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If A+B+C=0, then prove that `Det[[1,cosC,cosB],[cosC,1,cosA],[cosB,cosA,1]]=0` |
Answer» `=1[1-cos^2A]-cosc[cosc-cosAcosB]+cosB[cosAcosB-cosB]` `=1-cos^2A-cos^2C+cosAcosBcosC+cosAcosBcosC-cos^2B` `=1-[cos^A+cos^2B+cos^2C]+2coaAcosBcosC` `=1-[cos^2A+cos^2B+(cos(A+B)^2)+2cosAcosB[A+B]]` `=1-cos^2A-cos^2B-cos^2Acos^2B-acos^2b-2cosAcosBsin^2Asin^2B+2cosAcosBsinAsinB+2cos^2Acos^2B-2cosAcosBsinAsinB` `=1-cos^2A-cos^2B-sin^2Asin^2B+cos^2Acos^2B` `=sin^2A-cos^2B-sin^2Asin^2B+cos^2Acos^2B` `=sin^2Acos^2B-cos^2Bsin^2A` `LHS=0=RHS` |
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