Ask Me Help Desk - Verify that sinx=cosx=(2sin^2x-1)/9sinx-cosx)?

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verify that sinx=cosx=(2sin^2x-1)/9sinx-cosx)?

What you wrote is this:

$ \sin x \ = \ \cos x \ = \frac {2 sin^2 x} {9 \sin x - \cos x} $

Is that really what you meant?

I meant this (sinx+cosx)=(2sin^2x-1)/(sinx-cosx)?

Comment on ebaines's post

I meant (sinx+cosx)=(2sin^2x)/(sinx-cosx)

Quote:

Originally Posted by Katelyn45

I ment this (sinx+cosx)=(2sin^2x-1)/(sinx-cosx)?

OK:

$ \sin x + \cos x = \frac { \sin^2x-1}{\sin x - \cos x} $

Here's a hint to get you started: multiply through by the denominator of the right hand side. This turns the left hand side into

$ (\sin x + \cos x )(\sin x - \cos x) = \sin^2 x - cos^2 x $

Now think about the various identities for cos (2x). OK, that's enough of a hint - can you take it from here?

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