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Can you prove the identity: sin^2x+cos^4x=cos^2x+sin^4x ?
I'm usually pretty good at these questions, but this one has me stumped! This is as far as I have gotten:
Left Side:
=sin^2x+cos^4x
=sin^2x+(1-sin^2x)cos^2x
=sin^2x+cos^2-sin^2xcos^2x
And this is all I got, anything else I try does not equal the right side.
If you have any other ways of going though with this, please feel free to use that! I just wanted to just start anyone off!
Thank you!
2 Answers
- RU MattLv 67 years agoFavorite Answer
sin^2x + cos^4x = cos^2x + sin^4x
Use Pythagorean Identity (PI) to break down cos^4x:
sin^2x + (1 - sin^2x)^2 = cos^2x + sin^4x
Expand by FOILing:
sin^2x + 1 - 2sin^2x + sin^4x = cos^2x + sin^4x
Use PI again to re-write the sine terms:
(1 - cos^2x) + 1 - 2(1 - cos^2x) + sin^4x = cos^2x + sin^4x
Distribute:
1 - cos^2x + 1 - 2 + 2cos^2x + sin^4x = cos^2x + sin^4x
Collect terms:
cos^2x + sin^4x = cos^2x + sin^4x
Source(s): BA in Mathematics - grunfeldLv 77 years ago
LHS = sin^2 x + cos^4 x
LHS = sin^2 x + ( 1 - sin^2 x )^2
LHS = sin^2 x + 1 - 2 sin^2 x + sin^4 x
LHS = 1 - sin^2 x + sin^4 x
LHS = cos^2 x + sin^4 x
LHS = RHS
QED
Source(s): my brain