ON THE TRIGONOMETRY OP THE PARABOLA. 



97 



XXXVII. Since 



log u=u—u~^ —i(u- —u-'^) + ^(u^— u-^)—^(u^ —u-^), &c., 



and as 



while 



M-«-i = 2tan0, ti'—u~^=2tar\(d-^e), 

 M«— M-»=2 tan(e-L0-L0-Lto n terms), 



M=sec + tan d. 



We have therefore 



PV— PT 



logu=: =tan0-itan(0-L0) + l.tan(0-^0^0, &c.). (53) 



We may convert this into an expression for the arc of a circle by 

 changing -"-into +5 tan into-v^ — 1 sin, and the parabolic arc into the circular 

 arc multiplied by i/'—l. 



Hence, since PT in the circle is equal to 0, 



- = sin0-isin29 + isin3e— isin40, 



a formula given in Lacroix, ' Traite du Calcul Differentiel et du Calcul 

 Integral,' torn. i. p. 94. 



XXXVIII. In the trigonometry of the circle, the sines and cosines of 

 multiple arcs may be expressed in terms of powers of the sines and cosines of 

 the simple arcs. Thus 



cos 29= 2cos=0— 1 



cos 30= ^cos^e— 3cos0 



cos 40= Scos-*©— 8 cos- 0+1 



cos 50= 16 cos' 0—20 cos^ + 5 cos 



cos 60= 32 cos« 0—48 cos* + 18 cos- - 1 



"1 



sin20=sin0(2cos0) 



sin30=sin0(4<cos2 0— 1) 



sin 40= sin (8 cos' 0—4 cos 0) 



sin 50=sin (16 cos* — 12 cos''0+ 1) 



sin 60=sin (32 cos' 0—32 cos' + 6 cos 0). 



Hence in the trigonometry of the parabola, 

 sec(0-L0)=2sec'0— 1 

 sec (0-i-0-J-0)=4. sec' 0—3 sec 

 sec (0 -L -L -I- 0)= 8 sec* 0— 8 sec- + 1 

 sec(0-L-0-i-0-L0-i-0)=i6sec'0— 2Osec'0 + 5sec0 

 sec(0J-0-L0a-0-i-0-L0)=32sec«0-48sec*0+18sec20-l 



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

 tan (0-i-0-J-0)=tan (4 sec" 0—1) 

 tan (0-i- 0-1- 0-1-0)= tan (8 sec' 0—4 sec 0) 

 tan (0-1- 0-1- 0-1- 0-1-0)= tan (16 sec* 0-12 sec- 0+1) 

 tan (0^0-i-0-i-0-L0a-0)=tan (32 sec' 0— 32sec'0+6 sec 0) 



The preceding formulae may easily be verified. 

 1856. '^ ^ J ^ 



(54) 



M55) 



