Deriving the trigonometric addition formulae using area and cosine rule

Deriving the trigonometric addition formulae using area and cosine rule

Abstract

The trigonometric

addition formulae

are very important and useful in Mathematics.
They can be derived from geometric proof, relations from analytical geometry,

Euler formula

or relations from
vectorial

analysis

. In this work, we prove the sine

addition formula

by considering area. Using cosine
rule, we prove the cosine addition formula. Our proof is simpler because
it requires the knowledge of

area of triangles

and cosine rule which are easily understood by students.
From the addition formulae, we obtain the

difference

formulae by solving simultaneous equations.

Keywords

Trigonometric

addition

and difference formulae, area of triangles, cosine rule

Deriving the trigonometric addition formulae using area and cosine rule

Deriving the trigonometric addition formulae using area and cosine rule

1

Introduction

The sine and cosine addition and difference formulae are given by

sin

⁡

(

α

±

β

)

=

sin

⁡

α

⁢

cos

⁡

β

±

sin

⁡

β

⁢

cos

⁡

α

(1)

and

cos

⁡

(

α

±

β

)

=

cos

⁡

α

⁢

cos

⁡

β

∓

sin

⁡

α

⁢

sin

⁡

β

,

(2)

respectively. The sine and cosine addition and diference formulae can be obtained from geometric proofs ,
Euler’s formula, analytical geometry and vectorial analysis . Feldman uses the cosine rule
and

Pythagoras theorem

to get the cosine difference formula. In this work, we consider the area of triangles to obtain the
sine addition formula since the area of any

triangle

depends on the sine of the angle. Then, using the cosine rule twice, we obtain
the cosine addition formula. Finally, we obtain the trigonometric difference formulae from addition formulae by solving
two simultaneous equations.

2

Preliminaries

We use the important formulas and identities:

From Fig. (

1

), we have

Figure 1:

Triangle XYZ

area of

△

XYZ

=

1

2

⁢

x

⁢

y

⁢

sin

⁡

θ

and the cosine rule gives

z

2

=

x

2

+

y

2

-

2

⁢

x

⁢

y

⁢

cos

⁡

θ

.

The simple

trigonometric identities

are given by

sin

⁡

(

180

∘

-

θ

)

=

sin

⁡

θ

,

(3)

cos

⁡

(

180

∘

-

θ

)

=

-

cos

⁡

θ

,

(4)

sin

2

⁡

θ

+

cos

2

⁡

θ

=

1

.

(5)

3

Derivation of Addition Trigonometric Formulae

Figure 2:

Quadrilateral

ABCD

Fig. (

2

) shows a quadrilateral ABCD which consists of two right-angled triangles

A

⁢

B

⁢

C

and

A

⁢

C

⁢

D

.
In

△

⁢

A

⁢

B

⁢

C

,

C

⁢

A

^

⁢

B

=

α

,

A

⁢

C

=

s

. By simple

trigonometry

, we have

A

⁢

B

=

s

⁢

cos

⁡

α

,

B

⁢

C

=

s

⁢

sin

⁡

α

.

(6)

In

△

⁢

A

⁢

C

⁢

D

,

C

⁢

A

^

⁢

D

=

β

,

A

⁢

D

=

h

. We have the following

trigonometric equations

:

cos

⁡

β

=

s

h

,

C

⁢

D

=

h

⁢

sin

⁡

β

.

(7)

We also note that

B

⁢

C

^

⁢

D

=

180

∘

-

α

.

We can express the

area of quadrilateral

ABCD as:

Area of

⁢

△

⁢

A

⁢

B

⁢

C

+

Area of

⁢

△

⁢

A

⁢

C

⁢

D

=

Area of

⁢

△

⁢

A

⁢

B

⁢

D

+

Area of

⁢

△

⁢

B

⁢

C

⁢

D

1

2

⁢

(

A

⁢

B

)

⁢

(

B

⁢

C

)

+

1

2

⁢

(

A

⁢

C

)

⁢

(

C

⁢

D

)

=

1

2

⁢

(

A

⁢

B

)

⁢

h

⁢

sin

⁡

(

α

+

β

)

+

1

2

⁢

(

C

⁢

D

)

⁢

(

B

⁢

C

)

⁢

sin

⁡

(

180

∘

-

α

)

.

(8)

Substituting eqs. (

3

), (

6

) and (

7

) into eq. (

3

), we have, after simplifications,

1

2

⁢

s

2

⁢

sin

⁡

α

⁢

cos

⁡

α

+

1

2

⁢

h

⁢

s

⁢

sin

⁡

β

=

1

2

⁢

h

⁢

s

⁢

cos

⁡

α

⁢

sin

⁡

(

α

+

β

)

+

1

2

⁢

h

⁢

s

⁢

sin

⁡

β

⁢

sin

2

⁡

α

.

(9)

Dividing eq. (

9

) by

1

2

⁢

h

⁢

s

,

we obtain

s

h

⁢

sin

⁡

α

⁢

cos

⁡

α

+

sin

⁡

β

=

cos

⁡

α

⁢

sin

⁡

(

α

+

β

)

+

sin

⁡

β

⁢

sin

2

⁡

α

.

Simplifying and using eq. (

5

), we have

cos

⁡

α

⁢

sin

⁡

(

α

+

β

)

=

s

h

⁢

sin

⁡

α

⁢

cos

⁡

α

+

sin

⁡

β

⁢

(

1

-

sin

2

⁡

α

)

,

(10)

=

s

h

⁢

sin

⁡

α

⁢

cos

⁡

α

+

sin

⁡

β

⁢

cos

2

⁡

α

.

Dividing eq. (

10

) by

cos

⁡

α

and using eq. (

7

),
we finally obtain the sine addition formula given by eq. (

1

) with the

+

sign.

We next consider

△

⁢

B

⁢

C

⁢

D

. By using cosine rule and eqs. (

4

), (

6

) and (

7

),
we have

B

⁢

D

2

=

C

⁢

D

2

+

B

⁢

C

2

-

2

⁢

(

C

⁢

D

)

⁢

(

B

⁢

C

)

⁢

cos

⁡

(

180

∘

-

α

)

,

(11)

=

h

2

⁢

sin

2

⁡

β

+

s

2

⁢

sin

2

⁡

α

+

2

⁢

h

⁢

s

⁢

sin

⁡

β

⁢

sin

⁡

α

⁢

cos

⁡

α

.

Using cosine rule in

△

⁢

A

⁢

B

⁢

D

and eq. (

11

),
we obtain

cos

⁡

(

α

+

β

)

=

h

2

+

A

⁢

B

2

-

B

⁢

D

2

2

⁢

h

⁢

(

A

⁢

B

)

(12)

=

h

2

⁢

(

1

-

sin

2

⁡

β

)

+

s

2

⁢

(

cos

2

⁡

α

-

sin

2

⁡

α

)

-

2

⁢

h

⁢

s

⁢

sin

⁡

β

⁢

sin

⁡

α

⁢

cos

⁡

α

2

⁢

h

⁢

s

⁢

cos

⁡

α

.

Using eq. (

5

) and

s

2

=

h

2

⁢

cos

2

⁡

β

, eq. (

12

) reduces to

cos

⁡

(

α

+

β

)

=

h

2

⁢

cos

2

⁡

β

+

s

2

⁢

(

cos

2

⁡

α

-

sin

2

⁡

α

)

-

2

⁢

h

⁢

s

⁢

sin

⁡

β

⁢

sin

⁡

α

⁢

cos

⁡

α

2

⁢

h

⁢

s

⁢

cos

⁡

α

(13)

=

s

2

⁢

(

1

+

cos

2

⁡

α

-

sin

2

⁡

α

)

-

2

⁢

h

⁢

s

⁢

sin

⁡

β

⁢

sin

⁡

α

⁢

cos

⁡

α

2

⁢

h

⁢

s

⁢

cos

⁡

α

=

2

⁢

s

2

⁢

cos

2

⁡

α

-

2

⁢

h

⁢

s

⁢

sin

⁡

β

⁢

sin

⁡

α

⁢

cos

⁡

α

2

⁢

h

⁢

s

⁢

cos

⁡

α

=

s

h

⁢

cos

⁡

α

-

sin

⁡

α

⁢

sin

⁡

β

.

Substituting eq. (

7

) into eq. (

13

),
we finally obtain cosine addition formula given by eq. (

2

) with the

+

sign.
We point out the proof of the cosine addition formula is a generalization to that of Feldman
who used

s

=

h

=

1

.

4

Derivation of Trigonometric Difference Formulae

The difference formulas can be obtained by replacing

β

by

-

β

in eqs. (

1

)
and (

2

).

But we adopt a different approach. We set

γ

=

α

+

β

so that

β

=

γ

-

α

.

Then eqs. (

1

) and (

2

) reduce to

sin

⁡

(

γ

)

=

sin

⁡

α

⁢

cos

⁡

(

γ

-

α

)

+

cos

⁡

α

⁢

sin

⁡

(

γ

-

α

)

(14)

and

cos

⁡

γ

=

cos

⁡

α

⁢

cos

⁡

(

γ

-

α

)

-

sin

⁡

α

⁢

sin

⁡

(

γ

-

α

)

,

(15)

respectively. We obtain the trigonometric difference formulae by solving eqs. (

14

)
and (

15

) simultaneously.

(

14

)

×

cos

⁡

α

-

(

15

)

×

sin

⁡

α

yields

sin

⁡

γ

⁢

cos

⁡

α

-

cos

⁡

γ

⁢

sin

⁡

α

=

(

sin

2

⁡

α

+

cos

2

⁡

α

)

⁢

sin

⁡

(

γ

-

α

)

so that

sin

⁡

(

γ

-

α

)

=

sin

⁡

γ

⁢

cos

⁡

α

-

cos

⁡

γ

⁢

sin

⁡

α

(16)

using eq. (

5

).

Similarly,

(

14

)

×

sin

⁡

α

+

(

15

)

×

cos

⁡

α

results in

cos

⁡

(

γ

-

α

)

=

cos

⁡

γ

⁢

cos

⁡

α

+

sin

⁡

γ

⁢

sin

⁡

α

.

(17)

References

Title

Deriving the trigonometric addition formulae using area and cosine rule

Canonical

name

DerivingTheTrigonometricAdditionFormulaeUsingAreaAndCosineRule1

Date of creation

2013-03-11 19:55:01

Last modified on

2013-03-11 19:55:01

Owner

dkrbabajee (19083)

Last modified by

(0)

Numerical id

1

Author

dkrbabajee (0)

Entry type

Definition

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