proof of angle sum identities

proof of angle sum identities

We will derive the

angle sum identities

for the various

trigonometric
functions

here. We begin by deriving the identity for the sine by means
of a geometric

argument

and then obtain the remaining identities by
algebraic manipulation.

Theorem 1

.

sin

⁡

(

x

+

y

)

=

sin

⁡

(

x

)

⁢

cos

⁡

(

y

)

+

cos

⁡

(

x

)

⁢

sin

⁡

(

y

)

Proof.

Let us make the

restrictions

0

∘

<

x

<

90

∘

and

0

∘

<

y

<

90

∘

for the time being. Then we may draw a

triangle

A

⁢

B

⁢

C

such that

∠

⁢

C

⁢

A

⁢

B

=

x

and

∠

⁢

A

⁢

B

⁢

F

=

y

:

?

?

?

?

?

?

?

?

?

?

?

?

w

w

w

w

w

w

w

w

w

w

w

w

w

w

w

w

A

B

C

Since the angles of a triangle add up to

180

∘

, we must have

∠

⁢

B

⁢

C

⁢

A

=

180

∘

-

x

-

y

, so we have

sin

⁡

(

∠

⁢

B

⁢

C

⁢

A

)

=

sin

⁡

(

180

∘

-

x

-

y

)

=

sin

⁡

(

x

+

y

)

.

We now draw

perpendiculars

two different ways in order to derive ratios.
First, we drop a perpendicular

A

⁢

D

from

C

to

A

⁢

B

:

?

?

?

?

?

?

?

?

?

?

?

?

w

w

w

w

w

w

w

w

w

w

w

w

w

w

w

w

A

B

C

D

Since

A

⁢

C

⁢

D

and

B

⁢

C

⁢

D

are

right triangles

we have, by

definition

,

cot

⁡

(

∠

⁢

C

⁢

A

⁢

B

)

=

A

⁢

D

¯

/

C

⁢

D

¯

cot

⁡

(

∠

⁢

A

⁢

B

⁢

C

)

=

B

⁢

D

¯

/

C

⁢

D

¯

sin

⁡

(

∠

⁢

C

⁢

A

⁢

B

)

=

C

⁢

D

¯

/

A

⁢

C

¯

.

Second, we draw a perpendicular

A

⁢

E

form

A

to

B

⁢

C

. Depending on whether

x

+

y

<

90

∘

or

x

+

y

<

90

∘

the point

E

will or will not lie between

B

and

C

, as illustrated below. (There is also the case

x

+

y

=

90

∘

,
but it is trivial.)

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

Υ

Υ

Υ

Υ

Υ

Υ

Υ

¨

¨

¨

¨

¨

¨

¨

¨

¨

A

B

C

E

?

?

?

?

?

?

?

?

?

?

?

?

?

?

¨

¨

¨

¨

¨

¨

¨

w

w

w

w

w

w

w

w

w

w

w

w

w

w

w

w

A

B

C

E

Either way,

A

⁢

B

⁢

E

and

A

⁢

C

⁢

E

are right triangles, and we have, by definition,

sin

⁡

(

∠

⁢

B

⁢

C

⁢

A

)

=

A

⁢

E

¯

/

A

⁢

C

¯

sin

⁡

(

∠

⁢

A

⁢

B

⁢

C

)

=

A

⁢

E

¯

/

A

⁢

B

¯

.

Combining these ratios, we find that

sin

⁡

(

∠

⁢

B

⁢

C

⁢

A

)

/

sin

⁡

(

∠

⁢

A

⁢

B

⁢

C

)

=

A

⁢

B

¯

/

A

⁢

C

¯

.

To finish deriving the sum identity, we manipulate the ratios derived above
algebraically and use the fact that

A

⁢

D

¯

+

B

⁢

D

¯

=

A

⁢

B

¯

:

sin

⁡

(

x

+

y

)

=

sin

⁡

(

∠

⁢

B

⁢

C

⁢

A

)

=

A

⁢

B

¯

⁢

sin

⁡

(

∠

⁢

A

⁢

B

⁢

C

)

/

A

⁢

C

¯

=

(

A

⁢

D

¯

+

B

⁢

D

¯

)

⁢

sin

⁡

(

∠

⁢

A

⁢

B

⁢

C

)

/

A

⁢

C

¯

=

C

⁢

D

¯

⁢

(

cot

⁡

(

∠

⁢

C

⁢

A

⁢

B

)

+

cot

⁡

(

∠

⁢

A

⁢

B

⁢

C

)

)

/

sin

⁡

(

∠

⁢

A

⁢

B

⁢

C

)

⁢

A

⁢

C

¯

=

sin

⁡

(

∠

⁢

C

⁢

A

⁢

B

)

⁢

sin

⁡

(

∠

⁢

A

⁢

B

⁢

C

)

⁢

(

cos

⁡

(

∠

⁢

C

⁢

A

⁢

B

)

sin

⁡

(

∠

⁢

C

⁢

A

⁢

B

)

+

cos

⁡

(

∠

⁢

A

⁢

B

⁢

C

)

sin

⁡

(

∠

⁢

A

⁢

B

⁢

C

)

)

=

sin

⁡

(

∠

⁢

C

⁢

A

⁢

B

)

⁢

cos

⁡

(

∠

⁢

A

⁢

B

⁢

C

)

+

cos

⁡

(

∠

⁢

C

⁢

A

⁢

B

)

⁢

sin

⁡

(

∠

⁢

A

⁢

B

⁢

C

)

=

sin

⁡

(

x

)

⁢

cos

⁡

(

y

)

+

cos

⁡

(

x

)

⁢

sin

⁡

(

y

)

To lift the restriction on the range of

x

and

y

, we use the identities
for

complements

and

negatives

of angles.

∎

Entry under construction

Title

proof of angle sum identities

Canonical

name

ProofOfAngleSumIdentities

Date of creation

2013-05-28 14:35:29

Last modified on

2013-05-28 14:35:29

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

15

Author

rspuzio (6075)

Entry type

Proof

Classification

msc 43-00

Classification

msc 51-00

Classification

msc 42-00

Classification

msc 33B10

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