Question 635024
Hello,
The problem that I have is that I need to verify the identity:
         sin^4x+cos^4x=1-2cos^2x+2cos^4x
I started with the right hand side and I tried to group the two cosine identities
=1-2cos^2x+2cos^4x
=1-2cos^2x(1-cos^2x)
using the Pythagorean identity I got
=1-2cos^2x(sin^2x)
I'm not sure if I can use the Pythagorean identity on the first part making it
=(2sin^2x)(sin^2x)
But if I do, I'm not getting any closer to the left hand side. 
Please help, or tell me if I'm at least on the right track.
Thank you 


{{{sin^4(x) + cos^4(x) = 1 - 2cos^2(x) + 2cos^4(x)}}}


{{{sin^4(x) = 1 - 2cos^2(x) + 2cos^4(x) - cos^4(x)}}}


{{{sin^4(x) = 1 - 2cos^2(x) + cos^4(x)}}}


Since {{{sin^2(x) = 1 - cos^2(x)}}}, then {{{sin^4(x) = (1 - cos^2(x))^2}}}


We now have: {{{highlight_green((1 - cos^2(x))^2 = 1 - 2cos^2(x) + cos^4(x))}}}


When the left-side is FOILED, you end up with the right-side.


OR


{{{sin^4(x) + cos^4(x) = 1 - 2cos^2(x) + 2cos^4(x)}}}


{{{sin^4(x) = 1 - 2cos^2(x) + 2cos^4(x) - cos^4(x)}}}


{{{sin^4(x) = 1 - 2cos^2(x) + cos^4(x)}}}


{{{2cos^2(x) - cos^4(x) = 1 - sin^4(x)}}}


{{{cos^2(x)(2 - cos^2(x)) = 1 - sin^4(x)}}}


{{{cos^2(x)(2 - cos^2(x)) = (1 - sin^2(x))(1 + sin^2(x))}}}


Since {{{1 - sin^2(x) = cos^2(x)}}}, then: {{{cos^2(x)(2 - cos^2(x)) = (1 - sin^2(x))(1 + sin^2(x))}}}  becomes: {{{cos^2(x)(2 - cos^2(x)) = cos^2(x)(1 + sin^2(x))}}}


This means that {{{2 - cos^2(x) = 1 + sin^2(x)}}}, or {{{sin^2(x) + cos^2(x) = 2 - 1}}}


{{{highlight_green(sin^2(x) + cos^2(x) = 1)}}} (TRUE PYTHAGOREAN IDENTITY)


Send comments and “thank-yous” to “D” at  MathMadEzy@aol.com