H2 Mathematics - Trigonometry

8/4/2019 H2 Mathematics - Trigonometry

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H2 Mathematics – Essentials for Trigonometry

1 Objective

This set of materials has been developed for students to acquire just-in-time skills in

trigonometry (which is not a formal topic in the syllabus), which may be applied in other

areas of H2 Mathematics, e.g. summation of series, differentiation, integration, complexnumbers, etc.

You should spend at least 3 hours in understanding the concepts & applying them.

Reference books:

Pure Mathematics by L Bostok & S Chandler

New Additional Mathematics by Ho Soo Thong & Khor Nyak Hiong

2 Basic concepts

2.1 Trigonometric ratios

adjacent sidecos

hypothenuse

OQ

OPθ = =

opposite sidesin

hypothenuse

PQ

OPθ = =

opposite sidetan

adjacent side

PQ

OQθ = =

2.2 Signs of trigonometric ratios in the four quadrants

2nd quadrant

only sinθ positive (S)

1st quadrant

all 3 ratios positive (A)

3rd quadrant

only tanθ positive (T)

4th quadrant

only cosθ positive (C)

θ



Page 2 of 12

2.3 Trigonometric identities

sintan

cos

θ θ

θ =

1cot

tan

θ θ

=

1sec

cosθ

θ =

1cosec

sinθ

θ =

Note: denominator in

each identity cannot be

zero.

Pythagorean identity: 2 2sin cos 1θ θ + = -- (*)

Dividing (*) throughout by 2sin θ , we obtain2 21 cot cosecθ θ + =

Dividing (*) throughout by2

cos θ , we obtain2 2tan 1 secθ θ + =

2.4 Useful relationships

Negative angles: ( )cos cosθ θ − =

( )sin sinθ θ − = −

( )tan tanθ θ − = −

Complementary angles: cos sin2

π θ θ

− =

sin cos2

π θ θ

− =

tan cot2

π

θ θ

− =

cot tan2

π θ θ

− =

For example,1

sin sin6 6 2

π π − = − = −

Note: π radians = 180



Page 3 of 12

2.5 Special angles

θ 0 6

π

4

π

3

π

2

π

π 3

2

π

2π

sinθ 0 1

2

1

2

3

2

1 0 1− 0

cosθ 1 3

2

1

2

1

2

0 1− 0 1

tanθ 0 1

3

1 3 undefined 0 undefined 0

3 Compound angle identities

For any 2 angles A and B, A B+ , A B− and 2 A B+ are called compound angles.

We can prove that: ( )sin sin cos cos sin x y x y x y+ = + -- (1) and

( )cos cos cos sin sin x y x y x y+ = − -- (2)

If you are interested, you may wish to refer to the geometry proofs of (1) and (2) at:

http://www.acts.tinet.ie/compoundanglesandcalcu_668.html

From (1), we replace y with y− to obtain ( )( ) ( )sin sin x y x y+ − = −

( ) ( )sin cos cos sin x y x y= − + −

sin cos cos sin x y x y= − -- (3)

From (2), we use the same approach to obtain ( )( ) ( )cos cos x y x y+ − = −

cos cos sin sin x y x y= + -- (4)

Consider (1) ÷ (2): ( )sin cos cos sin

tancos cos sin sin

x y x y x y

x y x y

++ =

−

Dividing each term in the numerator and denominator by cos cos x y ,

( )

sin cos cos sincos cos

tancos cos sin sin

cos cos

x y x y x y

x y x y x y

x y

+

+ =−

tan tan

1 tan tan

x y

x y

+=

−-- (5)



Page 4 of 12

From (5), we replace y with y− to obtain ( )( ) ( )tan tan x y x y+ − = −

tan tan

1 tan tan

x y

x y

−=

+-- (6)

Worked example 1: Without using a calculator, show that3 1

cos752 2

° −= . Similarly,

find (a) cos15° , (b) tan105° , (c) ( )sin 15°− .

( )cos 75 cos 30 45° ° °= + [ Note: 30

°& 45

°are special angles ]

cos 30 cos 45 sin 30 sin 45° ° ° °

= − [ Applying formula (2) ]

3 1 1 1

2 22 2= × − ×

3 12 2

−= (shown)

(a) ( )cos15 cos 45 30° ° °= −

cos 45 cos 30 sin 45 sin 30° ° ° °

= + [ Applying formula (4) ]

1 3 1 1

2 22 2= × + ×

3 1

2 2

+=

(b) ( )tan105 tan 60 45° ° °

= +

tan 60 tan 45

1 tan 60 tan 45

° °

° °

+=

−[ Applying formula (5) ]

3 1

1 3 1

+=

− ×

3 1

1 3

+=

−



Page 5 of 12

(c) ( ) ( )sin 15 sin 30 45° ° °− = − [ Applying formula (3) ]

sin 30 cos 45 cos 30 sin 45° ° ° °

= −

1 1 3 1

2 22 2= × − ×

1 32 2−=

4 Double angle identities

We make use of (1), (2), (5) in Section 3 to obtain double angle formulae sin 2 x , cos2 x ,

tan 2 x respectively.

From (1), we replace y with x to obtain ( )sin sin 2 x x x+ =

sin cos cos sin x x x x= +

2 sin cos x x= -- (7)

From (2), we use the same approach to obtain ( )cos cos 2 x x x+ =

cos cos sin sin x x x x= − 2 2cos sin x x= −

21 2sin x= − 2

2cos 1 x= − -- (8)

From (5), we use the same approach to obtain ( )tan tan 2 x x x+ =

tan tan1 tan tan

x x x x+=

−

2

2tan

1 tan

x

x=

−-- (9)



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Worked example 2: Without using a calculator, find the values of sin 2 A , cos2 A and

tan 2 A if (a)3

sin5

A = and A is acute, (b)1

cos2

A = − and A is obtuse,

(c) cot 2 A = and 90 270 A° °

< < , (d)2

sec3

A = and 90 0 A° °

− < < .

We will use formulae (7), (8), (9) for each part of this example.

(a)3

sin5

A = , so4

cos5

A = and3

tan4

A = from the diagram

3 4 24sin 2 2sin cos 2

5 5 25 A A A∴ = = × × =

2 2

2 2 4 3 7cos 2 cos sin

5 5 25 A A A

∴ = − = − =

22

3 32

2 tan 244 2tan271 tan 73

1164

A A

A

×

∴ = = = =−

−

(b)1

cos2

A = − , so3

sin2

A = and tan 3 A = − from the diagram


2 2 2 A A A∴ = = × × − = −

22


2 2 2 A A A

∴ = − = − − = −

( )22

2 tan 2 3tan 2 3

1 tan 1 3

A A

A

× −∴ = = =

−− −

A

3

4

5

2

−1

3 A



Page 7 of 12

(c)1

cot 2 tan2

A A= ⇒ = . Since tan A is positive, A has to lie in the third quadrant.

So,1

sin5

A = − and2

cos5

A = − from the diagram

1 2 4sin 2 2sin cos 255 5

A A A∴ = = × − × − =

2 2


55 5 A A A

∴ = − = − − − =

22

12

2 tan 42tan21 tan 31

12

A A

A

×

∴ = = =−

−

(d)2 3

sec cos23

A A= ⇒ = ; since 90 0 A° °

− < < , A is in the fourth quadrant.

So1

sin2

A = − and1

tan3

A = −

(Try drawing the diagram yourself to obtain the ratios)


2 2 2

A A A∴ = = × − × = −

2 2


2 2 2 A A A

∴ = − = − − =

22

12

2tan 3tan 2 3

1 tan 11

3

A A

A

× −

∴ = = = −−

− −

−2

5−1

A



Page 8 of 12

By replacing x with2

xin (7), (8), (9) we can obtain expressions for sin x , cos x , tan x

respectively. These are half-angle identities.

For example, sin 2 sin 2sin cos

2 2 2

x x x x

= =

Similarly, 2 2 2 2cos cos sin 1 2sin 2cos 12 2 2 2 2

x x x x x= − = − = −

and2

2tan2tan

1 tan2

x

x x

=

−

.

By considering ( )sin 2 x x+ and the double angle formula for sine, we can obtain an

expression for sin 3 x in terms of sin x as well as cos3 x in terms of cos x . These aremultiple angle identities.

For example, ( )sin 2 sin 3 x x x+ =

sin cos 2 cos sin 2 x x x x= +

( ) ( )2sin 1 2sin cos 2sin cos x x x x x= − +

3 2sin 2sin 2sin cos x x x x= − +

( )3 2sin 2sin 2sin 1 sin x x x x= − + −

33sin 4sin x x= −

By a similar approach, ( ) 3cos 2 cos3 4cos 3cos x x x x x+ = = −

Try to apply the approach and see for yourself.



Page 9 of 12

5 Factor formulae

We make use of (1), (2), (3), (4) in Section 3 to obtain 4 factor formulae (or product-to-

sum identities):

Consider (1) + (3):

( ) ( )sin sin sin cos cos sin sin cos cos sin x y x y x y x y x y x y+ + − = + + −

2 sin cos x y= -- (10)

Consider (1) − (3): ( ) ( )sin sin 2cos sin x y x y x y+ − − = -- (11)

Consider (2) + (4):

( ) ( )cos cos cos cos sin sin cos cos sin sin x y x y x y x y x y x y+ + − = − + +

2 cos cos x y= -- (12)

Consider (2) − (4), ( ) ( )cos cos 2sin sin x y x y x y+ − − = − -- (13)

Identities (10), (11), (12), (13) should be used when a given product is to be changed to a

sum or difference.

For example, to express 2cos7 cos 2θ θ as a sum we use (12) to give

( ) ( )cos 7 2 cos 7 2 cos9 cos5θ θ θ θ θ θ + + − = + .

Suppose in (10), we let x y p+ = and x y q− = . Then

2

p q x

+= and

2

p q y

−= .

Substitute these into the identity to obtain sin sin 2sin cos2 2

p q p q p q

+ −+ = -- (12)

We use the same approach for (11) to derive sin sin 2cos sin2 2

p q p q p q

+ −− = -- (13)

Similarly, cos cos 2cos cos2 2

p q p q p q

+ −+ = -- (14)

and cos cos 2sin sin2 2

p q p q p q+ −

− = − -- (15)



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Identities (12), (13), (14), (15) are best used when a sum or difference is to be expressed

as a product. We call these sum-to-product identities.

For example, to express sin 6 sin4θ θ − as a product we would use (13) to obtain

6 4 6 42cos sin 2cos5 sin

2 2

θ θ θ θ θ θ

+ −= .

Worked example 3: Prove thatsin sin

tancos cos 2

A B A B

A B

+ +≡

+. If A, B and C are the angles of

a triangle, deduce thatsin sin

cotcos cos 2

A B C

A B

+=

+.

sin sinLHS

cos cos

A B

A B

+≡

+

[ Applying formula (12) for the numerator & formula (14) for the denominator ]

2sin cos2 2

2cos cos2 2

A B A B

A B A B

+ −

≡+ −

tan2

A B+≡

Since A, B and C are angles in a triangle, A B C π + + =

2 2 2

A B C π +⇒ + =

2

A B+ ⇒

and

2

C are complementary

tan tan cot2 2 2 2

A B C C π + ∴ = − =

[ Recall the useful relationships in Section 2.4 ]



Page 11 of 12

Worked example 4: Prove cos cos3 cos5 cos 7 4cos 4 sin 2 sinθ θ θ θ θ θ θ − − + ≡ − .

LHS cos cos3 cos5 cos 7θ θ θ θ ≡ − − +

( ) ( )cos7 cos cos5 cos3θ θ θ θ ≡ + − + [ Applying formula (14) twice ]

7 7 5 3 5 32cos cos 2cos cos

2 2 2 2

θ θ θ θ θ θ θ θ + − + −≡ −

2cos 4 cos3 2cos 4 cosθ θ θ θ ≡ −

( )2cos 4 cos3 cosθ θ θ ≡ −

3 32cos 4 2sin sin

2 2

θ θ θ θ θ

+ − ≡ −

[ Applying formula (15) ]

4 cos 4 sin 2 sinθ θ θ ≡ −

6 Self-attempt Exercises

Apply compound angle identities to solve Q1 & Q2.

Q1 Without using a calculator, evaluate:

(a) cos80 cos20 sin80 sin20° ° ° °+ , (b) sin 37 cos 7 cos 37 sin 7° ° ° °

− ,

(c) sin165°, (d) tan75

°.

[ Answers: (a)1

2, (b)

1

2, (c) ( )

16 2

4− , (d) 2 3+ ]

Q2 Prove ( )

( )

sintan tan

cos cos

y x y x

x x y+ − ≡

+.

Apply double angle identities to solve Q3 & Q4.

Q3 Given that cos cθ = and that θ is acute, express in terms of c,

(a) cos2θ , (b) sin2θ , (c) tan 2θ , (d) sin2

θ .

[ Answers: (a) 22 1c − , (b) 22 1c c− , (c)2

2

2 1

2 1

c c

c

−

−, (d)

1

2

c−]

Q4 Provecos sin

sec 2 tan 2cos sin

A A A A

A A

++ ≡

−.



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