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Demonstrate equality (sinx)^6 + (cosx)^6 = 1 - 3(six*cosx)^2
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You should use the following special product, such that:
`(a + b)^3 = a^3 + b^3 + 3ab(a + b)`
Considering `a = sin^2 x` and `b = cos^2 x` yields:
`(sin^2 x + sin^2 x)^3 = (sin^2 x)^3 + (cos^2 x)^3 + 3sin^2 x*cos^2 x(sin^2 x + cos^2 x)`
You need to use Pythagorean identity, such that:
`sin^2 x + sin^2 x = 1`
Replacing `1` for `sin^2 x + sin^2 x` in the expansion above, yields:
`1^3 = sin^6 x + cos^6 x + 3sin^2 x*cos^2 x*1`
Keeping `sin^6 x + cos^6 x` to one side, yields:
`1 - 3sin^2 x*cos^2 x = sin^6 x + cos^6 x`
The last line represents exactly the identity you need to prove, hence, testing if the given expression holds, yields that `sin^6 x + cos^6 x = 1 - 3sin^2 x*cos^2 x` represents a valid statement.
Posted by sciencesolve on September 23, 2013 at 5:32 PM (Answer #1)
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