SOLUTION: The identity of: sin^4x+cos^4x= (sin^2x+cos^2x)(sin^2x+cos^2x) sin^2x+cos^2=1 This is the answer I come up with, it is not one of the options available as an answer. The a

Algebra ->  Trigonometry-basics -> SOLUTION: The identity of: sin^4x+cos^4x= (sin^2x+cos^2x)(sin^2x+cos^2x) sin^2x+cos^2=1 This is the answer I come up with, it is not one of the options available as an answer. The a      Log On


   

Question 106492: The identity of:
sin^4x+cos^4x=
(sin^2x+cos^2x)(sin^2x+cos^2x)
sin^2x+cos^2=1
This is the answer I come up with, it is not one of the options available as an answer.
The answers given are
1. -2sin^2xcos^2x
2. 1+2sin^2x-2sin^4x
3. 1+3sin^3x-2sin^2x
4. 1-2sin^2x+2sin^4x
5. 0
Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
check your factoring...

y=sin^4x+cos^4x ... subtracting 2sin^4x
...y-2sin^4x=cos^4x-sin^4x ... factoring
...y-2sin^4x=(cos^2x+sin^2x)(cos^2x-sin^2x)=cos^2x-sin^2x ... adding 2sin^2x
...y-2sin^4x+2sin^2x=cos^2x+sin^2x=1 ... adding (2sin^4x-2sin^2x)
...y=1+2sin^4x-2sin^2x