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In any triangle `A B C ,`prove that: ` a^3sin(B-C)+b^3sin(C-A)+c^3sin(A-B)=0` |
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Answer» By the sine rule, we have `a/("sin A")=b/("sin B")=c/("sin C")="k (sky)"` `rArr" "a=ksinA, b-ksinBandc=ksinC.` `:." "a^(3)sin(B-C)=k^(3)sin^(3)Asin(B-C)` `=k^(3)sin^(2)AsinAsin(B-C)` `=k^(3)sin^(2)A[sin{pi-(B+C)}sin(B-C)]" "[becauseA+(B+C)=pi]` `=k^(3)sin^(3)A[sin(B+C)sin(B-C)]` `=k^(3)sin^(2)A(sin^(2)B-sin^(2)C).` Similarly, `b^(3)sin(C-A)=k^(3)sin^(2)B(sin^(2)C-sin^(2)A)`. And, `c^(3)sin(A-B)=k^(3)sin^(2)C(sin^(2)A-sin^(2)B).` `:." "a^(3)sin(B-C)+b^(3)sin(C-A)+c^(3)sin(A-B)` `=k^(3)sin^(2)A(sin^(2)B-sin^(2)C)+k^(3)sin^(2)B(sin^(2)C-sin^(2)A)+k^(3)sin^(2)C(sin^(2)A-sin^(2)B)` `=0` |
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