Spherical trigonometry

1

Spherical Trigonometry

SOLO HERMELIN

Updated 12.11.12http://www.solohermelin.com

http://www.solohermelin.com/

2

SOLO

TABLE OF CONTENT Spherical Trigonometry

History

Sine and Cosine Laws in Spherical Triangles

Half Angles FormulasDelambre-Gauss Equations

Napier’s Equations

References

Applications: Flight on Earth Great Circles

3

SOLO

A1

B1

C1

B

a

bc

C

A

History Spherical triangles were studied by early Greek mathematicians such as Menelaus of Alexandria, who wrote a book on spherical triangles called “Sphaerica “and developed Menelaus' Theorem. E. S. Kennedy, however, points out that while it was possible in ancient mathematics to compute the magnitudes of a spherical figure, in principle, by use of the table of chords and Menelaus' theorem, the application of the theorem to spherical problems was very difficult in practice

Further advances were made in the Islamic world. In order to observe holy days of the Islamic calendar for which timings were determined by phases of the moon, astronomers initially used Menalaus' method to calculate the place of the moon and stars, though this method proved to be clumsy and difficult. It involved setting up two intersecting right triangles; by applying Menelaus' theorem it was possible to solve one of the six sides, but only if the other five sides were known. To tell the time from the sun's altitude, for instance, repeated applications of Menelaus' theorem were required. For medieval Islamic astronomers, there was an obvious challenge to find a simpler trigonometric method.


4

SOLO

History (continue - 1)

In the early 9th century, the Persian mathematician Muhammad ibn Mūsā al-Khwārizmi was a pioneer in spherical trigonometry and wrote a treatise on the subject. In the 10th century, another Persian mathematician, Abū al-Wafā' al-Būzjāni, established the angle addition formulas, e.g. for sin(a + b), and discovered the sine formula for spherical trigonometry

Abū ʿAbdallāh Muḥammad ibn Mūsā al-Khwārizmī[

(c. 780 – c. 850)

Abū al-Wafāʾ, Muḥammad ibn Muḥammad ibn Yaḥyā ibn Ismāʿīl

ibn al-ʿAbbās al-Būzjānī (940 –998)

Here, a, b, and c are the angles at the centre of the sphere subtended by the three sides of the triangle, and α, β, and γ are the angles between the sides, where angle α is opposite the side which subtends angle a, and so forth

A1

B1

C1

B

a

bc

C

A


http://en.wikipedia.org/wiki/File:Buzjani,_the_Persian.jpg

5

SOLO

History (continue - 2)

A1

B1

C1

B

a

bc

C

A

Al-Jayyani (989–1079), an Arabic mathematician in the Islamic Iberian Peninsula, wrote what some consider the first treatise on spherical trigonometry, circa 1060, entitled The book of unknown arcs of a sphere,[6] in which spherical trigonometry was brought into its modern form. Al-Jayyani's book "contains formulae for right-angle triangles, the general law of sines and the solution of a spherical triangle by means of the polar triangle". This treatise later had a "strong influence on European mathematics", and his "definition of ratios as numbers" and "method of solving a spherical triangle when all sides are unknown" are likely to have influenced Regiomontanus.

In the 13th century, Persian mathematician Nasīr al-Dīn al-Tūsī was the first to treat trigonometry as a mathematical discipline independent from astronomy, and he further developed spherical trigonometry, bringing it to its present form.[7] He listed the six distinct cases of a right-angled triangle in spherical trigonometry. In his On the Sector Figure, he also stated the law of sines for plane and spherical triangles, and discovered the law of tangents for spherical triangles

Khawaja Muhammad ibn Muhammad

ibn Hasan Tūsī (1201 -1274 )

Abū ʿAbd Allāh Muḥammad ibn Muʿādh al-Jayyānī[1]

(989– 1079


Return to Table of Content

6

SOLO

Assume three points on a unit radius sphere, defined by the vectors

CBA 1,1,1

A1

B1

C1

B

a

bc

C

A

Figure 1: Spherical Triangle

The three great circles passing trough those three points define a spherical triangle with

- three spherical triangle verticesCBA ,,

- three spherical triangle side anglescba ˆ,ˆˆ

- three spherical triangle angles defined by the angles between the tangents to the great circles at the vertices.

ˆ,ˆˆ

The extreme cases are when the three vertices are on the same great circle:

• When exists one diameter on the great circle such that all three vertices are on the same side, then

180ˆ&0ˆˆ • When doesn’t exists one diameter on the great circle such that all three vertices are on the same side

180ˆˆˆ Therefore 570ˆˆˆ180


Sine and Cosine Laws in Spherical Triangles

7

SOLO

A1

B1

C1

B

a

bc

C

A

c

BAˆsin

11

b

CAˆsin

11

Those angles can be defined using unit vectors as follows

CBA 1,1,1

180ˆ,ˆ,ˆ0

11ˆsin&11ˆcos

11ˆsin&11ˆcos

11ˆsin&11ˆcos

cba

BAcBAc

ACbACb

CBaCBa

0ˆsin&0ˆsin&0ˆsin

ˆsinˆsin

1111ˆsin&

ˆsinˆsin

1111ˆcos

ˆsinˆsin

1111ˆsin&

ˆsinˆsin

1111ˆcos

ˆsinˆsin

1111ˆsin&

ˆsinˆsin

1111ˆcos

cba

ab

BCAC

ab

BCAC

ca

ABCB

ca

ABCB

bc

CABA

bc

CABA

180ˆ,ˆ,ˆ0

1

2

Spherical TrigonometrySine and Cosine Laws in Spherical Triangles

8

SOLO

A1

B1

C1

B

a

bc

C

A

We will use the following vector identities

wvuvwuwvu

vuwuwvwvu

to perform the following computations

3

4

abcBCACBACCBCACBCACab

cabABCBACBBABCBABCBca

bcaCABACBAACABACABAbc

ˆcosˆcosˆcos1111111111111111ˆsinˆsinˆcos



1

1

1

5

We obtained the Laws of Cosines for Spherical Triangle Sides

6


ˆcosˆsinˆsinˆcosˆcosˆcos



babac

cacab

cbcba

9

SOLO

A1

B1

C1

B

a

bc

C

A

ab

abc

ca

cab

bc

bca

ˆsinˆsin

ˆcosˆcosˆcosˆcos

ˆsinˆsin


ˆsinˆsin


6

Let compute

ab

abccba

ab

abcabcab

ab

abcabc

ab

abc

ˆsinˆsin

ˆcosˆcosˆcos2ˆcosˆcosˆcos1

ˆsinˆsin

ˆcosˆcosˆcos2ˆcosˆcosˆcosˆcos1ˆcos1

ˆsinˆsin


ˆsinˆsin

ˆcosˆcosˆcos1ˆcos1ˆsin

22

222

22

22222

22

2222

22

from which

abccbaab ˆcosˆcosˆcos2ˆcosˆcosˆcos1ˆsinˆsinˆsin 222 7

The plus sign is used since, by definition, 0ˆsin,ˆsin,ˆsin ab


10

SOLO

A1

B1

C1

B

a

bc

C

A

ˆsin,ˆsin,ˆsinFrom the definitions of, given by (1) and (2), we have

BACACACCBACBCACab

ACBABCBBACBABCBca

CBACABAACBACABAbc

111111111111111ˆsinˆsinˆsin



0

0

0

But since

BACACBCBA 111111111

we have

abccba

abcabc


ˆsinˆsinˆsinˆsinˆsinˆsinˆsinˆsinˆsin

222

by dividing this equation by we obtain the Law of Sines for Spherical Triangle

bca ˆsinˆsinˆsin

cba

abccba

cba ˆsinˆsinˆsin


ˆsin

ˆsinˆsin

ˆsinˆsin

ˆsin 222

8

9


11

SOLO

A1

B1

C1

B

a

bc

C

A

10

Let compute the expression

bca

bcacbaa

bca

bcacbaa

bca

bcacaabbcbcaaa

ab

abc

ca

cab

bc

bca

ˆsinˆsinˆsin

ˆcosˆcosˆcos2ˆcosˆcosˆcos1ˆcos

ˆsinˆsinˆsin

ˆcosˆcosˆcos2ˆcosˆcosˆsinˆcos

ˆsinˆsinˆsin

ˆcosˆcosˆcosˆcosˆcosˆcosˆcosˆcosˆcosˆcosˆcosˆsinˆcosˆsin

ˆsinˆsin

ˆcosˆcosˆcosˆsinˆsin


ˆcosˆcosˆcosˆcosˆcosˆcos

2

222

2

222

2

22222

But from (8)

abcaabccba ˆsinˆsinˆsinˆsinˆsinˆsinˆcosˆcosˆcos2ˆcosˆcosˆcos1 222

Substituting this in (10) gives ˆsinˆsinˆcosˆcosˆcosˆcos a

Finally we can write the Laws of Cosines for Spherical Triangle Angles.

11



c

b

a




12

SOLO

A1

B1

C1

B

a

bc

C

A

Half Angles Formulas

From (6)

bc

bcaˆsinˆsin


Using the half angle formulas from trigonometry, we have

bc

bcba

ccba

bc

cbacba

bc

acb

bc

bcabc

ˆsinˆsin

ˆ2

ˆˆˆsinˆ

2

ˆˆˆsin

ˆsinˆsin2

2

ˆˆˆsin

2

ˆˆˆsin2

ˆsinˆsin2

ˆcosˆˆcos

ˆsinˆsin2


2

ˆcos1

2

ˆsin 2

If we define2

ˆˆˆ:ˆ

cbap

we obtain bc

bpcpˆsinˆsin

ˆˆsinˆˆsin

2

ˆsin

12


13

SOLO

A1

B1

C1

B

a

bc

C

A

Half Angles Formulas (continue – 1)

13

In the same way

bc

app

bc

acbcba

bc

cba

bc

bcabc

ˆsinˆsin

ˆˆsinˆsin

ˆsinˆsin2

2

ˆˆˆsin

2

ˆˆˆsin2

ˆsinˆsin2

ˆˆcosˆcos

ˆsinˆsin2


2

ˆcos1

2

ˆcos2

bc

appˆsinˆsin

ˆˆsinˆsin

2

ˆcos

or

From (12) and (13)

app

bpcpˆˆsinˆsin

ˆˆsinˆˆsin

2

ˆtan

14

2

ˆˆˆ:ˆ

cbap


14

SOLO

A1

B1

C1

B

a

bc

C

A

Half Angles Formulas (continue – 2)and

bc

cpbpappˆsinˆsin

ˆˆsinˆˆsinˆˆsinˆsin

2

ˆcos

2

ˆsin2ˆsin

2

ˆˆˆ:ˆ

cbap

from this equation and (8) we obtain

abccba

cpbpappbc


ˆˆsinˆˆsinˆˆsinˆsinˆsinˆsinˆsin

222

15


15

SOLO

A1

B1

C1

B

a

bc

C

A

Half Angles Formulas (continue – 3)In the same way from (11)

16

ˆsinˆsin


a

we can write

ˆsinˆsin

ˆ2

ˆˆˆcos

2

ˆˆˆcos

ˆsinˆsin2

2

ˆˆˆcos

2

ˆˆˆcos2

ˆsinˆsin2

ˆˆcosˆcosˆsinˆsin2


2

ˆcos1

2

ˆsin 2

aa

If we define2

ˆˆˆ:ˆ

we obtain

ˆsinˆsin

ˆˆcosˆcos

2

ˆsin

a17


16

SOLO

A1

B1

C1

B

a

bc

C

A

Half Angles Formulas (continue – 4)Also

18

ˆsinˆsin

ˆˆcosˆˆcosˆsinˆsin2

2

ˆˆˆcos

2

ˆˆˆcos2

ˆsinˆsin2

ˆˆcosˆcosˆsinˆsin2


2

ˆcos1

2

ˆcos2

aa

from which

ˆsinˆsin

ˆˆcosˆˆcos

2

ˆcos

a

From (17) and (18)

ˆˆcosˆˆcos

ˆˆcosˆcos

2

ˆtan

a

19


17

SOLO

Half Angles Formulas (continue – 5)

Summarize


ˆsinˆsin

ˆˆcosˆˆcos

2

ˆcos

a

ˆˆcosˆˆcos

ˆˆcosˆcos

2

ˆtan

a

ˆsinˆsin

ˆˆcosˆcos

2

ˆsin

a

bc

appˆsinˆsin

ˆˆsinˆsin

2

ˆcos

app

bpcpˆˆsinˆsin

ˆˆsinˆˆsin

2

ˆtan

bc

bpcpˆsinˆsin

ˆˆsinˆˆsin

2

ˆsin

ˆsinˆsin

ˆˆcosˆˆcos

2

ˆcos

b

ˆˆcosˆˆcos

ˆˆcosˆcos

2

ˆtan

b

ˆsinˆsin

ˆˆcosˆcos

2

ˆsin

b

cpp

bpapˆˆsinˆsin

ˆˆsinˆˆsin

2

ˆtan

ˆsinˆsin

ˆˆcosˆˆcos

2

ˆcos

c

ˆˆcosˆˆcos

ˆˆcosˆcos

2

ˆtan

c

ˆsinˆsin

ˆˆcosˆcos

2

ˆsin

c

bpp

apcpˆˆsinˆsin

ˆˆsinˆˆsin

2

ˆtan

ac

bppˆsinˆsin

ˆˆsinˆsin

2

ˆcos

ac

apcpˆsinˆsin

ˆˆsinˆˆsin

2

ˆsin

ba

cppˆsinˆsin

ˆˆsinˆsin

2

ˆcos

ba

bpapˆsinˆsin

ˆˆsinˆˆsin

2

ˆsin

2

ˆˆˆ:ˆ

2

ˆˆˆ:ˆ

cbap

18

SOLO

A1

B1

C1

B

a

bc

C

A

Half Angles Formulas (continue – 6)And

20

ˆsinˆsin

ˆˆcosˆˆcosˆˆcosˆcos

2

ˆcos

2

ˆsin2ˆsin

aaa

ˆˆcosˆˆcosˆˆcosˆcosˆsinˆsinˆsin a

or

By cyclic substitution of the angles we obtain


ˆsinˆsinˆsinˆsinˆsinˆsinˆsinˆsinˆsin

cba

ˆsinˆsinˆsinLet divide by this equation to recover the Law of Sines (9)

cba

cpbpapp

cba

abccba

cba

ˆsinˆsinˆsin

ˆˆsinˆˆsinˆˆsinˆsin

ˆsinˆsinˆsin



ˆsinˆsinˆsinˆsin

ˆsinˆsin

ˆsinˆsin

ˆsin

152229

21

22



19

SOLO

A1

B1

C1

B

a

bc

C

A

Delambre-Gauss Equations

Jean Baptiste Joseph chevalier Delambre

(1749 –1822)

Using (12) and (13) we can write

2

ˆsin

2

ˆcos

2

ˆcos

2

ˆsin

2

ˆˆsin

ca

cpap

cb

app

ca

bpp

bc

bpcpˆsinˆsin

ˆˆsinˆˆsin

ˆsinˆsin

ˆˆsinˆsinˆsinˆsin

ˆˆsinˆsinˆsinˆsin

ˆˆsinˆˆsin

2

ˆˆsin

2

ˆˆˆcos2

2

ˆˆcos

2

ˆˆˆsin2

2

ˆcos

ˆsin

1ˆˆsinˆˆsin

ˆsinˆsin

ˆˆsinˆsinˆsin

1

babap

babap

capbp

ba

cpp

c

2

ˆcos

2

ˆsin

2

ˆˆsin

2

ˆcos

2

ˆcos

2

ˆˆcos

2

ˆˆsin

2

ˆcos2

2

ˆˆcos

2

ˆsin2

2

ˆcos

2

ˆcos

2

ˆsin2

1

c

ba

c

ba

bac

bac

cc


Carl Friedrich Gauss

(1777 – 1855)

20

SOLO

A1

B1

C1

B

a

bc

C

A

Delambre-Gauss Equations (continue – 1)


(1749 –1822)

Therefore

2

ˆcos

2

ˆˆcos

2

ˆcos

2

ˆˆsin

bac

23

2

ˆcos

2

ˆˆsin

2

ˆsin

2

ˆˆsin

bac

24



(1777 – 1855)

The Delambre-Gauss Formulas were first discovered by Delambre in 1807 (published in 1809) and rediscovered independently by Gauss.

Delambre: Director of Paris Observatory (1804 – 1822)Gauss : Director of Gőttingen Observatory (1807 – 1855)

21

SOLO

A1

B1

C1

B

a

bc

C

A

Delambre-Gauss Equations (continue – 2)


(1749 –1822)

25

Also

2

ˆsin

2

ˆsin

2

ˆcos

2

ˆcos

2

ˆˆcos

ca

cpap

bc

bpcp

ca

bpp

cb

appˆsinˆsin

ˆˆsinˆˆsinˆsinˆsin

ˆˆsinˆˆsinˆsinˆsin

ˆˆsinˆsin

ˆsinˆsin

ˆˆsinˆsin

2

ˆˆsin

2

ˆcos2

2

ˆˆcos

2

ˆsin2

2

ˆsin

ˆsin

1ˆˆsinˆsin

ˆsinˆsin

ˆˆsinˆˆsinˆsin

1

cp

c

cp

c

ccpp

ba

apbp

c

2

ˆsin

2

ˆsin

2

ˆˆsin

2

ˆsin

2

ˆcos

2

ˆˆcos

2

ˆˆsin

2

ˆcos2

2

ˆˆcos

2

ˆsin2

2

ˆsin

2

ˆcos

2

ˆsin2

1

c

ba

c

ba

bac

bac

cc

Therefore

2

ˆsin

2

ˆˆcos

2

ˆcos

2

ˆˆcos

bac

2

ˆsin

2

ˆˆsin

2

ˆsin

2

ˆˆcos

bac

26




(1777 – 1855)

22

SOLO

A1

B1

C1

B

a

bc

C

A

Napier’s Equations [4]

John Napier of Merchiston (1550 –1617)

Divide (23) by (25)2

ˆcot

2

ˆˆcos

2

ˆˆcos

2

ˆˆtan

ba

ba

to obtain

28

Divide (24) by (26)2

ˆcot

2

ˆˆsin

2

ˆˆsin

2

ˆˆtan

ba

ba

to obtain

27


2

ˆˆsin

2

ˆˆsin

2

ˆcot

2

ˆˆtan

ba

ba

2

ˆˆcos

2

ˆˆcos

2

ˆcot

2

ˆˆtan

ba

ba

23

SOLO

A1

B1

C1

B

a

bc

C

A

Napier’s Equations (continue)

Divide (24) by (23)

to obtain

29

Divide (26) by (25)

to obtain

30

2

ˆˆtan

2

ˆtan

2

ˆˆsin

2

ˆˆsin bac

2

ˆˆtan

2

ˆtan

2

ˆˆcos

2

ˆˆcos bac


John Napier of Merchiston (1550 –1617)


2

ˆˆsin

2

ˆˆsin

2

ˆtan

2

ˆˆtan

c

ba

2

ˆˆcos

2

ˆˆcos

2

ˆtan

2

ˆˆtan

c

ba

24

SOLO

Napier’s Rules for Right Angles Spherical Triangles

Let γ = 90 T , i,e, a Right Angle Spherical Triangle.Napier arranged the quantities as in the Right Circle . Any of t6he parts of this Circle is called a Middle Part, the two neighboring parts are called Adjacent Parts, and the two remaining parts are Opposite Parts.

ˆcosˆcosˆtanˆtanˆsin





coaccobco

bacococco

bcoccoaco

coccocoab

ccococoba


I

Ecuator

1R

2R

Ex

Ey

Ez

1

2

TrajectoryGreat Circcle

1 2

A

B'B

1R

2R1

2

O

A

B

Ca

b

c

Earth Center

North Pole

90ˆ

b a

coco

cco ˆ

co = complement

ccco

co

co

ˆ90ˆ

ˆ90ˆ

ˆ90ˆ

Napier’s Rules for Right Angles Spherical Triangles are

• The sine of any Middle Part equals the product of the tangents of the Adjacent Parts.

• The sine of any Middle Part equals the product of the cosines of the Opposite Parts.

25

SOLO

Napier’s Rules for Right Angles Spherical Triangles

Let γ = 90 T , i,e, a Right Angle Spherical Triangle.Napier arranged the quantities as in the Right Circle . Any of t6he parts of this Circle is called a Middle Part, the two neighboring parts are called Adjacent Parts, and the two remaining parts are Opposite Parts.

ˆsinˆcosˆcotˆtanˆcos

ˆcosˆcosˆcotˆcotˆcos

ˆcosˆsinˆcotˆtanˆcos

ˆsinˆsinˆcotˆtanˆsin

ˆsinˆsinˆcotˆtanˆsin

acb

bac

bca

cab

cba



I

Ecuator

1R

2R

Ex

Ey

Ez

1

2


1 2

A

B'B

1R

2R1

2

O

A

B

Ca

b

c

Earth Center

North Pole

90ˆ

b a

coco

cco ˆ

co = complement

ccco

co

co

ˆ90ˆ

ˆ90ˆ

ˆ90ˆ

Napier’s Rules for Right Angles Spherical Triangles are

• The sine of any Middle Part equals the product of the tangents of the Adjacent Parts.

• The sine of any Middle Part equals the product of the cosines of the Opposite Parts.

The final result is

26

Navigation SOLO

I

Ecuator

1R

2R11,

Ex

Ey

Ez

1

222 ,

,

12 TrajectoryGreat Circcle

1 2

0

Flight on Earth Great Circles

The Shortest Flight Path between two points 1 and 2 on the Earth is on the Great Circles (centered at Earth Center) passing through those points.

The Great Circle Distance between two points 1 and 2 is ρ.The average Radius on the Great Circle is a = (R1+R2)/2

a kmNmNma 852.11deg/76.60/

1

2111 ,, R

222 ,, RR – radiusϕ - Latitudeλ - Longitude

27

Navigation SOLO

I

Ecuator

1R

2R11,

Ex

Ey

Ez

1

222 ,

,


1 2

0

1

2


The Great Circle Distance between two points 1 and 2 is ρ.

a

1

2111 ,, R


212121 cos90sin90sin90cos90cos

/coscos

a

From the Law of Cosines for Spherical Triangles

or

212121 coscoscossinsin/cos a

2121211 coscoscossinsincos a

The Initial Heading Angle ψ0 can be obtained using theLaw of Cosines for Spherical Triangles as follows

a

a

/sincos

/cossinsincos

1

120

2

222

22221

coscoscossinsin1cos

coscoscossinsinsinsincos

The Heading Angle ψ from the Present Position (R,ϕ,λ) to Destination Point (R2,ϕ2,λ2)

28

Navigation SOLO

I

Ecuator

1R

2R11,

Ex

Ey

Ez

1

222 ,

,


1 2

0

1

2


The Distance on the Great Circle between two points 1 and 2 is ρ.

1

2111 ,, R


The Time required to travel along the Great Circle between points 1 and 2 is given by

22

2121211 coscoscossinsincos

yxHoriz

HorizHoriz

VVV

V

a

Vt

2121211 coscoscossinsincos a

29

Navigation SOLO

I

Ecuator

1R

2R

Ex

Ey

Ez

1

2


1 2

1R

2R1

2

O

A

B

Ca

b

c

c

Earth Center

North Pole

A

B'B90

P


1

2111 ,, R

222 ,, R

If the Aircraft flies with an Heading Error Δψ we want to calculate the Down Range Error Xd and Cross Range Error Yd, in the Spherical Triangle APB.

R – radiusϕ - Latitudeλ - Longitude

Using the Law of Cosines for Spherical Triangle APB we have

aaYd /sin

90sin

/sin

sin

2/sin/sin

/cos/cos/cos0ˆcos 21

90ˆ RRa

aYaX

aYaXaP

dd

ddP

Using the Law of Sines for Spherical Triangle APB we have

aY

aaX

dd /cos

/coscos 1

sin/sinsin 1 aaYd


30

SOLO

References

http://en.wikipedia.org/wiki/

[1] Lass H, “Vector and Tensor Analysis”, McGraw-Hill, 1950, pp.25-27

[2] Wertz, J. editor, “Spacecraft Attitude Determination andControl”,

Appendix A, D.Reidel Publishing Company, 1985[3] “ASM Handbook of Engineering Mathematics”, American Society for

Metals, 1983, pp.86-89

[4] Tuma, J.J., “Engineering Mathematics Handbook”, 3 Edition, McGraw-Hill, pp.34-35



http://mathworld.wolfram.com/SphericalTrigonometry.html

April 13, 2023 31

SOLO

TechnionIsraeli Institute of Technology

1964 – 1968 BSc EE1968 – 1971 MSc EE

Israeli Air Force1970 – 1974

RAFAELIsraeli Armament Development Authority

1974 – 2013

Stanford University1983 – 1986 PhD AA

Spherical trigonometry

Science

Transcript of Spherical trigonometry