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Asked by yarku

2sin(180°+x)sin(90°+x)/cost^4x-sin^4x
9 years ago

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Answered by Steve
2sin(180°+x)sin(90°+x)/cos^4x-sin^4x
2(-sinx)(cosx)/(cos^2x-sin^2x)(cos^2x+sin^2x)
-sin2x/cos2x
-tan2x
9 years ago
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2sin(180°+x)sin(90°+x)/cost^4x-sin^4x

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