SOLUTION: How to prove this following identity? (sin2x+cos2x)/(2cosx+sinx-2(cos^3x+sin^3x))=cosecx

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Question 976230: How to prove this following identity?
(sin2x+cos2x)/(2cosx+sinx-2(cos^3x+sin^3x))=cosecx

Answer by FrankM(1040) About Me  (Show Source):
You can put this solution on YOUR website!

First, the sum of cubes
a%5E3%2Bb%5E3+=+%28a-b%29%28a%5E2-ab%2Bb%5E2%29
This applies to the denominator

but you see the sin squared and cos squared add to 1?

simplify denominator a bit more to get:


Note that cos2%28x%29=2cos%5E2%28x%29-1%29
and if we factor a sin out of the first two terms in denominator we get:

and can make the substitution:

Next we have
sin%282x%29=2sin%28x%29cos%28x%29
so let's factor sin out of the later part of denominator

and make the substitution

now, factor out sin in the denominator:
+%28sin%282x%29%2Bcos%282x%29%29%2F%28sin%28x%29+%28cos%282x%29%2Bsin%282x%29+%29%29+
And after canceling numerator/denominator all we have left is
1%2Fsin%28x%29
which is cosecant.
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.