Solve Cos(x)=-2: Step-by-Step Guide

[GreenPrint]

Homework Statement

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.

Homework Equations

In work below

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]

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.

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

 

[eumyang]

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

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:

[STRIKE]cos(i ln(2 + √3)),[/STRIKE]

-cos(-i ln(2 + √3)).

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

[STRIKE]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. [/STRIKE]

-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 x=2\pi n +\pi +ln(2+\sqrt{3})/i

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]

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),

we can immediately use cos(π + A) = cosπcosA - sinπsinA = -cosA, and 1/i = -i, to get:

cos(i ln(2 + √3)),

and then since cos(-A) = cosA, and we can always add any whole multiple of 2π, the general solution would be:

cos(nπ ± 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.

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]

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),

we can immediately use cos(π + A) = cosπcosA - sinπsinA = -cosA, and 1/i = -i, to get:

cos(i ln(2 + √3)),

and then since cos(-A) = cosA, and we can always add any whole multiple of 2π, the general solution would be:

cos(nπ ± 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.

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 x=\pi(2n+1)+ln(2+\sqrt{3})/i

Sorry to be so harsh hehe

 

[GreenPrint]

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 x=2\pi n +\pi +ln(2+\sqrt{3})/i

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.

hmmm interesting your right... so just multiply threw by i? let's see...

 

[GreenPrint]

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

 

[Mentallic]

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?

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]

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

Nope, what I mean is to multiple the fraction \frac{ln(2+\sqrt{3})}{i} 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:

<br /> \cos(z) \equiv \frac{e^{i \, z} + e^{-i \, z}}{2}, \; z \in \mathbb{C}<br />

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

<br /> z = \cos(w) = \frac{e^{i \, w} + e^{-i \, w}}{2}<br />

<br /> e^{2 \, i \, w} - 2 \, z \, e^{i \, w} + 1 = 0<br />

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

<br /> e^{i \, w} = z + (z^{2} - 1)^{1/2}<br />

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

<br /> w = \arccos(z) \equiv = -i \, \log{\left(z + (z^{2} - 1)^{1/2}\right)}<br />

 

[GreenPrint]

ok so your also saying that
pi (2n + 1) - i ln(2 + sqrt(3) ) = - i ln (2 + sqrt(3) ) I don't see how?

 

[GreenPrint]

By definition, as you previously did:

<br /> \cos(z) \equiv \frac{e^{i \, z} + e^{-i \, z}}{2}, \; z \in \mathbb{C}<br />

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

<br /> z = \cos(w) = \frac{e^{i \, w} + e^{-i \, w}}{2}<br />

<br /> e^{2 \, i \, w} - 2 \, z \, e^{i \, w} + 1 = 0<br />

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

<br /> e^{i \, w} = z + (z^{2} - 1)^{1/2}<br />

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

<br /> w = \arccos(z) \equiv = -i \, \log{\left(z + (z^{2} - 1)^{1/2}\right)}<br />

I don't understand what you are trying to do here?

 

[Mentallic]

Hang on, let's stop at x=\pi(2n+1)-iln(2+\sqrt{3})

This is correct, now don't change anything drastically. Just use the rules you know. This can be simplified a bit further. Use the log rule that -ln(x)=ln(x^{-1}) and then rationalize the denominator inside the log function. It's not changing the answer, just making things look a little more neat I suppose

 

[Dickfore]

If you use the definition, you will have:

<br /> -i \, \log{(-2 + \sqrt{3})}<br />

<br /> -i \, log{(-2 - \sqrt{3})}<br />

Then, you need the definition of the complex logarithm. It is the inverse function of the complex exponential Let z = \rho \, e^{i \, \theta} and w = u + i \, v and z = \exp(w). Using the properties of the exponential function:

<br /> \rho \, e^{i \, \theta} = e^{u} \, e^{i v}<br />

This means that:
<br /> \rho = e^{u} \Rightarrow u = \ln{\rho}, \; \rho &gt; 0<br />

<br /> v = \theta + 2 \, n \, \pi, \; n \in \mathbb{Z}<br />

So, a complex logarithm is a multiple valued function (with infinite number of branches):

<br /> w = \log{z} \equiv = \ln{\rho} + i \, \left(\theta + 2 \, n \, \pi), \; z = \rho \, e^{i \, \theta}, \; \rho &gt; 0, \; n \in \mathbb{Z}<br />

 

[GreenPrint]

ok then

pi(2n + 1) + i ln(2 - sqrt(3) )
hmmm

 

[Dickfore]

I get a different solution with the imaginary part with the opposite sign. Also, there is another branch of the square root that you hadn't considered.

 

[GreenPrint]

I don't understand why
pi(2n + 1) + i ln(2 - sqrt(3) ) = i ln(2 - sqrt(3) )
i get that it has branches but
pi(2(5) + 2) + i ln(2 - sqrt(3) ) does not equal i ln(2 - sqrt(3) )

 

[Dickfore]

If you look at my definition of the arc cosine, you will see that your right hand sides are not in accordance with it.

 

[GreenPrint]

-i ln ( -2 + sqrt(3) )
raised to -1
i ln( (-2 + sqrt(3))^-1 )
simplified
i ln( 1/(-2 + sqrt(3)) )
Multiplied by conjerget
i ln( (-2 - sqrt(3))/( (-2 + sqrt(3))(-2 - sqrt(3)) ) )
simplify
i ln( (-2 - sqrt(3))/(4 + 2sqrt(3) - 2sqrt(3) - 3)
i ln( (-2 - sqrt(3))/(4 - 3)
i ln( (-2 - sqrt(3) )/1)
?

 

[GreenPrint]

oh wow let me see here let me reread

 

[tiny-tim]

Hi GreenPrint!

(what happened to that π and √ i gave you? )

Mentallic was right … I got a minus in the wrong place.

I've edited my last post to correct it.

 

[Dickfore]

Don't SIMPLIFY anything! Just follow the formulas. Also, why don't you try clicking on some of my formulas to see what the latex code looks like. It helps a lot in terms of readability of your equations.

 

[GreenPrint]

wow ok the thing confused there is that the value that I solved for was actually a side length and you are solving for an angle now? That's kind of werid but I made sense fo the formulas after realizing that... my only question is why didn't you use +/- and just used +