cos(x)=-2 DEFINED?

GreenPrint

1. The problem statement, all variables and given/known data

I was trying to solve that problem ignoring what people say about it not being defined and got a number strangely. I showed step by step so you can tell me what I did wrong as I don't think I did anything wrong what so ever.



2. Relevant equations

In work below

3. The attempt at a solution

e^(ix) = cosx + i sinx = cisx
e^(-ix) = cosx - i sinx = cis(-x)
cis(x) + cis(-x) = cosx + i sin x + cosx - i sinx = 2cosx
from which
cosx = (cis(x) + cis(-x) )/2
from which I set it equal to -2 and began to solve
cosx = (cis(x) + cis(-x) )/2 = -2
Multipled by 2 on both sides
cis(x) + cis(-x) ) = -4
Multipled by cisx on both sides
cis^2(x) + 1 = -4cis(x)
set equal to zero by adding -4cis(x) to both sides
cis^2(x) + 4cis(x) + 1 = 0
solved the quadratic for cis(x)
(-4 +/- sqrt(4^2-4(1)))/2
= -2 +/- sqrt(16 - 4)/2
= -2 +/- sqrt(12)/2
= -2 +/- (2sqrt(3))/2
= -2 +/- sqrt(3)
cis(x) = -2 +/- sqrt(3) = e^(ix)
solved for x took natural log of both sides
ln(-2 +/- sqrt(3)) = ln( e^(ix) )
took out exponents and used ln(e) = 1
ln(-2 +/- sqrt(3)) = ix
divided through by i
ln(-2 +/- sqrt(3)) /i = x
simpified for +/-
for +
ln(-2 + sqrt(3)) /i = x
for -
ln(-2 - sqrt(3)) /i
factored out negative one
ln(-(2 + sqrt(3))/i
used the fact that ln(xy) = lnx + lny
( ln(-1) + ln(2 + sqrt(3)) )/i
used the fact that ln(-1) = i pi
( i pi + ln(2 + sqrt(3)) )/i
canceled out the i
pi + ln((2 + sqrt(3)) )/i = x

eumyang

(Post removed. Completely misunderstood the question.)

CompuChip

I wouldn't say you did anything wrong, rather, you are talking about a different function.
People who tell you that cos(x) = -2 has no solution, are talking about the function from [0, 2pi] to [-1, 1] and then they are right.
If you consider the function defined from C to C (in general) by 1/2(cis(x) + cis(-x)) then you will indeed obtain the answer you got.

[edit] Actually I can kinda agree with you. The definition of the exponential on the complex plane can be rigorously given (in several ways, just avoid the one with sin(x) and cos(x) of course :D) and then defining cos(x) and sin(x) as being the restrictions of (exp(ix) +/- exp(-ix))/2 to the real number is rather reasonable.

GreenPrint

Actually I don't know I just have been told... so I treid it for myself and got a number... so therefore it does exist no?

GreenPrint

What do you mean from C to C what is C?

eumyang

C = set of complex numbers

tiny-tim

Hi GreenPrint!

(have a pi: π and a square-root: √ and try using the X² icon just above the Reply box )

Starting with your final answer, which is cos(π + ln(2 + √3)/i),

(and since cos(-A) = cosA, and we can always add any whole multiple of 2π, and since1/i = -i, the general solution would be: cos((2n+1)π ± i ln(2 + √3)),

we can immediately use cos(π + A) = cosπcosA - sinπsinA = -cosA, to get:

cos(i ln(2 + √3)),

-cos(-i ln(2 + √3)).

ok, let's check that answer, using cosx = (e^ix + e^-ix)/2:

cos(i ln(2 + √3)) = (e^{-ln(2 + √3)} + e^{ln(2 + √3)})/2

= (1/(2 + √3) + (2 + √3))/2 = ((2 - √3) + (2 + √3))/2 = 2.

hmm … something seems to have gone wrong.

-cos(i ln(2 + √3)) = -(e^{-ln(2 + √3)} + e^{ln(2 + √3)})/2

= -(1/(2 + √3) + (2 + √3))/2 = -((2 - √3) + (2 + √3))/2 = -2.

GreenPrint

So if I'm not specifically told we are in the set of real numbers or it's not implied any where or have never been told in class that we will always be in the set of real numbers then indeed I had to solve for cosx = -2 ??? So am I justified in solving for it because it's a summer assignment and so I've never been told to always assume we are in the set of real numbers so I should leave that as a solution on my paper and put "Assuming we are also in set of complex of numbers.." or something?

Mentallic

C means in the complex numbers. When people say it's not defined, they mean it's not defined in the real numbers. In the same way one asks where does the graph y=x²+1 and the x-axis cut each other, well, not at any real number, but it does in the complex numbers.

Now, just to get a bit picky, the log function in the complex plane is multi-valued, so the answer is actually [tex]x=2\pi n +\pi +ln(2+\sqrt{3})/i[/tex]

But you can do better than this. Can you manipulate the above to be in the form x=a+ib?
Rather than dividing by i, it's more conventional to multiply by i.

GreenPrint

hmmm let me look at that... also is that n symbol another symbol for pi or is it suppose to be pi just wounder I'd rather use n for pi then the other symbol...

Mentallic

No no no! He had the right answer, you multiplied by i when he divided by i!
Oh and it wouldn't work for n=2,4,6... rather the formula is [tex]x=\pi(2n+1)+ln(2+\sqrt{3})/i[/tex]

Sorry to be so harsh hehe

GreenPrint

hmmm interesting your right... so just multiply threw by i? lets see...

GreenPrint

pi i (2n + 1) + ln(2 + sqrt(3) )
?

Mentallic

I delved into the complex numbers back in high school when we were learning about quadratics. My teacher got really pissy at me
But ever since I would just act like a smart-xxx on my assignments and get into way more detail than required, just for the laughs. So be my guest!

Mentallic

Nope, what I mean is to multiple the fraction [tex]\frac{ln(2+\sqrt{3})}{i}[/tex] by i in both the numerator and denominator so nothing changes. The denominator will become a real number

GreenPrint

well that would give i^2 in the denomenator or simply -1 and i in the numerator

-i ln(2+ sqrt(3))
?
don't see how that is a real number hmmm

Dickfore

By definition, as you previously did:

[tex]
\cos(z) \equiv \frac{e^{i \, z} + e^{-i \, z}}{2}, \; z \in \mathbb{C}
[/tex]

So, one can try and find the inverse function [itex]w = \arccos(z)[/itex] by solving with respect to [itex]w[/itex]:

[tex]
z = \cos(w) = \frac{e^{i \, w} + e^{-i \, w}}{2}
[/tex]

[tex]
e^{2 \, i \, w} - 2 \, z \, e^{i \, w} + 1 = 0
[/tex]

This is a quadratic equation with respect to [itex]e^{i \, w}[/itex], with a solution:

[tex]
e^{i \, w} = z + (z^{2} - 1)^{1/2}
[/tex]

where the square root has two branches. Taking the logarithm (an infinitely valued complex function):

[tex]
w = \arccos(z) \equiv = -i \, \log{\left(z + (z^{2} - 1)^{1/2}\right)}
[/tex]