158 MR WALLACE'S Formula for finding the Logarithms 



Log cos x = log tan x log sin x, 



therefore 



A log cos x = A log tan a: A log sin x, 



and hence, again, 



( cosx cos (x+ Ax) 'A log sin x. 

 But from formula (A) art. 1 . we have also 



cos x cos (x + Ax) ( A log cos x) = sin x sin (x + AX) A log sin x, 



Hence, by putting the second members of these equations equal 

 to one another, and transposing, we find, 



{cos x cos (x + A x) + sin x sin (x + A x) } -A log sin x = cos xcos (x + Ax) A log tan x. 



The expression which forms the coefficient of A log sin x is 

 by the calculus of sines equivalent to cos A x, therefore 



cos A x (A log sin x) cos x cos (x + AX) -A log tan x . (8) 



By a mode of proceeding in all respects similar to the above, 

 we easily find 



cos A x ( A log cos x) = sin x sin (x + A x) A log tan x .' (9) 



As the omission of the factor cos A x will not affect the first 

 two terms of the series, which express the values of A log sin #, 

 and A log cos x, we may leave it out, and have more simply 



A log sin x = cos x cos (x + A x) A log tan x ; 



V- ' ' ; ' "'< E ) 



A log cos x = sin x sin (x + A x) A log tan x ; 



8. The degree of accuracy of these approximations may be 

 estimated by comparing the series which express the exact va- 

 lues of A log sin x, A log cos #, and A log tan oc . 



We have already found, art. 1, that 



1 (A* 1 ) 2 cos x (AX)* 

 Alogsm* = m ^ .Aj-j^-g- + ^- SBjt _ &c. 



( sin x 1 (A x)* sin x A x s 1 



A log cos x = m i - . A x + z -- - + 3 -- - + &c. } ; 

 (. cos x cos 1 x 2 cos 3 x 3 j 



