InterviewSolution
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InterviewSolution
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Prove that:`sin^2A=cos^2(A-B)+cos^2B-2cos(A-B)cosAcosBdot` |
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Answer» `R.H.S. = cos^2(A-B)+cos^2B-2cos(A-B)cosAcosB` `=cos^2B+cos(A-B)[cos(A-B)-2cosAcosB]` `=cos^2B+cos(A-B)[cosAcosB+sinAsinB-2cosAcosB]` `=cos^2B+cos(A-B)[-cosAcosB+sinAsinB]` `=cos^2B-cos(A-B)[cosAcosB-sinAsinB]` `=cos^2B-cos(A-B)[cos(A+B)]` Now, we know, `cos(A-B)cos(A+B) = cos^2A-sin^2B` So, it becomes, `=cos^2B-cos^2A+sin^2B` `=cos^2B+sin^2B-cos^2A` `=1-cos^2A` `=sin^2A = L.H.S.` |
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