1'LAXi: TRIGONOMETRY. 59 



So to transform cos (/? - y) cos (y - a) cos (a - /3), 



cos (y - a) cos (a - 0) = J {cos (y - /3) + cos (j8 + y - 2a)} ; 



.'. cos 03 - y) cos (y - a) cos (a - ft) 



= i {I + cos 2 (y-/3) + cos 2 (y-a) + cos 2 (a-/?)}. 



2G7. Solve the equations ; 



(1) 2 sin a; sin3a;=l, 



(2) cos x cos 3x = cos 2x cos 6, 



(3) sin 5x cos 3x = sin 9a? cos 7a^ 



(4) sinOa-f-sin 5# + 2sin 2 x = 1, 



(.") ) cos ax cos ta = cos (a + c) x cos (b + c) x, 



(G) 4 sin a sin /3 sin x = sin 2a -f sin 2/3 + sin 2#, 



(7) cos a 4- cos (x - a) = cos (x- p) + cos (x + p- a), 



(8) 2 sin* 2x cos 2x = sin* 3#, 



(9) 2 cot 2a- tan 2s =3 cot 3a, 



(10) 8cos* = -^- + 



sin a; cos a; 



(11) sin 2x + cos 2a; -f sin a; - cos a; = 0, 



(12) (1 + sin *)(! - 2 sin ) = (1 - cos a) (1 + 2 cos a)', 



= 

 sin ft cos (a -I- ) tan a ' 



(14) cos2a;-l-2cosa?cosa-2cos2a=l, 



(15) sin a cos 3a; - 3 sin 3a cos x 4- sin 4a 4- 2 sin 2a * 0, 



(16) COHa + 8 J = l. 

 cos a; sin a; 



If 



sin (3x + y) = cos (2a; 4- 2y) ; 



thru will x = (5m-3n) - + -^ 



y ; or a; - 

 y-(5n-3m)^~ 



tegers. 



