View Full Version : Verifiying trig id's
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
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.