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) About Me  (Show Source):
You can put this solution on YOUR website!
We will use the identities

cos(2q) = 1-2sinē(q)
sin(2q) = 2sin(q)cos(q)
cosē(q) = 1-sinē(q)

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

1-2sinē(2x) - 4[1-2sinē(x)] + 3

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

The numbers combine to 0, so we have:

-2[2sin(x)cos(x)]ē + 8sinē(x)

-2[4sinē(x)cosē(x)] + 8sinē(x)

-8sinē(x)cosē(x) + 8sinē(x)

-8sinē(x)[1-sinē(x)] + 8sinē(x)

-8sinē(x)+8sin4(x) + 8sinē(x)

8sin4(x)

Edwin