equals 3 (A1)

A1 N2

Note: If no working shown, award N1 for

.

[3 marks]

sin2θ + cos2θ = 1

cos2θ = 913

cosθ = ± 3√13

cosθ = √ 913

cosθ = − 3√13

3√13

22 + x2 = 13

cosθ = − 3√13

3√13

13b. Find .tan2θ

[5 marks] Markschemecorrect substitution into

(seen anywhere) (A1)

e.g.

correct substitution into (seen anywhere) (A1)

e.g.

,

,

valid attempt to find (M1)

e.g.

,

correct working A1

e.g.

,

,

A1 N4

Note: If students find answers for which are not in the range

, award full FT in (b) for correct FT working shown.

[5 marks]

sin 2θ

2 ( ) (− )2√13

3√13

cos2θ

(− )2− ( )23

√132

√13

2(− )2− 13

√13

1 − 2( )22

√13

tan2θ

2( )(− )2√13

3√13

(− )2−( )2

3√13

2√13

2(− )23

1−(− )223

(2)(2)(−3)

13

−913

413

− 12

(√13)2

−11813

− 1213

513

tan2θ = − 125

cosθ

[−1, 1]

14a.

The diagram below shows a plan for a window in the shape of a trapezium.

Three sides of the window are long. The angle between the sloping sides of the window and the base is

, where .

Show that the area of the window is given by .

2 mθ0 < θ < π

2

y = 4 sin θ + 2 sin 2θ[5 marks] Markschemeevidence of finding height, h (A1)

e.g. ,

evidence of finding base of triangle, b (A1)

e.g. ,

attempt to substitute valid values into a formula for the area of the window (M1)

e.g. two triangles plus rectangle, trapezium area formula

correct expression (must be in terms of ) A1

e.g.

,

attempt to replace by

M1

e.g.

AG N0

[5 marks]

sin θ = h

22 sin θ

cosθ = b

22 cosθ

θ

2 ( × 2 cosθ × 2 sin θ) + 2 × 2 sin θ12

(2 sin θ)(2 + 2 + 4 cosθ)12

2 sin θ cosθ

sin 2θ

4 sin θ + 2(2 sin θ cosθ)

y = 4 sin θ + 2 sin 2θ

14b. Zoe wants a window to have an area of. Find the two possible values of

.

Markschemecorrect equation A1

e.g. ,

evidence of attempt to solve (M1)

e.g. a sketch,

,

A1A1 N3

[4 marks]

5 m2

θ

y = 54 sin θ + 2 sin 2θ = 5

4 sin θ + 2 sin θ − 5 = 0

θ = 0.856(49.0∘)θ = 1.25(71.4∘)

14c. John wants two windows which have the same area A but different values of .

Find all possible values for A .

θ

[4 marks]

[7 marks] Markschemerecognition that lower area value occurs at

(M1)

finding value of area at (M1)

e.g.

, draw square

(A1)

recognition that maximum value of y is needed (M1)

(A1)

(accept ) A2 N5

[7 marks]

θ = π

2

θ = π

2

4 sin( ) + 2 sin(2 × )π

2

A = 4

A = 5.19615…

4 < A < 5.204 < A < 5.19

15a. Show that .

Markschemeattempt to substitute

for (M1)

correct substitution A1

e.g.

AG N0

[2 marks]

4 − cos2θ + 5 sin θ = 2sin2θ + 5 sin θ + 3

1 − 2sin2θ

cos2θ

4 − (1 − 2sin2θ) + 5 sin θ

4 − cos2θ + 5 sin θ = 2sin2θ + 5 sin θ + 3

15b. Hence, solve the equation for

.

Markschemeevidence of appropriate approach to solve (M1)

correct working A1

e.g. ,

,

correct solution (do not penalise for including (A1)

A2 N3

[5 marks]

4 − cos2θ + 5 sin θ = 00 ≤ θ ≤ 2π

(2 sin θ + 3)(sin θ + 1)(2x + 3)(x + 1) = 0

sin x =−5±√1

4

sin θ = −1sin θ = − 3

2

θ = 3π

2

[2 marks]

[5 marks] 16a.

The following diagram shows the triangle ABC.

The angle at C is obtuse,,

and the area is .

Find .

Markschemecorrect substitution into the formula for the area of a triangle A1

e.g. ,

attempt to solve (M1)

e.g. ,

() (A1)

A1 N3

[4 marks]

AC = 5 cmBC = 13.6 cm20 cm2

ACB

× 5 × 13.6 × sin C = 2012

× 5 × h = 2012

sin C = 0.5882…sin C = 8

13.6

C = 36.031…∘

ACB = 144∘

16b. Find AB.

Markschemeevidence of choosing the cosine rule (M1)

correct substitution A1

e.g.

A1 N2

[3 marks]

(AB)2 = 52 + 13.62 − 2(5)(13.6)cos143.968…

AB = 17.9

17a.

The straight line with equation makes an acute angle

with the x-axis.

Write down the value of .

y = x34

θ

tanθ

[4 marks]

[3 marks]

[1 mark] Markscheme (do not accept

) A1 N1

[1 mark]

tanθ = 34

x34

17b. Find the value of

(i) ;

(ii) .

Markscheme(i)

,

(A1)(A1)

correct substitution A1

e.g.

A1 N3

(ii) correct substitution A1

e.g.

,

A1 N1

[6 marks]

sin 2θ

cos2θ

sin θ = 35

cosθ = 45

sin 2θ = 2 ( ) ( )35

45

sin 2θ = 2425

cos2θ = 1 − 2( )235

( )2− ( )24

535

cos2θ = 725

18a.

Let and

.

Find

.

Markscheme (A1)

A1 N2

[2 marks]

f(x) = cos 2x

g(x) = 2x2 − 1

f ( )π

2

f ( ) = cosππ

2

= −1

18b. Find

.(g ∘ f)( )π

2

[6 marks]

[2 marks]

[2 marks] Markscheme

(A1)

A1 N2

[2 marks]

(g ∘ f)( ) = g(−1)π

2

(= 2(−1)2 − 1)

= 1

18c. Given that can be written as

, find the value of k, .

Markscheme

A1

evidence of (seen anywhere) (M1)

A1 N2

[3 marks]

(g ∘ f)(x)cos(kx)k ∈ Z

(g ∘ f)(x) = 2(cos(2x))2 − 1(= 2cos2(2x) − 1)

2cos2θ − 1 = cos2θ

(g ∘ f)(x) = cos4x

k = 4

19a.

The diagram below shows a quadrilateral ABCD with obtuse angles and.

AB = 5 cm, BC = 4 cm, CD = 4 cm, AD = 4 cm ,

, , .

Use the cosine rule to show that .

Markschemecorrect substitution A1

e.g. ,

AG

[1 mark]

BAC = 30∘

ABC = x∘

AC = √41 − 40 cosx

25 + 16 − 40 cosx

52 + 42 − 2 × 4 × 5 cosx

AC = √41 − 40 cosx

[3 marks]

[1 mark] 19b. Use the sine rule in triangle ABC to find another expression for AC.

Markschemecorrect substitution A1

e.g. ,

(accept ) A1 N1

[2 marks]

=ACsinx

4sin30

AC = 4 sin x12

AC = 8 sin x4sinx

sin30

19c. (i) Hence, find x, giving your answer to two decimal places.

(ii) Find AC .

Markscheme(i) evidence of appropriate approach using AC M1

e.g. , sketch showing intersection

correct solution,

(A1)

obtuse value (A1)

to 2 dp (do not accept the radian answer 1.94 ) A1 N2

(ii) substituting value of x into either expression for AC (M1)

e.g.

A1 N2

[6 marks]

8 sin x = √41 − 40 cosx

8.682…111.317…

111.317…

x = 111.32

AC = 8 sin 111.32

AC = 7.45

19d. (i) Find y.

(ii) Hence, or otherwise, find the area of triangle ACD.

[2 marks]

[6 marks]

[5 marks] Markscheme(i) evidence of choosing cosine rule (M1)

e.g.

correct substitution A1

e.g.

,

,

A1 N2

(ii) correct substitution into area formula (A1)

e.g. ,

area A1 N2

[5 marks]

cosB = a2+c2−b2

2ac

42+42−7.452

2×4×47.452 = 32 − 32 cosy

cosy = −0.734…

y = 137

× 4 × 4 × sin 137128 sin 137

= 5.42

20. Solve , for

.

Markschemeevidence of substituting for

(M1)

evidence of substituting into (M1)

correct equation in terms of (seen anywhere) A1

e.g. ,

evidence of appropriate approach to solve (M1)

appropriate working A1

e.g. ,

,

correct solutions to the equation

e.g. ,

, ,

(A1)

A1 N4

[7 marks]

cos2x − 3 cosx − 3 − cos2x = sin2x

0 ≤ x ≤ 2π

cos2x

sin2x + cos2x = 1

cosx

2cos2x − 1 − 3 cosx − 3 = 12cos2x − 3 cosx − 5 = 0

(2 cosx − 5)(cosx + 1) = 0(2x − 5)(x + 1)

cosx = 3±√49

4

cosx = 52

cosx = −1x = 5

2x = −1

x = π

[7 marks] 21a.

Let .

Show that .

Markschemechanging

into A1

e.g.

simplifying A1

e.g ,

AG N0

[2 marks]

f(x) = sin3x + cos3x tan x, < x < ππ2

f(x) = sin x

tanxsinxcosx

sin3x + cos3x sinxcosx

sin x(sin2x + cos2x)sin3x + sin x − sin3x

f(x) = sin x

21b. Let . Show that

.

Markschemerecognizing

, seen anywhere (A1)

evidence of using double angle identity , seen anywhere (M1)

evidence of using Pythagoras with M1

e.g. sketch of right triangle,

(accept

) (A1)

A1

AG N0

[5 marks]

sin x = 23

f(2x) = − 4√59

f(2x) = sin 2x

sin(2x) = 2 sin xcosx

sin x = 23

sin2x + cos2x = 1

cosx = −√53

√53

f(2x) = 2 ( ) (− )23

√53

f(2x) = − 4√59

[2 marks]

[5 marks] 22a.

The following diagram shows a semicircle centre O, diameter [AB], with radius 2.

Let P be a point on the circumference, with

Find the area of the triangle OPB, in terms of .

Markschemeevidence of using area of a triangle (M1)

e.g.

A1 N2

[2 marks]

POB = θ

θ

A = × 2 × 2 × sin θ12

A = 2 sin θ

22b. Explain why the area of triangle OPA is the same as the area triangle OPB.

MarkschemeMETHOD 1

(A1)

A1

since R1

then both triangles have the same area AG N0

METHOD 2

triangle OPA has the same height and the same base as triangle OPB R3

then both triangles have the same area AG N0

[3 marks]

POA =π − θ

area ΔOPA = 2 × 2 × sin(π − θ)12

(= 2 sin(π − θ))

sin(π − θ) = sin θ

22c.

Let S be the total area of the two segments shaded in the diagram below.

Show that .S = 2(π − 2 sin θ)

[2 marks]

[3 marks]

[3 marks] Markschemearea semicircle

A1

A1

M1

AG N0

[3 marks]

= × π(2)212

(= 2π)

area ΔAPB = 2 sin θ + 2 sin θ

(= 4 sin θ)

S = area of semicircle − area ΔAPB(= 2π − 4 sin θ)

S = 2(π − 2 sin θ)

22d. Find the value of when S is a local minimum, justifying that it is a minimum.

MarkschemeMETHOD 1

attempt to differentiate (M1)

e.g.

setting derivative equal to 0 (M1)

correct equation A1

e.g. ,

,

A1 N3

EITHER

evidence of using second derivative (M1)

A1

A1

it is a minimum because

R1 N0

OR

evidence of using first derivative (M1)

for (may use diagram) A1

for (may use diagram) A1

it is a minimum since the derivative goes from negative to positive R1 N0

METHOD 2

is minimum when is a maximum R3

is a maximum when (A2)

A3 N3

[8 marks]

θ

= −4 cosθdS

−4 cosθ = 0cosθ = 04 cosθ = 0

θ = π

2

S ′′(θ) = 4 sin θ

S ′′ ( ) = 4π

2

S ′′ ( ) > 0π

2

θ < ,S ′(θ) < 0π

2

θ > ,S ′(θ) > 0π

2

2π − 4 sin θ

4 sin θ

4 sin θ

sin θ = 1

θ = π

2

[8 marks] 22e. Find a value of for which S has its greatest value.

MarkschemeS is greatest when

is smallest (or equivalent) (R1)

(or ) A1 N2

[2 marks]

θ

4 sin θ

θ = 0π

23a. Given that and

, find

.

Markschemeevidence of choosing the formula

(M1)

Note: If they choose another correct formula, do not award the M1 unless there is evidence of finding

correct substitution A1

e.g.

,

A1 N2

[3 marks]

cosA = 13

0 ≤ A ≤ π

2cos2A

cos2A = 2cos2A − 1

sin2A = 1 − 19

cos2A = ( )2−1

389

cos2A = 2 × ( )2− 11

3

cos2A = − 79

23b. Given that and

, find

.

sin B = 23

≤ B ≤ ππ

2cosB

[2 marks]

[3 marks]

[3 marks] MarkschemeMETHOD 1

evidence of using (M1)

e.g.

,

(seen anywhere),

(A1)

A1 N2

METHOD 2

diagram M1

for finding third side equals (A1)

A1 N2

[3 marks]

sin2B + cos2B = 1

( )2+ cos2B = 12

3

√ 59

cosB = ±√ 59

(= ± )√53

cosB = −√ 59

(= − )√53

√5

cosB = −√53

24a.

The expression can be expressed in the form

.

Find the value of a and of b .

Markschemerecognizing double angle M1

e.g. ,

, A1A1 N3

[3 marks]

6 sin x cos xa sin bx

3 × 2 sin xcosx

3 sin 2x

a = 3b = 2

24b. Hence or otherwise, solve the equation , for

.

6 sin xcosx = 32

≤ x ≤π

2

[3 marks]

[4 marks] Printed for GEMS INTERNATIONAL SCHOOL AL KHAIL

International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

Markschemesubstitution

M1

A1

finding the angle A1

e.g. ,

A1 N2

Note: Award A0 if other values are included.

[4 marks]

3 sin 2x = 32

sin 2x = 12

π

6

2x = 5π

6

x = 5π

12