LOGARITHMIC TABLES.] 



MATHEMATICS. TRIGONOMETRY. 



639 



= 1 and sin. S=S. 



we have cos. 

 .'. log. sin. (0 + )-log. sin. 0=log. ^1 + cotan. .08-- 



Now log. (1 + x) - M. ( x _ ^ + ^_ &c. 

 .'. log. sin. (9 + 8) log. sin. 9. 



M. 



M. 



. 6 - 



omitting 



=M. cotan. 



. ). 

 / 



. 



sin. 2 



Now if 5 be an angle of n" then 8 the circular measure 

 of this angle, equals n. sin. 1*. 



.'. L. sin. (0 + n*) L. sin. 0=M n ain. 1* cotan. 



M n sin. 1' 

 - X 



In like manner we may prove that 



L. tan. (0 + n") L. tan. 



(l nsin. 1" tan. 0\ 

 Now suppose that our tables are calculated for angles 



that differ by so small an angle $", that we can neglect 

 the second term of Equation (60) so that 



L. sin. (0 + ") L. sin. 6 = M S sin. 1" cotan. 0. 



Also, if n < S, a fortiori, 



L. sin. (9 + n") L. sin. = M n sin. 1* cotan. 0. 



Now, for any value of 6 the Tables give us 



L. sin. (9 + &") L. sin. 6 call this A 



A = M S sin 1" cotan. 0. 



.'. L. sin. (0 + n") L. sin. = - A 



i.e., when two angles are nearly equal, difference between 

 log. sines is proportional to the difference of angles. 



-^ is the difference corresponding to 1* 



o 



This 



can easily be multiplied by n. Also L. sin. is given 

 by the Tables, and hence we can find L. sin. (0 -j- n") 

 and conversely having given L. sin. (0 -f- n"), where 

 onlv L. sin. is given in tables, we can easily find the 

 n seconds. 



If the numerical values of formula (60) are taken for 

 different angles, it will be found that in order to make 

 the difference between two consecutive L. sines, given in 

 the Tables proportional to the number of seconds, we 

 must have = I" from up to 1. 30'., and S = 10" 

 from 1. SO 1 up to 5, and S = 60" from 5 up- 

 ward the log. sines being calculated to 7 places of 

 decimals.* 



The practical mode of employing the Tables will be 

 readily understood from the following examples : 



Thus, 



(1). Find L. sin. (15 11' 16'- 5). 



By tables L. Bin. 15 11' 



Do. diff. l'=77, 55 X 



6' = 



9-4181495 

 775,5 



4ur>, :5 

 38,7 



L. sin. (15 11' 16"' 6) - 9-4182775 

 (2). Find L. sin. (3 19* 37* ' 4). 



By tables L. ran. (3 s 19' 30") - 87634252 



* . . . ~ - - AA n . (.* ___ 2534 1 



4 _ 144', 9 



Do. diff, 1' - 362.3 



(3). 



Find 0. 



IfL. sin. 

 L. sin. 

 L. sin. (35 33'; 



L. sin. (3 W 37'- 4) = 8-70ati'J31 



Diff. 1" 



20-45 x 30 

 2 



.'. = 35. 33'. 32".8. 



NVe proceed in precisely the same manner with the 

 logarithms of the other trigonometric functions, with 

 this exception, that what is true of the L. sines and L. 

 tangents of small angles, is true of the L. cosines and L. 

 cotangents of angles which are nearly equal to 90. 

 For, since sin. = cos. (90 0) and tan. = cotan. 

 (90 0) then if be small, 9<r is nearly equal 

 to 90. 



(71). Another way of treating the L. tines of small angles. 



If the tables give the log. sines, (to., for intervals of 

 1' throughout, and not the refined tables for small angles 

 before spoken of, we can find the accurate value of a 

 small angle whose L. sine is given, and vice vertd, by the 

 following process. 



If be small, so that we can omit 0* s , <tc., we have 

 (Art 60) 



97645827 

 97644849 

 978 

 893-5 

 84-5 

 68-9 

 25-6 * 

 23-00 



sin. 



-6' 



&OOS. = 1 H- 



Hence 



sin. 

 



" = 1- 



Now suppose to contain n" 



.'. "^ e = cog. ifl 



n sm. V 



2 

 n sin. 1 



cos. 



, log. sin. 0-log. n. -log. sin. 1" = 5- log. cos. 



1 



10 



L. sin. 0-log. n-L. sin. 1" =*-^.L.cos. 3 



8 



o-k- cos. -f- L. sin. 1* -- 5- 



.'.L. sin. 0=log. n 



& L. sin. V = 4-6855749. 



.'. L. sin. = log. n + \ L. cos -f 1-3522416 



and log. n 



L. sin. - -. L. cos. - 1-3522416 







It is to be observed, that when 6 is very small, cos. 

 changes very slowly. Hence in the term L. cos. we 



See Tables at page 660, el leq. 



