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paradox xyz cool Grade: 11
        

prove that:


cos2Acos2B + sin2(A - B) - sin2(A + B) = cos2(A + B)

6 years ago

Answers : (2)

vikas askiitian expert
510 Points
										

sin2(A-B) - sin2(A+B) =sin(A+A)sin(-B-B)                                         [using formula ,sin2a-sin2b=sin(a-b)sin(a+b)]


                               =sin(-2B)sin2A=-sin2Asin2B


now LHS becomes


 =cos2Acos2B-sin2Asin2B                        [ using formula ,cosacosb - sinasinb =cos(a+b)]


=cos(2A+2B)=cos2(A+B) = RHS 


hence proved      

6 years ago
madhu voleti
44 Points
										Cos2ACos2B+Sin^2(A-B)-Sin^2(A+B)=Cos2(A+B)

Solution:-
L.H.S:-Cos2ACos2B+Sin^2(A-B)-Sin^2(A+B)
=Cos2ACos2B+Sin(A-B+A+B)*Sin(A-B-A-B) ["a^2-b^2=(a+b)(a-b)"]
=Cos2ACos2B+Sin2A*Sin(-2B)
=Cos2ACos2B-Sin2ASin2B
=Cos2(A+B) ["Cos(A+B)=CosACosB-SinASinB"]

Thank You!!!!!!!!!!!!!!!!!!!!!!
Wish You A Happy Learning Trignometry........
6 years ago
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