ON THE TRIGONOMETRY OF THE PARABOLA. 85 



XXV. Let ^ sec (j) cf(p=p, \ sec x<'^X— 9 ' then as 



fsec<i>rfw=f sec ^rf^+ f secy <^X> ^^^ (5)' 

 \ sec (od<i)=p + q, and w=(^-'-x* 



Hence it' f be the amplitude which gives the tangential difference =p, and 

 X the amplitude which gives the tangential difference =g, ^-^x i^ the am- 

 plitude which will give the tangential difference =p + q. In the same way 

 we might show, that if ;// be the angle which gives this difference =r, 

 (0-i-X-J-i//) is the angle which will give this difference =p + g+r. 



Let a be the amplitude of the number A, and p its logarithm ; /3 the 

 amplitude of the number B, and q its logarithm ; r the amplitude of the 

 number C, and c its logarithm. Then 



A=seca + tana, B=sec/3+ tan/3, C = secy+tany, 

 and log A=p, log B=q, log C=r, or 



p+q + r= log A + log B + log C. 

 We have also 



ABC = (sec a + tan a)(sec ^ + tan j3)(sec y + tan y) 



= sec (a-i-/3-Ly) + tan (a-^/3-^y). 



Now as j9 is the logarithm of sec a+ tan a, q the logarithm of sec /3+ tan /3, 

 r the logarithm of sec y+ tan y, 



p + q+r is the log of sec(a-L/3-^y)+ tan(a-L/3-i-y), or of A B C, 



as shown above. We may therefore conclude that 



log (ABC)=log A + log B + log C (29) 



XXVI. If e be the angle which gives the difference between the parabolic 

 arc and its subtangent equal to ?n, (e-'-e) is the angle which will give this 

 difference equal to 2m, (e-^e-'-e) is the angle which will give this difference 

 equal to 3m, and so on to any number of angles. Hence, in the circle, if ^ 

 be the angle which gives the circular arc equal to the radius, 2^ is the angle 

 which will give an arc equal to twice the radius, and so on for any number 

 of angles. This is of course self-evident in the case of the circle, but it is 

 instructive to point out the complete analogy which holds in the trigonome- 

 tries of the circle and of the parabola. 



Hence the amplitude which gives the difference between the parabolic arc 

 and its subtangent equal to the semiparameter is given by the simple equation 



sece'+tane'=e2 (30) 



And more generally, if e" be the amplitude which gives the difference between 

 the parabolic arc and its subtangent equal to n times the modulus, we shall 



^^^^ sece''+tane'' = e» (31) 



In the same way it may be shown that if e^ be the angle which gives the 



difference between the parabolic arc and its subtangent equal to -th part the 



modulus, we shall have i^ 



sece^+tane^=e" (32) 



Let the difference be equal to one-half the modulus, then w=2, and 

 ^ec €(+ tan e^=e . 



