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herz · Mar 20, 2008, 08:31 AM

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!

galactus · Mar 20, 2008, 09:09 AM

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

Is this it?:

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

herz · Mar 20, 2008, 09:14 AM

Yes that is the priblem

herz · Mar 20, 2008, 10:21 AM

Originally Posted by galactus

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

galactus · Mar 20, 2008, 01:00 PM

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.