OPERATIVE ROOTS OF THE CIRCLE-FUNCTION 119 



ratios contained in the formulae just given. Thus we find the following 

 operative properties of K: 



K^ = cos 0.0 + sin o . K^ = o 

 K-^= cos {- 1) .0 + sin {-!) .K^= - K, 

 Kl =x cos 2 . O + sin 2 . K^ ^^ - O 

 Kl = cos2.0 + sin ^.K^=~ K^ 



isT* = cos 4 . O + sin 4 . iiT^ = o (i) 



Kl = cos 6 .0 + sin 6 . K^ K' ^ ~ cos 6 . o - sin 6 . K, 



K\ + ^ = cosd .K^~sine .0 K^-^ =^ cos 6 . K^ + sin 6 . O (2) 



K9±^== - K^ K6 ±« = K^ (3) 



K^^ + K-^ = 2cosd.O Kl-K-^ = 2 sin 6 . K^ 



K'-'+ K'-' = 2 sine.o Kl-^-K'-' = 2cose.K^ (4) 



(See however 4*21). 



Here the value oi K\'^^ has been' obtained from trigonometrical properties 

 of the sine and cosine ; for we have assumed that 



K\ + ^ = cos {i-\-6) .0 + sin {1 + 6) . K^= - sin 6 .0 + cos 6 . K^. 



But we may obtain the same result by direct operation ; for 



if^ + '= Js:^ = [cos^. O + sine.K^]K^ = cosd.K^ + sin 3. K) 



= cos 6 .Kr — sin 6 .O; (5) 



for, as shown above, Kl= - O. For another example, 



Kl^'=K,^'Kl = [ - 0][Kl] = -Kl: 

 hnt Blso Kl^"" = [cos 6. Q + sin 6. K^'\K^±''=- cos 6.0 + sin 6. K\^\ (6) 

 and K\^'' = K'^ox^K- \ both of which = - i?,. Of course 



O0 = 4> and <^0 = </>. 



3*2. By operating with it either on zero or on y, we can break up a 

 /^-operation into two parts which resemble the scalar and vector parts of a 

 quaternion. For 



Klo = [cos e .0 + sin6 . K^]o = cos6 .0 + sin . Vr^ - o^ = r sin 6 ; 

 K^r = [cos 6 .0 + sine . K;\r = cos6 .r ^ sin 6 . Vr^'ITT^ = r cos 6. (i) 

 Hence we may write 



r.Kl^^Klr.O + K^o.K^; (2) 



and also fKl + '* = Kir .K] + K^o .K\ + '' 



= Kir.Kl + K]o . K' + ' = rK''^ + \- (3) 



and rK^^ + ^o = K^^r . i^Tjo + K^,o . K^r - y^ 5^^ (^ 4. y)^ 



and vKl + V = K^y . K]y - K^o . K^o = y^ cos {6 + y) : (4) 



for K\ + 'o = Kly. and K\+\ = isT^ + ^o. 



Put slightly differently, 



Klir sin 6) = JsTJK^o = K\ [y sin y) ^ r sin {6 + y). 



Ky cos 6) =- KlK^/ = Kl {y cos y) - y cos {6 + y). (5) 



