Prove #Cos^2A+Sin^2Acos2B=cos^2B+sin^2Bcos2A#?

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Answer 1 · 2018-03-09T16:53:46

#LHS=cos^2A+sin^2A*cos2B#

#=1/2[2cos^2A+2sin^2A*cos2B]#

#=1/2[1+cos2A+(1-cos2A)*cos2B#

#=1/2[1+cos2A+cos2B-cos2A*cos2B#

#=1/2[{1+cos2B}+{cos2A(1-cos2B)}]#

#=1/2[2cos^2B+2sin^2B*cos2A]#

#=cos^2B+sin^2B*cos2A=RHS#

Prove #Cos^2A+Sin^2A*cos2B=cos^2B+sin^2B*cos2A#?