document.write( "Question 1144138: 2sin(3x)cos(3x)(2cos^2 3x-1)
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Algebra.Com's Answer #765143 by ikleyn(52937) $\"\"$ $\"About$

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document.write( "The product of the first three factors (terms) is\r\n" );
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document.write( "    2*sin(3x)*cos(3x) = sin(6x).\r\n" );
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document.write( "The fourth term is\r\n" );
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document.write( "    2*cos^2(3x)-1 = cos(6x).\r\n" );
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document.write( "Therefore\r\n" );
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document.write( "    2*sin(3x)*cos(3x)*(2cos^2 3x-1) = sin(6x)*cos(6x) = (1/2)*sin(12x).    ANSWER\r\n" );
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