98 REPORT — 1856. 



If we add in the above series any two corresponding secants and tangents, 

 the sum will be an integral power of sec + tan 8. 



Thus sec(d-^e) + tan (6 -^6)= (sec 6 + tan 0)% 



Again, since in the circle 



cos 0= cos 

 2cos2 0=cos20 + l 

 4;cos^0=cos30 + 3cos0 



8 cos^ 0= cos 40 + 4 cos 20+1 

 and [-(56) 



sin 0=sin 



2sin2 0=— COS20 + 1 



4 sin3 0=— sin 30+3 sin 



8 sin^0=cos 40—4 COS 20 + 3. 



Hence in parabolic trigonometry, 

 sec0=sec 

 2sec-0=sec(0-^0) + l 

 4 sec5 0= sec (0-^0-^0)+ sec 30 

 8sec*0=sec(0^0^0^0)+4sec(0-«-0) + l 1,(5Y) 



tan 0=tan 

 2tan-0=sec(0-L0)-l 

 4 tan' 0=tan (0-l0-i-0)-3 tan 

 8 tan* 0=sec (0^0^0-i-0)-4 sec (0-l0) + 3. 

 XXXIX. The roots of the expression 



2;2»_2a2;» + l=0 (58) 



maybe represented under the form cos A+ ^~1 sin A, when a is Zmthan 1. 

 This has long been known. It is not difficult to show that when a is greater 

 than 1, the roots may be exhibited under the form 



sec A + tan A (59) 



Since a is greater than 1, let a=sec0, and let be divided into n angles 0, 

 connected by the relation 



J.0 j-^ ^(j) &c. =0; (60) 



and it has been shown in (6) that 



sec(<p^(j>^(t>-^<l> to n<i>) + tan ((j>->-(p^<p -^f to n(p) = (sec <p + tan (j>y. 



Let sec ^ + tan ^=M, then 2sec0=M' + M~S 

 and therefore 2 sec0=2sec (<^J-0^./.-^to «.^)=m» + m-". 

 Substitute this value of 2 sec in (58), and we shall have 



Z^n _ (m1 + ?f-«)2» + 1=0, 



or resolving into factors, 



(5-0 ("""""0=" ^''^ 



