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Name : ____________________________ ( ) Date : ________________

Class : ____________________ Duration: 30 minutes

There are 3 questions in this test. Answer all questions on writing paper(s).

1. Prove that 1 cos2 sin 2

cot1 cos2 sin 2

. [3]

2

2

2

2

1 cos2 sin 2LHS

1 cos2 sin 2

1 2cos 1 2sin cos --- M1cao (Double Angle)

1 1 2sin 2sin cos

2cos 2sin cos

2sin 2sin cos

2cos cos sin --- M1 (Common Factor

2sin sin cos

)

cos --- M1 (simplifying)

sin

cot

RHS (proved)

CHUNG CHENG HIGH SCHOOL SEC 4 ADDITIONAL MATHEMATICS

Class Test 1 Trigonometric Identities and Formulae

Proofs in Plane Geometry Marking Scheme

25

25 25 25

20

_____________________

Parent’s Signature

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2. The diagram shows a quadrilateral ABCD in which BAD, AED, AFC and BCD are right

angles, ABC = , where 0 90 , AD = 4 cm and AB = 8 cm.

(a) Show that the perimeter, P cm of the quadrilateral ABCD is given by

12sin 4cos 12P . [3]

o o

o

o o

cos8

8cos M1cao (for finding , , , , or )

Since 180 90 (sum of s of )

90

90 90 (complementary s)

M1cao (

BF

BF BF AF CF AE EF DC

BAF

DAE

For finding )

sin4

4sin

4sin (opposite sides of rectangle are equal)

sin cos8 4

4cos8sin

8sin 4cos

(opposite sides of rectangle are equal)

Perimeter of

DAE

DE

DE

FC DE

AF AE

AEAF

EF AF AE

DC

ABCD AB

8 8cos 4sin 8sin 4cos 4 M1 (sum to find perimeter of ABCD)

12sin 4cos 12 (shown) (AG)

BF FC CD AD

B

A

C F

E D

4 cm

8 cm

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(b) Express P in the form sin 12R where R is positive and is acute. [3]

2 2 2 2 2

2

12sin 4cos 12

For 12sin 4cos sin( ),

sin( ) (sin cos cos sin )

cos 12 and sin 4

(sin cos ) 12 2 M1cao

160

4 10 or 4 10(rejected as

P

R

R R

R R

R

R

R

0)

sin 4

cos 12

1tan M1cao

3

18.434 (3 dec. pl.)

4 10 sin( 18.4 ) 12 (1 dec. pl.) A1

R

R

R

P

(c) Find the maximum value of P and the corresponding value of . [3]

Max. value of 12 4 10

24.649

24.6 (3 sig. fig.) B1

When sin( 18.434 ) 1

18.434 90 M1

71.6 (1 dec. pl.) B1

P

(d) Find the value of when P = 18. [2]

18

4 10 sin( 18.434 ) 12 18

4 10 sin( 18.434 ) 6

6sin( 18.434 ) M1

4 10

18.434 28.316

9.9 (1 dec. pl.) A1

P

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3. In the diagram, A, B, C and D are points on the circle. The tangent at C meets AD produced at P.

The chords AC and BD intersect at Q and CD is the angle bisector of ACP .

Prove that

(a) ABQ is similar to DCQ , [2]

In and ,ABQ DCQ

( s in the same segment) M1

(vert. opp. s) M1

BAQ CDQ

AQB DQC

ABQ is similar to DCQ (AA similarity). (proved) (AG)

(b) DCAQDQAB , [1]

By similar triangles,

corr. sides. of similar s M1cao

(proved) (AG)

AB AQ

DC DQ

AB DQ AQ DC

(c) AD = CD. [3]

(given. bisector of ) M1

( s inalternate segment ) M1

ACD PCD CD ACP

PCD CAD

(base s of an isosceles )

is an isosceles triangle M1

ACD CAD

ADC

Therefore, AD = CD (proved) (AG)

A

B

C

D Q

P

−1 if statement is not provided.