Simplifying trig expression

[axrw]

I just wrote this up, but for some reason it didn't go through.


I have this expression that I am supposed to simplify:

(1 + sin 2x + cos 2x)/(1 + sin 2x - cos 2x)

I am getting:

2cot x * [(sin x + cos x)/(sin x + 2cos x)]

I put both in my graphing calc and they don't match, but I can't see another way of simplifying it.

Thanks for any help.

 

[galactus]

I just wrote this up, but for some reason it didn't go through.


I have this expression that I am supposed to simplify:

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

I am getting:

2cot x * [(sin x + cos x)/(sin x + 2cos x)]

I put both in my graphing calc and they don't match, but I can't see another way of simplifying it.

Thanks for any help.

Use the identities:

\(\displaystyle \L\\sin2x=2sinxcosx\)

\(\displaystyle \L\\cos2x=2cos^{2}x-1\)

Make the subs and get:

\(\displaystyle \L\\\frac{1+2sinxcosx+2cos^{2}x-1}{1+2sinxcosx-2cos^{2}x+1}\)

Simplify and factor out 2:

\(\displaystyle \L\\\frac{sinxcosx+cos^{2}x}{1+sinxcosx-cos^{2}x}\)

\(\displaystyle \L\\\frac{cosx(sinx+cosx)}{sinx(cosx+sinx)}\)

=\(\displaystyle \L\\\fbox{cot(x)}\)

 

[axrw]

Thank you galactus. I'm still not sure how you are supposed to keep all these identities in mind while solving or simplifying, but I'll keep practicing. Thank you very much for the help, I can see where I went wrong now.

 

[galactus]

That's something that comes with practice. Using the right identity and hammering it into shape. There are certain ones which are used more frequently than others. There is probably other ways to simplify this one.
A lot of times it helps to get them all in terms of sin, cos, or whatever.