View Full Version : Verify that sinx=cosx=(2sin^2x-1)/9sinx-cosx)?
Katelyn45
Mar 29, 2011, 02:31 PM
ebaines
Mar 29, 2011, 02:43 PM
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?
Katelyn45
Mar 29, 2011, 03:00 PM
I meant this (sinx+cosx)=(2sin^2x-1)/(sinx-cosx)?
Katelyn45
Mar 29, 2011, 03:07 PM
I meant (sinx+cosx)=(2sin^2x)/(sinx-cosx)
ebaines
Mar 30, 2011, 05:54 AM
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?