document.write( "Question 696412: cosA+cosB+cosC=3/2 ,then show that the triangle is equilateral . \n" ); document.write( "
Algebra.Com's Answer #429058 by Edwin McCravy(20056)\"\" \"About 
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document.write( "cos(A) + cos(B) + cos(C) = \"3%2F2\"\r\n" );
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document.write( "Let A=P+Q\r\n" );
document.write( "and B=P-Q\r\n" );
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document.write( "(with hopes of showing that Q=0)\r\n" );
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document.write( "Then A+B = 2P and \r\n" );
document.write( "     A-B = 2Q\r\n" );
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document.write( "A+B+C = 180°\r\n" );
document.write( "    C = 180°-(A+B)\r\n" );
document.write( "    C = 180°-2P\r\n" );
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document.write( "Substituting in the given equation: \r\n" );
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document.write( "cos(P+Q)+cos(P-Q)+cos(180°-2P) = \"3%2F2\"\r\n" );
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document.write( " cos(P+Q) + cos(P-Q) - cos(2P) = \"3%2F2\"\r\n" );
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document.write( "We write an identity for each term on the left \r\n" );
document.write( "and add the three equations:\r\n" );
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document.write( "cos(P+Q) = cos(P)cos(Q)-sin(P)sin(Q)\r\n" );
document.write( "cos(P+Q) = cos(P)cos(Q)+sin(P)sin(Q)\r\n" );
document.write( "-cos(2P) = -[2cosē(P)-1]\r\n" );
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document.write( "       \"3%2F2\" = 2cos(P)cos(Q)-2cosē(P)+1\r\n" );
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document.write( "Clear the fraction:\r\n" );
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document.write( "       3 = 4cos(P)cos(Q)-4cosē(P)+2\r\n" );
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document.write( "Rearrange the equation with 0 on the right:\r\n" );
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document.write( "4cosē(P) - 4cos(Q)cos(P) + 1 = 0\r\n" );
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document.write( "That is a quadratic equation in cos(P).\r\n" );
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document.write( "Its solution(s) must be real so the\r\n" );
document.write( "discriminant must be ≧ 0\r\n" );
document.write( "              bē-4ac ≧ 0\r\n" );
document.write( "        16cosē(Q)-16 ≧ 0\r\n" );
document.write( "       16(cosē(Q)-1) ≧ 0\r\n" );
document.write( "           cosē(Q)-1 ≧ 0\r\n" );
document.write( "             cosē(Q) ≧ 1\r\n" );
document.write( "The square of a cosine is never > 1\r\n" );
document.write( "Therefore\r\n" );
document.write( "             cosē(Q) = 1\r\n" );
document.write( "              cos(Q) = ą1\r\n" );
document.write( "             Q = 0° or 180°\r\n" );
document.write( "           A-B = 2Q \r\n" );
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document.write( "A and B cannot differ by 2(180°) or 360° so they \r\n" );
document.write( "must differ by 0°, so Q=0°, so\r\n" );
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document.write( "             A-B=0\r\n" );
document.write( "               A=B\r\n" );
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document.write( "Now if we interchange B and C throughout the\r\n" );
document.write( "above, we will get A=C.  So the three angles are \r\n" );
document.write( "equal and thus the triangle is equilateral.\r\n" );
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document.write( "Edwin
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