SOLUTION: prove that cos4x-4cos2x+3=8sin^4x

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Question 800689: prove that cos4x-4cos2x+3=8sin^4x
Answer by Edwin McCravy(20065)   (Show Source): You can put this solution on YOUR website!
We will use the identities

cos(2q) = 1-2sin²(q)
sin(2q) = 2sin(q)cos(q)
cos²(q) = 1-sin²(q)

cos(4x) - 4cos(2x) + 3

1-2sin²(2x) - 4[1-2sin²(x)] + 3

1 - 2[2sin(x)cos(x)]² - 4 + 8sin²(x) + 3

The numbers combine to 0, so we have:

-2[2sin(x)cos(x)]² + 8sin²(x)

-2[4sin²(x)cos²(x)] + 8sin²(x)

-8sin²(x)cos²(x) + 8sin²(x)

-8sin²(x)[1-sin²(x)] + 8sin²(x)

-8sin²(x)+8sin4(x) + 8sin²(x)

8sin4(x)

Edwin

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