document.write( "Question 1122126: (cos 6x + 6 cos 4x + 15 cos 2x + 10)/
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Algebra.Com's Answer #738232 by ikleyn(52909)\"\" \"About 
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document.write( "(cos(6x) + 6cos(4x)  + 15cos(2x) + 10)/(cos(5x) + 5cos(3x) + 10cos(x)) \r\n" );
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document.write( "Consider the numerator and re-group it in this way\r\n" );
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document.write( "Numerator = cos(6x) + 6*cos(4x)  + 15*cos(2x) + 10 = \r\n" );
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document.write( "          = (cos(6x) + cos(4x)) + (5*cos(4x) + 5*cos(2x)) + (10*cos(2x) + 10)    (*)\r\n" );
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document.write( "Next, use the basic Trigonometry formula cos(a) + cos(b) = \"2%2Acos%28%28a%2Bb%29%2F2%29%2Acos%28%28a-b%29%2F2%29%29\".  You will get\r\n" );
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document.write( "              cos(6x) + cos(4x) = 2*cos(5x)*cos(x), \r\n" );
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document.write( "              cos(4x) + cos(2x) = 2*cos(3x)*cos(x), \r\n" );
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document.write( "              cos(2x) + 1 = 2*cos(x)*cos(x).\r\n" );
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document.write( "Substitute it into the formula (*). Then you can continue (*) in this way\r\n" );
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document.write( "Numerator = 2*cos(5x)*cos(x) + 5*2*cos(3x)*cos(x) + 10*2*cos(x)*cos(x) = \r\n" );
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document.write( "          = 2*cos(x)*(cos(5x) + 5*cos(3x) + 10*cos(x))\r\n" );
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document.write( "Now notice that the long expression in the Numerator parentheses is exactly the denominator of the original formula.\r\n" );
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document.write( "Canceling common factors in the numerator and denominator, you get the final expression\r\n" );
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document.write( "     (cos(6x) + 6*cos(4x)  + 15*cos(2x) + 10) / (cos(5x) + 5*cos(3x) + 10*cos(x))  = 2*cos(x)\r\n" );
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document.write( "Answer.  (cos(6x) + 6*cos(4x)  + 15*cos(2x) + 10) / (cos(5x) + 5*cos(3x) + 10*cos(x)) = 2*cos(x).\r\n" );
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