1. Aug 26, 2010

### GreenPrint

1. The problem statement, all variables and given/known data
Factor completely
(cis(x) - cis(-x) )/(2i) + (1 - cis^2(-x))/(i + i cis^2(-x) )

Yes I know that it's just simply sin(x) + tan(x)
which equals sin(x) ( 1 + sec(x) )

but however if you write it out in it's proper form

(cis(x) - cis(-x) )/(2i) + (1 - cis^2(-x))/(i + i cis^2(-x) )

and try to factor this further you'll realize that you can factor out cis(x) and cis(-x) and what not from sine and cosine so leaving it as this

sin(x) ( 1 + sec(x) )

is wrong because you can factor further... I need help with this

http://www.wolframalpha.com/input/?i=factor+(e^(ix)-e^(-ix))/(2i)+(1+%2B+2/(+e^(ix)+%2B+e^(-ix)+)

the answer is that according to wolfram which is clearly more factored then

sin(x) ( 1 + sec(x) )

because it be factored into one fraction... which is also much more simple than three different ones =)... but regardless I need to factor that and get that answer and don't know how to...

I got down to here

-i( cis(x) (1/2 + cis(x) ) - (cis(-x)/2 + 1 ) )

so how do i get to this

-(i cis(-x) (-1+ cis(x) ) (1+ cis^3 (x) )/(2 (1+ cis^2(x)))

as you can see it's much more factored =) can you assist me to get here thanks

2. Relevant equations

3. The attempt at a solution

2. Aug 26, 2010

### GreenPrint

lol does anybody know how to do this?

3. Aug 26, 2010

### Staff: Mentor

Why do you think this is wrong? It's the same as what you got before. And really, you are not factoring, but instead are simplifying the original expression.
???
Why would you want to get there?
To simplify the original expression, replace the cis expressions, using cis(x) = cos(x) + i sin(x), and cis(-x) = cos(-x) + i sin(-x) = cos(x) - i sin(x). Here I'm using the identities cos(-x) = cos(x) and sin(-x) = - sin(x).

I was able to rewrite the original expression as sin(x) + tan(x), which you point out is equal to sin(x)(1 + sec(x)). Either of these would be good answers.

4. Aug 26, 2010

### GreenPrint

isn't this more factored because the sis function has been factored
like I would say that
-(i e^(-i x) (-1+e^(i x)) (1+e^(i x))^3)/(2 (1+e^(2 i x)))
was more factored than
sin(x)(1 + sec(x))
because you can factor out the sis function... ?
and I don't know how to get to
-(i e^(-i x) (-1+e^(i x)) (1+e^(i x))^3)/(2 (1+e^(2 i x)))
and when I tried I got to
-i( cis(x) (1/2 + cis(x) ) - (cis(-x)/2 + 1 ) )
and am stuck and was wanting to know how to get to the answer three lines above from there because I don't know how to factor further from that point =( and get that answer

5. Aug 26, 2010

### Staff: Mentor

Do you understand what factoring means?

When you factor an expression, you write it as a product of two or more expressions. For example, x2 + 5x + 6 can be factored to (x + 3)(x + 2).

6. Aug 26, 2010

### GreenPrint

Yes I do think so =)

see I put this in
http://www.wolframalpha.com/input/?i=factor+sin(x)+++tan(x)
and was like well that better equal

http://www.wolframalpha.com/input/?i=factor+(e^(ix)+-+e^(-ix))/(2i)+%2B+(1+-+e^(-2ix)+)/(i+%2B+i+e^(-2x))

hmm that is odd
looks like i forgot the to square it in the first post I'll edit the link

Last edited: Aug 26, 2010
7. Aug 26, 2010

### Staff: Mentor

It seems to me that you have this need to make things as complicated as possible, the exact opposite of simplification...

8. Aug 26, 2010

### GreenPrint

hmmm nope lets see here

Last edited: Aug 26, 2010
9. Aug 26, 2010

### Staff: Mentor

sin(x) + tan(x) is the simplest of the bunch. I have no idea whether any of the others are actually equal to sin(x) + tan(x).

10. Aug 26, 2010

### GreenPrint

hmmm so what is the most factored form? My head hurts =_=

11. Aug 27, 2010

### ehild

Divide both the numerator and the denominator of the second factor by cis(-x) and use that 1/cis(-x)=cis(+x).

ehild