document.write( "Question 976230: How to prove this following identity?
\n" ); document.write( "(sin2x+cos2x)/(2cosx+sinx-2(cos^3x+sin^3x))=cosecx
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Algebra.Com's Answer #598227 by FrankM(1040)\"\" \"About 
You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( "First, the sum of cubes \r
\n" ); document.write( "\n" ); document.write( "\"a%5E3%2Bb%5E3+=+%28a-b%29%28a%5E2-ab%2Bb%5E2%29\"\r
\n" ); document.write( "\n" ); document.write( "This applies to the denominator \r
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\n" ); document.write( "\n" ); document.write( "but you see the sin squared and cos squared add to 1? \r
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\n" ); document.write( "\n" ); document.write( "simplify denominator a bit more to get:\r
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\n" ); document.write( "\n" ); document.write( "Note that \"cos2%28x%29=2cos%5E2%28x%29-1%29\"\r
\n" ); document.write( "\n" ); document.write( "and if we factor a sin out of the first two terms in denominator we get: \r
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\n" ); document.write( "\n" ); document.write( "and can make the substitution:\r
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\n" ); document.write( "\n" ); document.write( "Next we have \r
\n" ); document.write( "\n" ); document.write( "\"sin%282x%29=2sin%28x%29cos%28x%29\"\r
\n" ); document.write( "\n" ); document.write( "so let's factor sin out of the later part of denominator\r
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\n" ); document.write( "\n" ); document.write( "and make the substitution\r
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\n" ); document.write( "\n" ); document.write( "now, factor out sin in the denominator:\r
\n" ); document.write( "\n" ); document.write( "\"+%28sin%282x%29%2Bcos%282x%29%29%2F%28sin%28x%29+%28cos%282x%29%2Bsin%282x%29+%29%29+\" \r
\n" ); document.write( "\n" ); document.write( "And after canceling numerator/denominator all we have left is \r
\n" ); document.write( "\n" ); document.write( "\"1%2Fsin%28x%29\"\r
\n" ); document.write( "\n" ); document.write( "which is cosecant.\r
\n" ); document.write( "\n" ); document.write( "There may be a faster or more elegant method, but it's key to see that the sum of two cubes had to be factored to start the process.
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