HELP!

I am stuck as to where to begin on this... any guidance? I am not seeing where to start.

1+sin2x / cosx + sinX = cos2x / cosx - sinx

Thanks in advance!

Without proper grouping symbols it's difficult to tell what you mean.

Is this it?:

\frac{1+sin(2x)}{cos(x)+sin(x)}=\frac{cos(2x)}{cos (x)-sin(x)}

Yes that is the priblem

Without proper grouping symbols it's difficult to tell what you mean.

Is this it?:

\frac{1+sin(2x)}{cos(x)+sin(x)}=\frac{cos(2x)}{cos (x)-sin(x)}

correct, this is the problem

We must verify the identity. Look at the right side.

We can take note that

cos(2x)=cos^{2}(x)-sin^{2}(x)

Notice this is the difference of two square, which can be written as

cos^{2}(x)-sin^{2}(x)=(cos(x)+sin(x))(cos(x)-sin(x))

Then we have \frac{(cos(x)+sin(x))(\sout{cos(x)-sin(x)})}{\sout{cos(x)-sin(x)}}

So, \frac{1+sin(2x)}{cos(x)+sin(x)}=cos(x)+sin(x)

1+sin(2x)=cos^{2}(x)+2sin(x)cos(x)+sin^{2}(x)

Remember that 2sin(x)cos(x)=sin(2x) and cos^{2}(x)+sin^{2}(x)=1

So, we have 1+sin(2x)=1+sin(2x)

And it is shown.