document.write( "Question 1023016: To construct a regular heptagon, one has to find the angle in degrees $\"x+=+%28360%2F7%29\"$ . Letting $\"u+=+2cosx\"$ , show it satisfies the equation $\"u%5E3+%2Bu%5E2-2u-1=0\"$ \n" ); document.write( "

Algebra.Com's Answer #638620 by ikleyn(52794) $\"\"$ $\"About$

You can put this solution on YOUR website!
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\n" ); document.write( "See http://mathforum.org/kb/message.jspa?messageID=5116464\r
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\n" ); document.write( "\n" ); document.write( "For your convenience, I copy and past it for you:\r
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document.write( "Can you take as given the fact that cos(2 pi/7) is the real part of\r\n" );
document.write( "a root of the 7th cyclotomic polynomial? That is,\r\n" );
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document.write( "x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 = 0\r\n" );
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document.write( "If so, then consider\r\n" );
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document.write( "cos(6y) + cos(5y) + ... + cos(y) + 1 = 0\r\n" );
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document.write( "For the specific value y = 2 pi/7,\r\n" );
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document.write( "cos(6y) = cos(y)\r\n" );
document.write( "cos(5y) = cos(2y)\r\n" );
document.write( "cos(4y) = cos(3y)\r\n" );
document.write( "cos(3y) = 4 (cos y)^3 - 3 cos(y)\r\n" );
document.write( "cos(2y) = 2 (cos y)^2 - 1\r\n" );
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document.write( "Letting x = cos(2 pi/7),\r\n" );
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document.write( "x + (2 x^2 - 1) + 2 (4 x^3 - 3x) + (2 x^2 - 1) + x + 1 = 0\r\n" );
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document.write( "Collecting like terms,\r\n" );
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document.write( "8 x^3 + 4 x^2 - 4x - 1 = 0\r\n" );
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document.write( "So (2 pi/7) is indeed a root of your equation, and thus also of the\r\n" );
document.write( "minimal polynomial x^3 + x^2 - 2x - 1.\r\n" );
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