document.write( "Question 1067660: Write 4 cos 2θ sin 4θ as a sum or difference. \n" ); document.write( "
Algebra.Com's Answer #682810 by Edwin McCravy(20055)\"\" \"About 
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document.write( "4cos(2θ)sin(4θ)\r\n" );
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document.write( "We notice that this has a product of a sine and a cosine\r\n" );
document.write( "and we remember that such a product appears in the formulas: \r\n" );
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document.write( "sin(A+B) = sin(A)cos(B) + cos(A)sin(B)\r\n" );
document.write( "sin(A-B) = sin(A)cos(B) - cos(A)sin(B)\r\n" );
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document.write( "So we substitute A=4θ and B=2θ in those\r\n" );
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document.write( "sin(4θ+2θ) = sin(4θ)cos(2θ) + cos(4θ)sin(2θ)\r\n" );
document.write( "sin(4θ-2θ) = sin(4θ)cos(2θ) - cos(4θ)sin(2θ)\r\n" );
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document.write( "or\r\n" );
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document.write( "sin(6θ) = sin(4θ)cos(2θ) + cos(4θ)sin(2θ)\r\n" );
document.write( "sin(2θ) = sin(4θ)cos(2θ) - cos(4θ)sin(2θ)\r\n" );
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document.write( "If we add those equations together, equal to equals,\r\n" );
document.write( "the last terms cancel and we get\r\n" );
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document.write( "sin(6θ) + sin(2θ) = 2sin(4θ)cos(2θ)\r\n" );
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document.write( "So our problem\r\n" );
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document.write( "4cos(2θ)sin(4θ) = 2[2sin(4θ)cos(2θ)] = 2[sin(6θ) + sin(2θ)]\r\n" );
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document.write( "or\r\n" );
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document.write( "2sin(6θ) + 2sin(2θ)\r\n" );
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document.write( "Edwin
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