document.write( "Question 1046271: If cos(A-B)+cos(B-C)+cos(C-A)= -3/2, prove that: cosA+cosB+cosC=sinA+sinB+sinC=0. \n" ); document.write( "
Algebra.Com's Answer #662021 by robertb(5830)\"\" \"About 
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Consider \r
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\n" ); document.write( "\n" ); document.write( "=\"3%2B+2%28cos%28A-B%29+%2B+cos%28B-C%29+%2B+cos%28C-A%29%29\".\r
\n" ); document.write( "\n" ); document.write( "But \"cos%28A-B%29%2Bcos%28B-C%29%2Bcos%28C-A%29=+-3%2F2\" ===> \"3%2B+2%28cos%28A-B%29+%2B+cos%28B-C%29+%2B+cos%28C-A%29%29+=+0\".\r
\n" ); document.write( "\n" ); document.write( "Since the values of A, B, and C are real, all the sine and cosine values are also real, and so, we conclude that \r
\n" ); document.write( "\n" ); document.write( "\"%28cosA%2BcosB%2BcosC%29%5E2+=+0\" and \"%28sinA%2BsinB%2BsinC%29%5E2+=+0\", hence\r
\n" ); document.write( "\n" ); document.write( "cosA+cosB+cosC = 0 and sinA+sinB+sinC = 0.
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