Trigo

Multiplying the given relations by sinB and cosB and subtracting we get
-√2(sinA cosB - cosA sinB) = sinBcosB (cos²B + sin² B)
so
sin(A-B) = - (12√2)sin2B ............................(1)
again squaring and adding the given relations, we get
2 x 1 = 1 + (cos⁶B + sin⁶B) + 2 ( cos⁴ B - sin⁴ B )
1 = 1.(cos⁴ B + sin⁴ B - cos²Bsin²B) + 2.1.(cos²B - sin²B)
1 = (1 - 3cos²Bsin²B) + 2cos2B
34(2sinBcosB)² = 2cos2B
3 sin²2B = 8cos2B
3 cos²2B + 8cos2B - 3 = 0
(cos2B+3)(3cos2B-1) = 0
cos 2B = 1/3 as cos 2B≠3
sin²2B = 1 -cos²2B = 1 - 1/9 = 8/9
sin 2B = ± 2√23

hence from (1)
sin ( A - B ) = - (12√2)sin2B = ±13