If 2cos⁡A=cos⁡B+cos3⁡B, and 2sin⁡A=sin⁡B−sin3⁡B then sin (A- B)=

2cos⁡A=cos⁡B+cos3⁡B----iand  2sin⁡A=sin⁡B−sin3⁡B----ii⇒ 2sin⁡Acos⁡B−2cos⁡Asin⁡B =sin⁡B−sin3⁡Bcos⁡B−cos⁡B+cos3⁡Bsin⁡B =−sin⁡Bcos⁡B⇒ sin⁡(A−B)=−122sin⁡2BNow squaring and adding Eqs. (i) and (ii), we get 2=cos2⁡B+sin2⁡B+cos6⁡B+sin6⁡B+2cos4⁡B−sin2⁡B⇒1=cos2⁡A+sin2⁡A3−3cos2⁡Asin2⁡Acos2⁡A+sin2⁡A+2 cos 2B⇒1=1−(3/4)sin2⁡2B+2cos⁡2B⇒−3sin2⁡2B+8cos⁡2B=0⇒3cos2⁡2B+8cos⁡2B−3=0⇒cos⁡2B=13⇒ sin⁡2B=±223 ∴sin⁡(A−B)=±13