Prosthaphaeresis
Amy S. Gordon
In partial fulfillment of the MAT Degree
Department of Mathematics
University of Nebraska-Lincoln
Dr. David Fowler - Advisor
July, 2011
Gordon – Expository 2
Prosthaphaeresis, pronounced pross thuh fair ee sis, is a little known method to multiply or
divide numbers using sine and cosine values of angles. It was used briefly during the late 16th
and early 17th centuries according to Wikipedia, which also states that the word prosthaphaeresis
comes from two Greek expressions, “prosthesi” meaning addition and “afairesi” meaning
subtraction. The method involves using the four trigonometric identities listed below as well as
tables of trigonometric or inverse trigonometric functions.
sin α sin β = ½ (cos(α – β) – cos(α + β))
cos α cos β = ½ (cos(α – β) + cos(α + β))
sin α cos β = ½ (sin(α + β) + sin(α – β))
cos α sin β = ½ (sin(α + β) – sin(α – β))
Prosthaphaeresis in Action
To get an understanding of how the method works a few examples are shown in steps:
Example 1:
•
9,421 x 628
Choose one of the following trigonometric identities:
sin α sin β = ½ (cos(α – β) – cos(α + β))
cos α cos β = ½ (cos(α – β) + cos(α + β))
sin α cos β = ½ (sin(α + β) + sin (α – β))
cos α sin β = ½ (sin(α + β) – sin(α – β))
•
using this one
Convert the two numbers being multiplied to decimal values. This will not change the
answer, only the decimal placement which can be restored later.
9,421 x 10-4 = (.9421) and 628 x 10-3 = (.628)
•
View the decimal values as cosine values (since that is the equation chosen) and look up
the angles which correspond to their values in a trig table.
Using the table in Appendix A find the angle that corresponds to the decimals;
cos-1(.9421) = 20° and the angle that corresponds to cos-1(.628) = 51°.
•
Substitute the angle values into the equation chosen in step 1 and again use the trig table
to find their equivalent values.
(To avoid a negative inside the first parentheses switch α and β which will not make a
difference in the answer since cos(20) x cos(51) will give the same answer as cos(51) x
cos(20))
Gordon – Expository 3
cos α cos β = ½ (cos(α – β) + cos(α + β))
cos(51) cos(20) = ½ (cos(51 - 20) + cos(51 + 20))
= ½ (cos(31) + cos(71))
= ½ (.8572 + .3256)
= ½ (1.1828)
= .5914
•
Restore the magnitude by replacing the decimal places that were taken out in step 2.
10-4 x 10-3 = 10-7 so we must put back 7 decimal places.
Therefore the answer to 9,421 x 628 is ≈ 5,914,000.
Switching the identity used to solve the problem will not change the result, which will be
shown in example 2. Here the same problem from example 1 will be used but with the identity
sin α cos β = ½ (sin (α + β) + sin (α – β)).
Example 2:
9,421 x 628
Again --- 9,421 x 10-4 = (.9421) and 628 x 10-3 = (.628)
•
Using the table in Appendix A it can be found that the angle corresponding to sin-1(.9421)
is 70° and the angle corresponding to cos-1(.628) is 51°.
•
Substitute the angle values into the equation above and again use the trig table to find
their corresponding values.
sin α cos β = ½ (sin(α + β) + sin(α - β))
sin (70) cos(51) = ½ (sin(70 + 51) + sin(70 - 51))
= ½ (sin(121) + sin(19))
121° must be translated into an angle measure between 0 - 90° which are the only angles
contained in trig tables. But knowledge of trigonometry and the unit circle relationship gives the
fact that every angle after 90° corresponds to an angle between 0 - 90°. Since 121° is 31° past
90°, the angle to which 121° corresponds in the 1st quadrant is 31° less than 90°, or 59°, and in
the second quadrant sine is positive. Thus:
= ½ (sin(59) + sin(19))
= ½ (.8572 + .3256)
= ½ (1.1828)
= .5914
The approximate answer is now obtained by multiplying .5914 by 107; 9421 x 628 ≈ 5,914,000.
Gordon – Expository 4
•
A note: I first used sin-1(.9421) as being equal to 71° rather than 70° since estimating
from the trig table that it was .001 closer to sin71° than it was to sin 70°, which gave an
answer of 5,950,000 and a error of about .0057 percent. I later changed it to 70° after
testing calculations on a calculator which gave sin-1(.9421) = 70° and in turn gave the
same answer as example 1.
This glitch sparked wonderment if all four of the identities would yield the exact same
answer, so again using the same problem and the other two identities I went ahead and verified
that yes, all four identities gave the same estimate for 9,421 x 628. The work of verifying this is
included in Appendix B with examples 3 and 4.
Looking at the result, 5,914,000, and comparing to the actual answer, 5,916,388 it can be
seen that the error is quite small.
(5,916,388 - 5,914,000) / 5,916,388 = .000403625 = .040 percent.
But will the error always be so small? To get a better idea more numbers needed to be
tested. The work is as follows…
sin-1(.8)sin-1(.2) = ½ (cos(α – β) – cos(α + β))
sin(53) sin(12) = ½ (cos(41) – cos(65))
= ½ (.7547 - .4226)
= ½ (.3321)
= .16605
The approximate answer is now obtained by multiplying .16605 by 102. Thus 8 x 2 is
approximately 16.605.
Example 5a: 8 x 2
Having an error of 3.8 percent on such a simple problem did not seem acceptable. I
debated about what would bring more accuracy to the problem. More precise cosine and sine
values seemed to be in order. Although a few tables with a couple more significant digits were
located on the web, I decided to use a calculator to look at digits beyond four or five decimal
places.
Example 5b: 8 x 2
sin-1(.8)sin-1(.2) = ½ (cos(α – β) – cos(α + β))
sin(53.130) sin(11.537) = ½ (cos(41.593) – cos(64.667))
= ½ (.74788 - .42788)
= ½ (.32)
= .16
Gordon – Expository 5
The approximate answer is again obtained by multiplying .16 by 102. So 8 x 2 is
approximately16 or in this case exactly 16.
Adding the decimal places from the calculator, the error was eliminated in this problem.
Since such a small change in significant digits impacted the accuracy with a simple problem,
would it make the same difference for a larger problem?
Example 6a: 345,269 x 7,081
(done only with trig table)
cos α sin β = ½ (sin(α + β) – sin(α – β))
cos-1(.345269) sin-1(.7081) = ½ (sin(α + β) – sin(α – β))
cos(70)sin(45) = ½ (sin (115 65) – sin (25))
= ½ (.9063 - .4226)
= ½ (.4837)
= .24185
The approximate answer is now obtained by multiplying this answer, .24185 by 1010. Thus
345,269 x 7,081 is approximately 2,418,500,000. The actual answer is 2,444,849,789 which will
give an error of 1.1 percent.
Example 6b: 345,269 x 7,081
(done with calculator)
cos α sin β = ½ (sin(α + β) – sin(α – β))
cos (.345269) sin-1(.7081) = ½ (sin(α + β) – sin(α – β))
cos(69.802)sin(45.081) = ½ (sin (114.883) – sin (24.721))
= ½ (.9071689 - .4182000)
= ½ (.4889689)
= .24448445
The approximate answer is again obtained by multiplying .24448445 by 1010. Therefore;
345,269 x 7,081 is approximately 2,444,844,500. The actual answer is 2,444,849,789 which will
give an error of 0.0002 percent.
-1
Even though this error was very small I wondered if it were possible to get no error…
Example 6c: 345,269 x 7,081
(calculator & more decimal places)
cos α sin β = ½ (sin(α + β) – sin(α – β))
cos (.345269) sin-1(.7081) = ½ (sin(α + β) – sin(α – β))
cos(69.80178261)sin(45.08053563) = ½ (sin (114.8823182) – sin (24.72124698))
= ½ (.9071739051 - .4182039469)
= ½ (.4889699582)
= .2444849791
The approximate answer is obtained by multiplying .2444849791 by 1010. Thus, 345,269 x
7,081 is approximately 2,444,849,791.
-1
Gordon – Expository 6
Here, a difference of 2 units makes the error: 0.000000082 percent, which is all but
negligible and the closest I could get to no error since was the maximum amount of decimal
places allowed on the calculator.
After performing the problems above with a calculator, to ensure more decimal places,
and reading a paper by Nicholas Rose, in which Rose reported that there were trigonometric
tables of up to fifteen decimal places, I am convinced that the 16th century method of
prosthaphaeresis was extremely accurate. By adding in only a few more decimal places to my
calculations with the calculator, the error reduced significantly and taking the decimal places to
the tenth place gave an error almost negligible. Another thought to improve accuracy is that
instead of the table having only degrees in whole number format, they could be figured to the
nearest tenth or hundredth of a degree.
The method can also be used to divide numbers. The process is in essence the same but
the anti-trigonometry tables are used; finding secant and cosecant angles values instead of sine
and cosine values.
Proof
The following is a proof of the identity cos α cos β = ½ (cos(α – β) + cos(α + β)) used in
the prosthaphaeresis method. The figures constructed are on a unit circle centered on orthogonal
axes centered at point A. AE and AC are radii of length 1. Intersecting segments at points D, G,
H, and I are perpendicular.
We aim to show that AG = cos α cos β
Fact: cosine =
𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝐴𝐹
1. Thus cos β = 𝐴𝐸 =
𝐴𝐹
1
= AF
2. Considering triangle AHE we see that
𝐴𝐻
𝐴𝐻
cos (α + β) = 𝐴𝐸 = 1 = AH
3. Considering triangle AGF, where AG will be referred to as x
and AF as r:
Gordon – Expository 7
𝑥
It can be seen that cos 𝛼 = 𝑟 or x = r cos α
From #1, cos β = AF = r
Therefore x = cos β cos α
We now aim to show that IF = sin α sin β
4. Fact: sine =
𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒
𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝐸𝐹
Thus sin β = 𝐴𝐸 =
𝐸𝐹
1
= EF
5. Note that IF is parallel to AD by construction.
∠EIF = 90° by construction & ∠IEF = 90° – ∠IFE, since the sum of angles in a triangle is 180°
∠IFA = 90° - ∠ IFE since they are complementary angles within the right angle EFA
By the transitive property of equality ∠IFA = ∠IEF, since they both equal 90° - ∠ IFE
∠IFA = ∠α since they are alternate interior angles and again by the transitive property of
equality ∠IEF = α
Therefore IF is the sine of ∠IEF and in turn sin α
6. Consider triangle EIF, where IF will be referred to as y and EF as r:
𝑦
It can be seen sin 𝐼𝐸𝐹 = 𝑟 and as shown in #5 ∠IEF = α thus y = r sin α
From #4 it is seen that sin β = r
Therefore y = sin β sin α
7. Now we must look at what has been shown thus far Looking at the picture we know:
AG(JK) = cos α cos β from #3
HG(LM) = sin α sin β from #6
AH(NO) = cos (α + β) from #2
Thus it can be seen that cos(α + β) = cos α cos β – sin α sin β
We now aim to show that
cos(α + β) = cos α cos β – sin α sin β is the same as cos α cos β = ½ [cos(α + β) + cos(α – β)]
Gordon – Expository 8
8. Since the cosine is an even function and the sine is an odd function, we can derive a
subtraction formula for the cosine.
cos (α – β) = cos (α + (-β))
= cos (α) cos (-β) – sin α sin (-β)
= cos (α) cos (β) – sin α (-sin (β))
= cos (α) cos (β) + sin α sin (β)
Therefore:
cos (α + β) = cos α cos β – sin α sin β
+ cos (α – β) = cos α cos β + sin α sin β
cos (α + β) + cos (α – β) = 2 (cos α cos β) or cos α cos β = ½ [cos(α + β) + cos(α – β)]
The other three identities are proved in similar fashion.
Origins of Prosthaphaeresis
In the 16th century a need arose for accurate and speedy calculations for mathematicians,
astronomers and cartographers who labored to solve involved spherical trigonometry problems.
The standard spherical trigonometry problem concerned triangles on the surface of a sphere
(earth) making one side an arc. These individuals were solving navigational, cartographic and
astronomic problems, which often required repeated multiplications and divisions of sine and
cosine values to get a single unknown. And the calculations were usually seven digit numbers
being multiplied together by hand. The repeated operational processes were long and had to be
repeated over and over again, which became tedious. Two areas in which prosthaphaeresis
became useful were in preparing ephemeral tables, which tell when and where certain objects
would be in the sky and in charting a course for a ship’s travels, both of which required very
accurate measurements.
To ease the tediousness of the repeated multiplications, mathematicians searched for
speedy alternatives, which brought prosthaphaeresis. Many individuals have been credited with
discovering, attributing to or being in dispute about discovering the method of prosthaphaeresis.
Each research lead gave a differing view on who was responsible for developing it. The
trigonometric identity for the product of two sine values, which prosthaphaeresis utilizes, first
appeared in a paper by Johannes Werner in 1510. Werner’s focus was spherical trigonometry and
his Wikipedia page makes the statement that some authors attribute him to inventing
prosthaphaeresis, but in his time they were called Werner’s formulas. The next mention of
Gordon – Expository 9
prosthaphaeresis that was found skipped ahead to 1582, when Christopher Clavius, a German
priest, mathematician and astronomer was said to have devised a clever method for speeding up
calculations. His Wikipedia page credits Clavius with showing how to use the trigonometric
identity cos α cos β = ½ [cos (α + β) + cos (α – β)] to make the work of long problems faster. In
an article written in 1999, Tycho Brahe, a Danish astronomer, was said to have used the method
to great effect, but further research showed Brahe was in dispute over claims to the
prosthaphaeretic method. The dispute was between Brahe and Nicolai Reymers Baer, aka Ursus.
Brahe claimed that he had played a key role in the mathematical discovery around 1580, whereas
Ursus claimed that that it had been the work of two other mathematicians, Paul Wittich and Jost
Burgi, whom Brahe had worked for. Since Brahe was a poor mathematician in his time, his
claims were dismissed in favor of the more wealthy prominent Ursus. An interesting side note is
a mention on a recent television show of a Tycho Brahe exhibit in a museum being stolen for its
hidden mathematical message. He would have relished the limelight. As for Ursus’ claims,
prosthaphaeresis is not mentioned, with confidence, in association with Wittich but the method is
mentioned with great connection to Burgi, who is said to have been a major contributor to
prosthaphaeresis before inventing logarithms. That statement is a contradiction to the better
known theory that logarithms were developed by John Napier. The website faults Burgi with
waiting too long to publish his findings, thus Napier’s/Briggs better timed book of logarithms
made the way to notoriety. The publication of logarithms in 1614 was in essence the end to the
prosthaphaeretic method.
Prosthaphaeresis vs. logarithms
The similarity between prosthaphaeresis and logarithms is noticeable. It is believed that
Napier learned of prosthaphaeresis from a friend who had visited Tycho’s observatory in 1590,
but his observation that exponents also have product to sum properties inspired his work on
logarithms. Napier’s logarithms utilized a table like prosthaphaeresis. The tables became books
filled with common logarithms from 0 – 100000, published by Briggs, who was Napier’s
assistant who assumed the task after Napier’s death. A small excerpt from the books can be seen
on page two of Appendix A. Using logarithms essentially require a search of the tables three
times and one addition to finish the problem.
Gordon – Expository 10
The logarithm of a number, as defined by Wikipedia, is an exponent by which a fixed
number called the base must be raised to in order to produce that number. While
prosthaphaeresis sped up the calculations of numbers it also had disadvantages, according to
author Brian Borchers, such as it required very accurate tables of cosine and secant functions,
and in the 16th century the degrees-minutes-seconds system of angle measurements was still in
use which made computing the sum and difference of two angles somewhat difficult. Also
prosthaphaeresis could not be used to compute exponents as mentioned earlier. Division was
somewhat inconvenient using the method according to Borchers; the process was the same
except the use of the inverse trigonometric tables were required.
An example comparing the two methods follows below:
Example 6a: 25.2 x 120 using prosthaphaeresis (with calc for more significant digits)
cos α sin β = ½ (sin(α + β) – sin(α – β))
cos-1(.252) sin-1(.12) = ½ (sin(α + β) – sin(α – β))
cos 75.404° sin 6.892° = ½ (sin(82.296) – sin(68.512))
= ½ (.99097 – .93049)
= ½ (.06048)
= .03024 x 105 = 3,024; (actual 3,024 which has error of 0 percent).
Example 6b: 25.2 x 120 using logarithms
Excerpt from table of logarithms ….
N
1.2
0
1
2
3
4
5
6
7
8
9
07918
08279
08636
08991
09342
09691
10037
10380
10721
11059
39794
39967
40140
40312
40483
40654
40824
40993
41162
41330
47712
47857
48001
48144
48287
48430
48572
48714
48855
48996
...
2.5
...
3.0
1. Rewrite the two numbers in scientific notation (A x 10B).
25.2 = 2.52 ×101 & 120 = 1.20 ×102
Gordon – Expository 11
2. Look up the values in the table of logarithms.
log(2.52) = .40140 and log(1.20) = .07918
3. Add the two values, if the number is greater than 1 you must subtract 1.
Compute .40140+.07918 =.48058 (don’t have to subtract 1)
4. Look in the body of the table for the number closest to the result and use the row and
column headers to determine what number it corresponds to. This is the value of “A” for the
product.
The value .48058 lies between .48001 and .48144, but it's closer to
.48001, which corresponds to the value 3.02
5. Add the two “B” values (the exponents from 1), plus 1 more if you subtracted 1 in step 3.
This is the value of “B” for the product.
Add the exponents 1 + 2 = 3. Didn’t have to add 1so total exponent is 3.
The product is therefore approximately 3.02 × 103 = 3,020. (The exact answer 3,024)
The prosthaphaeresis method yielded an exact answer and the logarithm method only had
a 0.13 percent error. Each method was fairly simple to perform once an understanding of how to
read the tables was mastered. After viewing pages and pages of logarithm tables a personal
preference for using trig tables surfaced since all their values were contained between the
numbers 0 to 90. But the inability to perform exponential calculations with prosthaphaeresis
made it clear why logarithms overtook the method. While it had some drawbacks
prosthaphaeresis served its purpose very well. It was most certainly a stepping stone to
logarithms and eventually the slide rule, which remained in use for hundreds of years until the
calculator came along. The drawback of not being able to calculate with exponents is what
caused the search for an even better, more versatile method to multiply, divide and deal with
exponents.
Gordon – Expository 12
Classroom Activity
Algebra Lesson Plan
Date: _____________
Objective: Students will be introduced to an early method of multiplication
which involves trigonometric and logarithmic tables.
Materials: Copy of trigonometry table, excerpt from logarithm table & example.
Bell work: Students will calculate the following in their notebooks with no calculator.
123,456 x 7,890,123
Procedures:
•
Discuss the bell work problem. Compare answers; try to figure out where they made
mistakes. Lead into discussion of math in the early ages without calculators.
•
Show power point highlighting the following …
a) How the right triangle ties into spherical trigonometry
b) Look at the unit circle and relate to coordinate plane
c) Show a trig table and log table
d) Discuss what a logarithm is
e) Look at some math that astronomers and cartographers performed
•
Example of multiplication problem solved using prosthaphaeresis (Appendix C).
a) Let students look over the problem and discuss in their group
b) Have students relate similar problem to the example and work through it
•
As a class, work the same example using the provided logarithm tables.
•
Wrap up the lesson by comparing and contrasting the four methods.
a) Working it out by hand
b) Prosthaphaeresis
c) Logarithms
d) Using a calculator
•
Assign 3 larger multiplication problems students must solve using the trig tables and
prosthaphaeresis.
Gordon – Expository 13
References
Henry Briggs. (last modified 10/30/02) Retrieved from Wikipedia:
http://cerebro.xu.edu/math/math147/02f/briggs/briggsintro.html. Retrieved on June 1, 2011.
Johannes Werner (n.d.) Retrieved from Wikipedia:
http://en.wikipedia.org/wiki/Johannes_Werner.
Retrieved on June 8, 2011.
Jost Burgi. (n.d.) Retrieved from Wikipedia: http://en.wikipedia.org/wiki/Jost_B%C3%BCrgi.
Retrieved on June 1, 2011.
Reading a Log Table. (n.d.) Retrieved from Wikispots:
http://obsoleteskills.wikispot.org/Reading_a_log_table. Retrieved on June 8, 2011.
Rose, Nicholas J. (2006) Prosthaphaeresis. http://www4.ncsu.edu/~njrose/pdfFiles/Prostha.pdf.
Retrieved on June 1, 2011.
Prosthaphaeresis. (n.d.) http://www.nmt.edu/~borchers/prost.pdf. Retrieved on June 1, 2011.
Prosthaphaeresis. (n.d.) Retrieved from Wikipedia:
http://en.wikipedia.org/wiki/Prosthaphaeresis. Retrieved on June 1, 2011.
Tycho Brahe and Mathematical Techniques. (1999)
http://www.hps.cam.ac.uk/starry/tychomaths.html. Retrieved on June 6, 2011.
Gordon – Expository 14
Appendix A
Trigonometric Table
Degrees
Radians
Sin
Cos
Tan
Degrees
Radians
Sin
Cos
Tan
0
0.00000
0.00000
1.00000
0.00000
46
0.80285
0.71934
0.69466
1.03553
1
0.01745
0.01745
0.99985
0.01746
47
0.8203
0.73135
0.682
1.07237
2
0.03491
0.0349
0.99939
0.03492
48
0.83776
0.74314
0.66913
1.11061
3
0.05236
0.05234
0.99863
0.05241
49
0.85521
0.75471
0.65606
1.15037
4
0.06981
0.06976
0.99756
0.06993
50
0.87266
0.76604
0.64279
1.19175
5
0.08727
0.08716
0.99619
0.08749
51
0.89012
0.77715
0.62932
1.2349
6
0.10472
0.10453
0.99452
0.10510
52
0.90757
0.78801
0.61566
1.27994
7
0.12217
0.12187
0.99255
0.12278
53
0.92502
0.79864
0.60182
1.32704
8
0.13963
0.13917
0.99027
0.14054
54
0.94248
0.80902
0.58779
1.37638
9
0.15708
0.15643
0.98769
0.15838
55
0.95993
0.81915
0.57358
1.42815
10
0.17453
0.17365
0.98481
0.17633
56
0.97738
0.82904
0.55919
1.48256
11
0.19199
0.19081
0.98163
0.19438
57
0.99484
0.83867
0.54464
1.53986
12
0.20944
0.20791
0.97815
0.21256
58
1.01229
0.84805
0.52992
1.60033
13
0.22689
0.22495
0.97437
0.23087
59
1.02974
0.85717
0.51504
1.66428
14
0.24435
0.24192
0.9703
0.24933
60
1.0472
0.86603
0.5
1.73205
15
0.2618
0.25882
0.96593
0.26795
61
1.06465
0.87462
0.48481
1.80405
16
0.27925
0.27564
0.96126
0.28675
62
1.0821
0.88295
0.46947
1.88073
17
0.29671
0.29237
0.9563
0.30573
63
1.09956
0.89101
0.45399
1.96261
18
0.31416
0.30902
0.95106
0.32492
64
1.11701
0.89879
0.43837
2.0503
19
0.33161
0.32557
0.94552
0.34433
65
1.13446
0.90631
0.42262
2.14451
20
0.34907
0.34202
0.93969
0.36397
66
1.15192
0.91355
0.40674
2.24604
21
0.36652
0.35837
0.93358
0.38386
67
1.16937
0.9205
0.39073
2.35585
22
0.38397
0.37461
0.92718
0.40403
68
1.18682
0.92718
0.37461
2.47509
23
0.40143
0.39073
0.9205
0.42447
69
1.20428
0.93358
0.35837
2.60509
24
0.41888
0.40674
0.91355
0.44523
70
1.22173
0.93969
0.34202
2.74748
25
0.43633
0.42262
0.90631
0.46631
71
1.23918
0.94552
0.32557
2.90421
26
0.45379
0.43837
0.89879
0.48773
72
1.25664
0.95106
0.30902
3.07768
27
0.47124
0.45399
0.89101
0.50953
73
1.27409
0.9563
0.29237
3.27085
28
0.48869
0.46947
0.88295
0.53171
74
1.29154
0.96126
0.27564
3.48741
29
0.50615
0.48481
0.87462
0.55431
75
1.309
0.96593
0.25882
3.73205
30
0.5236
0.5
0.86603
0.57735
76
1.32645
0.9703
0.24192
4.01078
31
0.54105
0.51504
0.85717
0.60086
77
1.3439
0.97437
0.22495
4.33148
32
0.55851
0.52992
0.84805
0.62487
78
1.36136
0.97815
0.20791
4.70463
33
0.57596
0.54464
0.83867
0.64941
79
1.37881
0.98163
0.19081
5.14455
34
0.59341
0.55919
0.82904
0.67451
80
1.39626
0.98481
0.17365
5.67128
35
0.61087
0.57358
0.81915
0.70021
81
1.41372
0.98769
0.15643
6.31375
36
0.62832
0.58779
0.80902
0.72654
82
1.43117
0.99027
0.13917
7.11537
Gordon – Expository 15
Degrees
Radians
Sin
Cos
Tan
Degrees
Radians
Sin
Cos
Tan
37
0.64577
0.60182
0.79864
0.75355
83
1.44862
0.99255
0.12187
8.14435
38
0.66323
0.61566
0.78801
0.78129
84
1.46608
0.99452
0.10453
9.51436
39
0.68068
0.62932
0.77715
0.80978
85
1.48353
0.99619
0.08716
11.4301
40
0.69813
0.64279
0.76604
0.83910
86
1.50098
0.99756
0.06976
14.3007
41
0.71558
0.65606
0.75471
0.86929
87
1.51844
0.99863
0.05234
19.0811
42
0.73304
0.66913
0.74314
0.90040
88
1.53589
0.99939
0.0349
28.6363
43
0.75049
0.682
0.73135
0.93252
89
1.55334
0.99985
0.01745
57.29
44
0.76794
0.69466
0.71934
0.96569
90
1.5708
1
6.1E-17
45
0.7854
0.70711
0.70711
1.00000
Logarithm Table Excerpt
1.000
1.001
1.002
1.003
1.004
1.005
1.006
1.007
1.008
1.009
1.010
1.011
1.012
1.013
1.014
1.015
1.016
1.017
1.018
1.019
1.020
1.021
1.022
0.00000000
0.00043408
0.00086772
0.00130093
0.00173371
0.00216606
0.00259798
0.00302947
0.00346053
0.00389117
0.00432137
0.00475116
0.00518051
0.00560945
0.00603795
0.00646604
0.00689371
0.00732095
0.00774778
0.00817418
0.00860017
0.00902574
0.00945090
1.10
1.11
1.12
1.13
1.14
1.15
1.16
1.17
1.18
1.19
1.20
1.21
1.22
0.0413927
0.0453230
0.0492180
0.0530784
0.0569049
0.0606978
0.0644580
0.0681859
0.0718820
0.0755470
0.0791812
0.0827854
0.0863598
2.00
2.01
2.02
2.03
2.04
2.05
2.06
2.07
2.08
2.09
2.10
2.11
2.12
2.13
2.14
2.15
2.16
2.17
2.18
2.19
2.20
2.21
2.22
0.3010300
0.3031961
0.3053514
0.3074960
0.3096302
0.3117539
0.3138672
0.3159703
0.3180633
0.3201463
0.3222193
0.3242825
0.3263359
0.3283796
0.3304138
0.3324385
0.3344538
0.3364597
0.3384565
0.3404441
0.3424227
0.3443923
0.3463530
3.00
3.01
3.02
3.03
3.04
3.05
3.06
3.07
3.08
3.09
3.10
3.11
3.12
3.13
3.14
3.15
3.16
3.17
3.18
3.19
3.20
3.21
3.22
0.4771213
0.4785665
0.4800069
0.4814426
0.4828736
0.4842998
0.4857214
0.4871384
0.4885507
0.4899585
0.4913617
0.4927604
0.4941546
0.4955443
0.4969296
0.4983106
0.4996871
0.5010593
0.5024271
0.5037907
0.5051500
0.5065050
0.5078559
4.00
4.01
4.02
4.03
4.04
4.05
4.06
4.07
4.08
4.09
4.10
4.11
4.12
4.13
4.14
4.15
4.16
4.17
4.18
4.19
4.20
4.21
4.22
0.6020600
0.6031444
0.6042261
0.6053050
0.6063814
0.6074550
0.6085260
0.6095944
0.6106602
0.6117233
0.6127839
0.6138418
0.6148972
0.6159501
0.6170003
0.6180481
0.6190933
0.6201361
0.6211763
0.6222140
0.6232493
0.6242821
0.6253125
Gordon – Expository 16
Appendix B
Example 3:
9,421 x 628
Again --- 9,421 x 10-4 = (.9421) and 628 x 10-3 = (.628)
•
Using table in Appendix A it can be found that the angle corresponding to sin-1(.9421) is
70° and the angle that corresponds to sin-1(.628) is 39°.
•
Substitute the angle values into the 3rd equation and again use the trig table to find the
corresponding values.
sin α sin β = ½ (cos(α – β) – cos(α + β)),
sin (70) sin(39) = ½ (cos(70 - 39) – cos(70 + 39))
= ½ (cos(31) – cos(109))
Again, 109° must be translated into an angle measure between 0 - 90°. Since 109° is 19°
past 90° the angle to which it corresponds in the 1st quadrant is 19° less than 90° or 71°, and in
the second quadrant cosine is negative thus;
= ½ (cos(31) – [-cos(71)])
= ½ (.8572 – [-.3256])
= ½ (1.1828)
= .5914
The approximate answer is now obtained by multiplying this answer, .5914 by 107. Therefore;
9421 x 628 is approximately 5,914,000.
Example 4:
9,421 x 628
cos-1 (.9421) sin-1 (.628) = ½ (sin(α + β) – sin(α – β)),
cos (20) sin(39) = ½ (sin(20 + 39) – sin(20 - 39))
It is not possible to avoid the negative since cos(20) x sin(39) will not give the same
result as sin(20) x cos(39). Thus a negative value must be used; sin(-19). And from knowledge of
unit circles the sin (-α) = - sin α, since it would put the angle into the fourth quadrant where sine
is negative.
= ½ (sin(59) – sin(-19))
= ½ (.8572 – (-.3256))
= ½ (1.1828)
= .5914
The approximate answer is now obtained by multiplying this answer, .5914 by 107. Therefore;
9421 x 628 is approximately 5,914,000.
Gordon – Expository 17
Appendix C
16 century multiplication
th
Study the following problem which is solved using prosthaphaeresis.
•
Choose one of the following trigonometric identities: (Choose the one in boldface)
1. sin α sin β = ½ (cos(α – β) – cos(α + β)) 2. cos α cos β = ½ (cos(α – β) + cos(α + β))
3. sin α cos β = ½ (sin(α + β) + sin (α – β)) 4. cos α sin β = ½ (sin(α + β) – sin(α – β))
•
Convert the two numbers being multiplied to decimal values, the decimals can be put
back later.
9,421 x 628
9,421 x 10-4 = (.9421) and 628 x 10-3 = (.628)
•
View the decimal values as cosine values since I choose #2 and look up the angles which
correspond to their values in a trig table.
Using the table find the angle that corresponds to cos(.9421) = 20° and the angle
that corresponds to cos(.628) = 51°.
•
Substitute the angle values into the equation chosen in step 1 and again use the trig table
to find their corresponding values.
cos α cos β = ½ (cos(α – β) + cos(α + β))
cos(20) cos(51) = ½ (cos(20 - 51) + cos(20 + 51))
= ½ (cos(31) + cos(71))
= ½ (.8572 + .3256)
= ½ (1.1828)
= .5914
•
Restore the decimal places that you took out in step 2.
10-4 x 10-3 = 10-7 so we must put back 7 decimal places, thus .5914 becomes 5,914,000.
The answer to 9,421 x 628 is ≈ 5,914,000.
TASK: Answer the following problem on a separate piece of paper using the trig table and the
above example as a guide; be prepared to share your work with others.
493 x 6,022