Time remaining:
-cos2x/(sinx-cosx)^2

label Calculus
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schedule 0 Hours
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I need assistance in simplifying this double angle identity; I have tried to use the cosine double angle in all its three forms and nothing seems to work. 

Apr 14th, 2015

= cos2x /   1 - 2cos x . Sin x

alternate form can be....

= cos2x /  (sin2x-1)


Apr 14th, 2015

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Apr 14th, 2015
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Apr 14th, 2015
Jun 26th, 2017
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