document.write( "Question 988624: prove that for any triangle ABC, where a,b, and c are respectively
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Algebra.Com's Answer #609197 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "\"%28a%2Acos%28A%29+-+b%2Acos%28B%29%29+%2F+%28b%2Acos%28A%29+-+a%2Acos%28B%29%29\"\"%22%22=%22%22\"\"cos%28C%29\"\r\n" );
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document.write( "The law of cosines: \r\n" );
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document.write( "\"a%5E2+=+b%5E2%2Bc%5E2-2ab%2Acos%28A%29\", \"b%5E2+=+a%5E2%2Bc%5E2-2ac%2Acos%28B%29\", \"c%5E2=b%5E2%2Ba%5E2-2ab%2Acos%28C%29\"\r\n" );
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document.write( "when solved for the cosines, give:\r\n" );
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document.write( "\"cos%28A%29=%28b%5E2%2Bc%5E2-a%5E2%29%2F%282bc%29\", \"cos%28B%29=%28a%5E2%2Bc%5E2-b%5E2%29%2F%282ac%29\", \"cos%28C%29=%28b%5E2%2Ba%5E2-c%5E2%29%2F%282ab%29\"\r\n" );
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document.write( "So the identity becomes to show that \r\n" );
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document.write( "\"%22%22=%22%22\"\"%28b%5E2%2Ba%5E2-c%5E2%29%2F%282ab%29\"\r\n" );
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document.write( "On the left side multiply numerator and denominator by LCD = 2abc\r\n" );
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document.write( "Cancel terms that add to zero and collect like terms:\r\n" );
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document.write( "\"+%28++a%5E2c%5E2-a%5E4-b%5E2c%5E2%2Bb%5E4+%29+%2F+%282ab%5E3-2a%5E3b%29+\"\r\n" );
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document.write( "Rearrange the terms of the numerator to factor by grouping, \r\n" );
document.write( "factor 2ab out of the denominator:\r\n" );
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document.write( "\"+%28+b%5E4-a%5E4-b%5E2c%5E2%2Ba%5E2c%5E2+%29+%2F+%282ab%28b%5E2-a%5E2%29%29+\"\r\n" );
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document.write( "Factor first two terms in numerator as the difference of squares.\r\n" );
document.write( "Factor -c² out of last two terms in numerator:\r\n" );
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document.write( "Factor (b²-a²) out of numerator:\r\n" );
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document.write( "\"+%28+%28b%5E2-a%5E2%29%28b%5E2%2Ba%5E2-c%5E2%29+%29+%2F+%282ab%28b%5E2-a%5E2%29%29+\"\r\n" );
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document.write( "Cancel (b²-a²)'s\r\n" );
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document.write( "\"+%28b%5E2%2Ba%5E2-c%5E2%29+%2F+%282ab%29+\"\r\n" );
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document.write( "That is the right side of the identity.\r\n" );
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document.write( "Edwin

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