In a triangle,prove that sin 3A sin(B-C)+sin 3Bsin(C-A)+sin3C(A-B)=0

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Saurabh Koranglekar · Answer

Dear student Please find the link below https://www.askiitians.com/forums/Trigonometry/in-a-triangle-abc-prove-that-sin3asin-b-c-sin3bsi_118942.htm Regards

Vikas TU · Answer

Dear student Dear student Please approach using ( Sin A/ a) = ( sin B / b ) = ( sin C / c) = k Expand sin 3A = 3ak - 4 ( ak)^3 If doubt still persists kindly revert for complete solution Regards

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In a triangle,prove that sin 3A sin(B-C)+sin 3Bsin(C-A)+sin3C(A-B)=0

In a triangle,prove that sin 3A sin(B-C)+sin 3Bsin(C-A)+sin3C(A-B)=0

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In a triangle,prove that sin 3A sin(B-C)+sin 3Bsin(C-A)+sin3C(A-B)=0

In a triangle,prove that sin 3A sin(B-C)+sin 3Bsin(C-A)+sin3C(A-B)=0

2 Answers

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