traced upon the Surface of the Sphere. 391 



This value of sin Q inserted in (5) gives 



d (p _ sin 2 (f> sin 2e { (cos 2a cos 2f) (cos 2e cos 2 (a rp (f>) ) }* 

 d *~ cos 2a cos 2e sin 2e sin ^) 



_ _l_sm9| cos 2a cos 2e ) 



But from (1) we have this (8) converted into 



sin 4 <p 



sn = 



... . , ,cos2e 

 sin 2 a> + sin* a> 



^ - 

 cos 2a cos 2e 



_ (cos 2a cos 2e) sin 2 (f> 

 ~ cos2a cos(2aip 



sm 



(9) 



If we resolve this with respect to $ instead of $, we shall get 



cos 2a cos 2 J/ cos 2e 



- - 



COt<P= 



sm 

 whence sin ^ and cos may also be obtained. 



2. 7%e perpendicular from any point upon the tangent to a circle. 



As this question refers to the magnitude of the perpendicular, but not 

 its position, the solution loses nothing of its generality by taking the great 

 circle through the centre of the circle for origin of 6, or, in other words, by 

 taking * = 0.* Then we have 



cos p = cos (f) cos X + sin sin X cos 6 ..................... (11) 



Differentiating, we have 



d (p sin X sin (f) sin 



dd cos cf> cos 6 sin X sin (p cos X 



* Indeed, had we taken (6 K) for the angle made by the radius -vector of the cur- 

 rent point in the circle and the first meridian, we should still eliminate the functions of 

 K at step ( 1 4), as we have actually done with 6, and the result is therefore exactly 

 the same. 



