Answer:
see below
Step-by-step explanation:
Prove : cos²(A-B)-sin²(A+B)=cos2Acos2B
LHS,
cos²(A-B) - sin²(A+B) (expand using double angle identities)
= [tex]\frac{1 + cos [2(A-B)]}{2} - \frac{1 - cos [2(A+B)]}{2}[/tex]
= [tex]\frac{1}{2} + \frac{1}{2} cos[2(A-B)] - \frac{1}{2} + \frac{1}{2} cos[2(A+B)][/tex]
= [tex]\frac{1}{2} cos[2(A-B)] + \frac{1}{2} cos[2(A+B)][/tex]
= [tex]\frac{1}{2} cos(2A-2B) + \frac{1}{2} cos(2A+2B)[/tex] (expand using sum/difference identities)
= [tex]\frac{1}{2} (cos2Acos2B + sin2Asin2B ) + \frac{1}{2} (cos2Acos2B - sin2Asin2B )[/tex]
= [tex]\frac{1}{2} cos2Acos2B + \frac{1}{2}sin2Asin2B + \frac{1}{2} cos2Acos2B - \frac{1}{2}sin2Asin2B[/tex]
= [tex]\frac{1}{2} cos2Acos2B + \frac{1}{2} cos2Acos2B[/tex]
= [tex]cos2Acos2B[/tex] (= RHS , Proven)
