Demonstrate the equality (sinx)^4+(cosx)^4=1-2(sinx*cosx)^2?
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There are a couple of ways to show this. I prefer to get both sides to be the same thing. For instance, the left side can be written as:
((sin x)^2 + (cos x)^2)^2 - 2(sin x)^2 (cos x)^2
Well, the bold part of the parenthesis is equal to 1. So:
1 - 2(sin x)^2 (cos x)^2 = 1 - 2 (sin x cos x)^2
We can combine the squares on the left for:
1 - 2 (sin x cos x)^2 = 1 - 2 (sin x cos x)^2
Thus, we have equality.
You need to move all the terms that contain `sin x` and `cos x` to the left side, such that:
`sin^4 x + 2sin^2 x*cos^2 x + cos^4 x = 1`
You need to use the following special product, such that:
`(a + b)^2 = a^2 + 2ab + b^2`
Considering `a = sin^2 x` and `b = cos^2 x` , yields:
`sin^4 x + 2sin^2 x*cos^2 x + cos^4 x = (sin^2 x + cos^2 x)^2`
Using the Pythagorean trigonometric identity, `sin^2 x + cos^2 x = 1` , yields:
`sin^4 x + 2sin^2 x*cos^2 x + cos^4 x =1^2 = 1`
Hence, the last line proves that the given identity `sin^4 x + cos^4 x = 1 - 2sin^2 x*cos^2 x` holds.
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