Can You Simplify This Trigonometric Expression Further?

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In summary, the conversation discusses factoring a complex expression and determining the most factored form. The original expression is simplified to sin(x) + tan(x) and compared to the simplified form obtained using the identities cis(x) = cos(x) + i sin(x) and cis(-x) = cos(x) - i sin(x). The conversation ends with a suggestion to divide the second factor by cis(-x) and simplify further.
  • #1
GreenPrint
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Homework Statement


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

Homework Equations





The Attempt at a Solution

 
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  • #2
lol does anybody know how to do this?
 
  • #3
GreenPrint said:

Homework Statement


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) )
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.
GreenPrint said:
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) )
?
GreenPrint said:
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
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
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
GreenPrint said:
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... ?
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
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:
  • #7
It seems to me that you have this need to make things as complicated as possible, the exact opposite of simplification...
 
  • #8
hmmm nope let's see here
 
Last edited:
  • #9
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
hmmm so what is the most factored form? My head hurts =_=
 
  • #11
Divide both the numerator and the denominator of the second factor by cis(-x) and use that 1/cis(-x)=cis(+x).

ehild
 

FAQ: Can You Simplify This Trigonometric Expression Further?

How do I factor a polynomial?

To factor a polynomial, you need to find its factors, which are numbers or expressions that can be multiplied together to give the original polynomial. To do this, you can use methods like factoring by grouping, the difference of squares, or the quadratic formula.

What is the difference between factoring and expanding?

Expanding is the process of multiplying out a factored expression, while factoring is the opposite process of finding the factors of an expression. In other words, factoring breaks down a polynomial into smaller parts, while expanding combines smaller parts into a larger whole.

How do I know when a polynomial is completely factored?

A polynomial is completely factored when it can no longer be broken down into any smaller parts. This means that all of its factors are prime numbers or irreducible expressions. You can check this by multiplying out the factors to see if they equal the original polynomial.

What are some common factoring techniques?

Some common factoring techniques include factoring by grouping, the difference of squares, the sum and difference of cubes, and the quadratic formula. Each technique is useful for factoring different types of polynomials, so it's important to familiarize yourself with each one.

Why is factoring important in math?

Factoring is important in math because it allows us to simplify complicated expressions and solve equations more easily. It also helps us find the roots and zeros of a polynomial, which are important in graphing and understanding the behavior of functions. Additionally, factoring is used in real-life applications, such as in finance and engineering.

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