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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 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
0 0
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ReportAnswer #1
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
answered by: nellicia
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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
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