\({sin(2x)-cos(x) \over 4sin^2x - 1} = {sin^2 x (cos x) + cos^2x \over 2 sin(x)+1}\)

It looks like the denominator for the right expression is a perfect square... but after that, I'm not sure what else to do!

Guest Feb 21, 2019

#1**+1 **

Lets see..

\({sin(2x)-cos(x) \over 4sin^2x - 1} = {sin^2 x (cos x) + cos^2x \over 2 sin(x)+1}\\ LHS\\ ={sin(2x)-cos(x) \over 4sin^2x - 1} \\ =\frac{2sinxcosx-cos(x) } {(2sinx - 1)(2sinx+1)} \\ =\frac{cosx(2sinx-1) } {(2sinx - 1)(2sinx+1)} \\ =\frac{cosx } {2sinx+1} \\ \)

\(RHS\\ ={sin^2 x (cos x) + cos^2x \over 2 sin(x)+1}\\ \ne LHS\)

I suspect this is an error on our part on on the part of your teacher or text.

Melody Feb 21, 2019