Trigonometric problem (aid)

(a)
sin(A+B)cos(A-B)
=1/2[sin(A+B+A-B)+sin(A+B-A+B)]
=1/2[sin(2A)+sin(2B)]
=sin AcosA+sinBcosB
(b)
sin2acos2a=cosa(1-si na)
sin2acos2a+sin2acos2a=cosa(1-sina)+cosa(1-sina)
sin4a=2cosa(1-sina)
2sin2acos2a=2cosa-sin2a
sin2a(2cos2a+1)=2cosa
2sinacosa(2cos2a+1)-2cosa=0
cosa(sina(2cos2a+1)-1)=0
cosa=0 or sina(2cos2a+1)=1
for cosa=0, a=90
for sina(2cos2a+1)=1
sina(2(1-2sin^2a)+1)-1=0
3sina-4sin^3a-1=0
sin3a=1
3a=90 or 450
a=30 or 150

b) sin2acos2a=cosa(1-sina)
sin2acos2a=cosa-sinacosa
sin2acos2a + (sin2a)/2 -cosa = 0
sin2a(1/2 + cos2a) -cosa =0
2sinacosa[(cos2a)/2 + (cosa)^2 ] -cosa =0
sinacosa[2(cosa)^2 -1 + 2(cosa)^2] - cosa =0
sinacosa[4 - 4(sina)^2 -1] -cosa = 0
sinacosa[3 - 4(sina)^2] - cosa = 0
cosa [ -4(sina)^3 +3sina -1] = 0
cosa (2sina - 1)[ -2(sina)^2 - sina + 1] = 0
cosa (2sina - 1)(-2sina + 1)(sina + 1) = 0
cosa = 0 or 2sina - 1 = 0 or -2sina + 1 = 0 or sina + 1 = 0
a=90 or a=30, 150 or a=30, 150 or a=270 (rej)
Therefore a = 30, 90, 150