document.write( "Question 1014364: If cosA+cosB+cosC=0, prove that cos3A+cos3B+cos3C=12cosAcosBcosC \n" ); document.write( "

Algebra.Com's Answer #630661 by ikleyn(52908) $\"\"$ $\"About$

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\n" ); document.write( "If cosA+cosB+cosC=0, prove that cos3A+cos3B+cos3C=12cosAcosBcosC
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document.write( "First of all, we need to prove that \r\n" );
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document.write( "if x + y + z = 0 then  = 3xyz \r\n" );
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document.write( "x + y + z = 0 \r\n" );
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document.write( "x + y = - z \r\n" );
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document.write( " =  \r\n" );
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document.write( " =  \r\n" );
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document.write( " = \r\n" );
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document.write( "Add 3xyz -3xyz (that is 0) to the right hand side \r\n" );
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document.write( " =  \r\n" );
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document.write( "Factor -3xy from the first three terms on the right hand side \r\n" );
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document.write( " = -3xy(x + y + z) + 3xyz \r\n" );
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document.write( " = -3xy(0) + 3xyz \r\n" );
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document.write( " = 3xyz \r\n" );
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document.write( "------- \r\n" );
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document.write( "Now let x = cosA, y = cosB, z = cosC \r\n" );
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document.write( "so x + y + z = 0 \r\n" );
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document.write( "Since cos3A =  , then cos3A =  \r\n" );
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document.write( "Similarly, cos3B =  \r\n" );
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document.write( "and cos3C =  \r\n" );
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document.write( "Add all three equations, we have: \r\n" );
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document.write( "cos3A+cos3B+cos3C =  \r\n" );
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document.write( "=  \r\n" );
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document.write( "= 4(3xyz) - 3(0) \r\n" );
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document.write( "= 12xyz \r\n" );
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document.write( "= 12cosAcosBcosC \r\n" );
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document.write( "Q.E.D \r\n" );
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document.write( "I brought it for you from https://answers.yahoo.com/question/index?qid=20150911203611AAaEjdD.\r\n" );
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