Sin^2x-cos^2x= 1/2 find value of sin^4x-cos^4x
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we know sin^2 x +cos^2 x = 1
therefore sin^2x +cos^2x=1
and sin^2x-cos^2x=1/2
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2sin^2 x= 1/2+1
sin^2 x =3/4
cos^2x = 1-3/4
=1/4
so sinx = √3/2
and cosx=1/2
so sin^4x-cos^4x
= (√3/2)^4-(1/2)^4
=1/2
therefore sin^2x +cos^2x=1
and sin^2x-cos^2x=1/2
___________________
2sin^2 x= 1/2+1
sin^2 x =3/4
cos^2x = 1-3/4
=1/4
so sinx = √3/2
and cosx=1/2
so sin^4x-cos^4x
= (√3/2)^4-(1/2)^4
=1/2
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