Solving arcsin(√2) with Hyperbolic Sin Function

[srfriggen]

Homework Statement

What is arcsin(√2)?

The Attempt at a Solution

sin^-1(√2)=a+bi

sin(a+bi)=√2

...expressing as hyperbolic sin function:

-i*sinh(-b+ai)=√2

sinh(-b+ai)=-(√2)/i

...using the definition of the sin hyperbolic function:

(e^-b+ai-e^b-ai)/2 = -(√2)/i

2e^-b+ai-2e^b-ai = -2√2*(1/i)

...converting using Euler's Formula:

2e^-b[cos(a)+isin(a)] - 2e^b[cos(-a)+i*sin(-a)] = -2√2*(1/i)

...using the fact that cos(a) and sin(a) are even and odd functions, respectively:

2e^-b[cos(a)+isin(a)] - 2e^b[cos(a)- i*sin(a)] = -2√2*(1/i)

...distributing, then rearranging terms into their Real and Imaginary parts:

(2e^-b-2e^b)cos(a) + (2e^-b+2e^b)sin(a)*i= -2√2*(1/i)


...Splitting this into two separate formulas, with the Reals part on the left equal to the Real part on the right, and the Imaginary part on the left equal to the imaginary part on the right:

{1} (2e^-b-2e^b)cos(a) = 2√2 ,

{2} (2e^-b+2e^b)sin(a) = -1

Now I need to solve for a and b...


Questions: for one, I am not sure if what I have is correct. I think it is, but perhaps I made a mistake.

If this is correct, I'm having a difficult time finding a and b. All previous example we did in class or the book had one equation equal to zero so it was clear where to start. A solution to {2} could be b=0, a=3pi/2, but that doesn't fit {1}. So am a bit stuck.


Thank you in advance for making it this far. I know there is a lot of algebra to this problem and I thank you for taking the time to consider this problem.

 

[vela]

Homework Statement

What is arcsin(√2)?

The Attempt at a Solution

sin^-1(√2)=a+bi

sin(a+bi)=√2

...expressing as hyperbolic sin function:

-i*sinh(-b+ai)=√2

sinh(-b+ai)=-(√2)/i

...using the definition of the sin hyperbolic function:

(e^-b+ai-e^b-ai)/2 = -(√2)/i

2e^-b+ai-2e^b-ai = -2√2*(1/i)

...converting using Euler's Formula:

2e^-b[cos(a)+isin(a)] - 2e^b[cos(-a)+i*sin(-a)] = -2√2*(1/i)

...using the fact that cos(a) and sin(a) are even and odd functions, respectively:

2e^-b[cos(a)+isin(a)] - 2e^b[cos(a)- i*sin(a)] = -2√2*(1/i)

...distributing, then rearranging terms into their Real and Imaginary parts:

(2e^-b-2e^b)cos(a) + (2e^-b+2e^b)sin(a)*i= -2√2*(1/i)


...Splitting this into two separate formulas, with the Reals part on the left equal to the Real part on the right, and the Imaginary part on the left equal to the imaginary part on the right:

{1} (2e^-b-2e^b)cos(a) = 2√2 ,

{2} (2e^-b+2e^b)sin(a) = -1

I didn't check the rest of your work, which seems a bit convoluted, but the righthand side you have does not equal $2\sqrt{2}-i$. It's pure imaginary.

Now I need to solve for a and b...


Questions: for one, I am not sure if what I have is correct. I think it is, but perhaps I made a mistake.

If this is correct, I'm having a difficult time finding a and b. All previous example we did in class or the book had one equation equal to zero so it was clear where to start. A solution to {2} could be b=0, a=3pi/2, but that doesn't fit {1}. So am a bit stuck.


Thank you in advance for making it this far. I know there is a lot of algebra to this problem and I thank you for taking the time to consider this problem.

 

[srfriggen]

I apologize if it seems convoluted. I am following the steps my professor used when doing a similar problem.

I copied down my notes incorrectly... the right hand side should be: -2√2*(1/i).

We are just learning about complex numbers now, so I wasn't sure what 1/i was equal to, but did some algebra and I think it equals -1.

So now the right hand side should look like: 2√2+0*i.

How does that look to you?

 

[vela]

Closer but no cigar. Remember $i^2 = -1$, so $i^4 = (i^2)^2 = (-1)^2 = 1$. That means $1/i = i^4/i = i^3 = (i^2)i = -i$. (Factors of $i$ are a pain. They're mistakes waiting to happen.)

Now that I've taken a closer look, I see the algebra's not that bad. It's just a little strange that you didn't start off with
$$\sin x = \frac{e^{ix} - e^{-ix}}{2i},$$ which you can derive from Euler's formula if you haven't seen it before.

 

[srfriggen]

I see... so a+bi can be written as: 0+(2√2)i

So it looks now like the equations I need to solve, for a and b, are:

{1} (2e^-b-2e^b)cos(a) = 0 ,

{2} (2e^-b+2e^b)sin(a) = 2√2


Factoring out 2's yields:

{1'} (e^-b-2e^b)cos(a) = 0 ,

{2'} (e^-b+2e^b)sin(a) = √2


I notice from {1'} is true when b=0.

Substituting b=0 into {2'} I get:

2sin(a)=√2

sin(a)=√2/2

a=pi/4


So a=pi/4 and b=0 is one solution.

It also looks like b=0 and 3pi/4 is a solution, so in general:

b=0, a=k2pi(pi/4), and b=0, a=k2pi(3pi/4) are the solutions.

 

[vela]

Well, apparently something went wrong. It's always a good idea to check if your answer makes sense. If $a=\pi/4$ and $b=0$, you have $z = a+bi = \pi/4$, and your solution therefore says that $\sin z = \sin \pi/4 = \sqrt{2}$. Clearly, that's not right.

 

[srfriggen]

are my equations for {1} and {2} at least correct? Or did I miss something else?

 

[vela]

You have an extra factor of 2 floating around in them.

 

[srfriggen]

I took those 2's out and made new equations {1'} and {2'}

 

[PeroK]

I see... so a+bi can be written as: 0+(2√2)i

So it looks now like the equations I need to solve, for a and b, are:

{1} (2e^-b-2e^b)cos(a) = 0 ,

{2} (2e^-b+2e^b)sin(a) = 2√2


Factoring out 2's yields:

{1'} (e^-b-2e^b)cos(a) = 0 ,

{2'} (e^-b+2e^b)sin(a) = √2

Something's gone wrong here. You haven't divided by 2 correctly! But, you've also got the factors wrong. If b = 0 you should have sin(a) = √2. But you have 2sin(a) = √2.

 

[vela]

That's not what I mean. Your {1'} and {2'} are also off by a factor of 2 (though in the case of {1'} it doesn't matter).

 

[srfriggen]

shoot, I typed it in incorrectly from my notebook. I see the extra 2.


So when b=0: (1+1)sin(a)=√2, or sin(a)=√2/2. Is this correct?

 

[vela]

No, it's not. Check your previous steps.

 

[PeroK]

shoot, I typed it in incorrectly from my notebook. I see the extra 2.


So when b=0: (1+1)sin(a)=√2, or sin(a)=√2/2. Is this correct?

No, you are out by a factor of 2. If b = 0, the equation should reduce to sin(a) = √2.

 

[srfriggen]

Ok, I re-worked the entire problem (using vela's advice to start with the definition for sin(x) and it made things simpler.

Now I end up with the last two formulas:

[1] (e^-b-e^b)cos(a)=0, and

[2] (e^-b+e^b)sin(a)=2√2

Working with [1] first I can set a=pi/2 to obtain a solution.

Substituting pi/2 into [2] I get:

e^-b+e^b=2√2.


Thoughts?

 

[PeroK]

If you look closely enough you might see a quadratic equation hiding in there!

 

[vela]

You can solve it in terms of inverse cosh. If that's not satisfactory, try using $x=e^b$ so that $1/x = e^{-b}$ and you get
$$x + \frac{1}{x} = 2\sqrt{2}.$$ It's straightforward to solve for $x$.

 

[srfriggen]

so one way to answer would be: b=cosh^-1(√2).

 

[haruspex]

e^-b+e^b=2√2.

Still wrong.
sin(z) = (e^iz+e^-iz)/(2i) = √2
Writing w = e^iz, (w + 1/w)/(2i) = √2

Edit: should be
sin(z) = (e^iz-e^-iz)/(2i) = √2
Writing w = e^iz, (w - 1/w)/(2i) = √2

 

[PeroK]

Still wrong.
sin(z) = (e^iz+e^-iz)/(2i) = √2
Writing w = e^iz, (w + 1/w)/(2i) = √2

sin(z) = (e^iz-e^-iz)/(2i) = √2
Writing w = e^iz, (w - 1/w)/(2i) = √2

But, in any case, it reduces to the same equation for e^b where z = a + ib and a = ∏/2.

 

[srfriggen]

So does this mean the work I've done is incorrect, or did I make it to the end without mistakes?

 

[jackmell]

May I ask why you're not just using the formula for arcsin:

$$\arcsin(z)=-i\log[iz+(1-z^2)^{1/2}]$$

I mean your thread title was to compute $\arcsin(\sqrt{2})$ and not to derive the formula so that's why I'm asking. And you do know it's infinitely-valued right? That's because for one, the square-root in the formula is two values and the log term is of course infinitely valued so it's actually doubly-infinitely-valued.

 

[srfriggen]

sin(z) = (e^iz-e^-iz)/(2i) = √2
Writing w = e^iz, (w - 1/w)/(2i) = √2

But, in any case, it reduces to the same equation for e^b where z = a + ib and a = ∏/2.

Does that mean the work I've done to this point is correct? Or is there a mistake in it?

I still just need to solve the equation x+1/x=2rad2?

 

[srfriggen]

You can solve it in terms of inverse cosh. If that's not satisfactory, try using $x=e^b$ so that $1/x = e^{-b}$ and you get
$$x + \frac{1}{x} = 2\sqrt{2}.$$ It's straightforward to solve for $x$.

so solving for x I get x=2√2±1.

Which means e^b=2√2±1

so b=ln(2√2±1).If this is correct please let me know. I know I need to go back and check other multiples of "a", such as pi*a.

 

[haruspex]

so solving for x I get x=2√2±1.

Which means e^b=2√2±1

so b=ln(2√2±1).


If this is correct please let me know. I know I need to go back and check other multiples of "a", such as pi*a.

The answer cannot be real. You're still losing a factor i. See my corrected post #20:
w² - 1/w = 2i√2
This will give you the same real part, but add an imaginary part.

 

[srfriggen]

The answer cannot be real. You're still losing a factor i. See my corrected post #20:
w² - 1/w = 2i√2
This will give you the same real part, but add an imaginary part.

"a" will be the real part. "b" will have an "i" tacked onto it. I left it out while doing the calculations but that's just how my professor did it in class.

 

[vela]

so solving for x I get x=2√2±1.

You messed up somewhere. You should get $x=\sqrt{2}\pm 1$.

 

[srfriggen]

You messed up somewhere. You should get $x=\sqrt{2}\pm 1$.

yes you're definitely correct. I see my mistake and just fixed it.

so b=ln\sqrt{2}±1*i?

 

[vela]

Right.

 

[srfriggen]

Right.

oh thank god!

thank you for your patience!

 

[vela]

Going back to your two equations
\begin{align*}
(e^b - e^{-b}) \cos a &= 0 \\
(e^b + e^{-b}) \sin a &= 2\sqrt{2}
\end{align*} you chose the $a=\pi/2$ solution to the first equation, which eventually led to two valid solutions. There are other solutions to that equation, namely b=0, as you noted earlier, and $a = 3\pi/2$. Can you see why they don't yield solutions to the problem?

 

[Curious3141]

I have to confess that I didn't scrutinise every single post. But the method used seems unnecessarily complicated.

I would just have gone with:

$\arcsin(\sqrt{2}) = a+bi$ where $a$ and $b$ are real.

so

$\sin(a+bi) = \sqrt{2}$

expanding the LHS and applying $\cos(bi) = \cosh(b)$ and $\sin(bi) = i\sinh(b)$,

$\sin(a)\cosh(b) + i\cos(a)\sinh(b) = \sqrt{2}$

equating real and imaginary parts,

$\sin(a)\cosh(b) = \sqrt{2}$ [eqn 1]

$\cos(a)\sinh(b) = 0$ [eqn 2]

from this, it should be quite easy to solve. I edited out the rest of the solution because complete solutions are not encouraged here. But if the OP has questions on how to proceed from here, he can ask.

EDIT: looking at Vela's last post, the OP did come up with the same simult. equations. But I think introducing the exponential form for them at this stage is an unnecessary complication. Much easier to see the solution leaving them as hyperbolic functions.

 

[srfriggen]

Going back to your two equations
\begin{align*}
(e^b - e^{-b}) \cos a &= 0 \\
(e^b + e^{-b}) \sin a &= 2\sqrt{2}
\end{align*} you chose the $a=\pi/2$ solution to the first equation, which eventually led to two valid solutions. There are other solutions to that equation, namely b=0, as you noted earlier, and $a = 3\pi/2$. Can you see why they don't yield solutions to the problem?

I do:

If b=0, then the original statement that arcsin(√2)=a, but a is a real number and the solution to this is not purely real.

If b=3\pi/2, then we have b=ln(-√2±2), which does not make sense in the case when it is negative.

So, the correct way to write out all of my solutions is:

arcsin(√2)=a+bi, where a=2pi multiples of \pi/2, b=ln(√2±1).

So a can equal pi/2, 5pi/2, 9pi/2, 13pi/2, 17pi/2..., or more generally, 4n(pi/2), where n is an integer.

 

[srfriggen]

Closer but no cigar. Remember $i^2 = -1$, so $i^4 = (i^2)^2 = (-1)^2 = 1$. That means $1/i = i^4/i = i^3 = (i^2)i = -i$. (Factors of $i$ are a pain. They're mistakes waiting to happen.)

Now that I've taken a closer look, I see the algebra's not that bad. It's just a little strange that you didn't start off with
$$\sin x = \frac{e^{ix} - e^{-ix}}{2i},$$ which you can derive from Euler's formula if you haven't seen it before.

Just realized, I can interpret 1/i as a -90 degree rotation of a magnitude 1 vector that lies purely on the Real line for this. So that means 1/i is -i, in the complex plane.

 

[vela]

I do:

If b=0, then the original statement that arcsin(√2)=a, but a is a real number and the solution to this is not purely real.

If b=3\pi/2, then we have b=ln(-√2±2), which does not make sense in the case when it is negative.

Your reasoning for $a=3\pi/2$ is flawed. When dealing with complex numbers, you can, in fact, take the log of a negative number, so you can't use that as a reason for ruling out that solution.

So, the correct way to write out all of my solutions is:

arcsin(√2)=a+bi, where a=2pi multiples of \pi/2, b=ln(√2±1).

So a can equal pi/2, 5pi/2, 9pi/2, 13pi/2, 17pi/2..., or more generally, 4n(pi/2), where n is an integer.

Your list is correct, but your formula isn't. The way you wrote it, $4n(\pi/2) = 2\pi n$, just produces multiples of $2\pi$. You want $a = \pi/2 + 2n\pi$.

 

[Curious3141]

Your reasoning for $a=3\pi/2$ is flawed. When dealing with complex numbers, you can, in fact, take the log of a negative number, so you can't use that as a reason for ruling out that solution.

I'll probably regret butting in when I haven't read every single post (as I said, I find the method used unduly convoluted and tedious), but aren't $a$ and $b$ defined a priori to be real?

 

[vela]

Oh, yeah. Never mind.

 

[srfriggen]

Your reasoning for $a=3\pi/2$ is flawed. When dealing with complex numbers, you can, in fact, take the log of a negative number, so you can't use that as a reason for ruling out that solution.

I was going to say the other way I can see showing it is flawed is by inspecting e^b+e^-b and seeing that if b is assumed to be real it cannot equal -2\sqrt{2}.

Is case closed on this one Vela?

 

[vela]

Yup!

 

[srfriggen]

Yup!

Thank you so much Vela (and all contributors on this post) for all of your help, support, patience, and steering me in the right direction while still making me work for the answer. I definitely learned a lot and feel really good that I understand the problem and I appreciate the help!

 

[srfriggen]

I'm back with one more like this. I've worked through it very hard and I think there should be much less back and forth, as I really get what I'm doing now. But something in this problem isn't working... perhaps there is no solution.

sin^-1(2i)= a+bi, find a and b (and and b are real numbers).

sin(a+bi)=2i


(e^i(a+bi)-e^-i(a+bi))/2i = 2i

e^-b+ai-e^b-ai = 4i²

e^-be^ai-e^be^-ai=-4

e^-b[cos(a)+isin(a)] - e^b[cos(-a)+isin(-a)]=-4

e^-b[cos(a)+isin(a)] - e^b[cos(a)-isin(a)]=-4


{1} (e^-b-e^b)cos(a)=-4

{2} (e^-b+e^b)sin(a)=0


For {2}, a=0 and a=∏ are solutions.

Using a=0 in {1}: e^-b+e^b=-4

Solving this quadratic I get: b=ln(2±√5), which is not a Real solution when b=ln(2-√5).

Similarly, letting a=∏ (or 2∏ multiples of ∏) I get b=ln(-2±\sqrt{5}), which is not always real as well.

From what I see, letting a=n∏, where n is an integer, is the only plausible solution for equations {1} and {2}, yet they do not yield answers.

I reworked the problem twice and am still getting the same answers.

 

[vela]

I'll assume your {1} and {2} are correct. In solving, you flipped a sign. It should be $e^{-b}-e^b = \pm 4$. The exponentials are subtracted.

 

[Curious3141]

I'm back with one more like this. I've worked through it very hard and I think there should be much less back and forth, as I really get what I'm doing now. But something in this problem isn't working... perhaps there is no solution.

sin^-1(2i)= a+bi, find a and b (and and b are real numbers).

sin(a+bi)=2i(e^i(a+bi)-e^-i(a+bi))/2i = 2i

e^-b+ai-e^b-ai = 4i²

e^-be^ai-e^be^-ai=-4

e^-b[cos(a)+isin(a)] - e^b[cos(-a)+isin(-a)]=-4

e^-b[cos(a)+isin(a)] - e^b[cos(a)-isin(a)]=-4{1} (e^-b-e^b)cos(a)=-4

{2} (e^-b+e^b)sin(a)=0For {2}, a=0 and a=∏ are solutions.

Using a=0 in {1}: e^-b+e^b=-4

Solving this quadratic I get: b=ln(2±√5), which is not a Real solution when b=ln(2-√5).

Similarly, letting a=∏ (or 2∏ multiples of ∏) I get b=ln(-2±\sqrt{5}), which is not always real as well.

From what I see, letting a=n∏, where n is an integer, is the only plausible solution for equations {1} and {2}, yet they do not yield answers.

I reworked the problem twice and am still getting the same answers.

Perhaps I could suggest that solution I was thinking of (and which I find easier). Don't introduce exponentials (yet). Leave things in hyperbolic form.

Following the method in my post, you will end up with:

$\sin{a}\cosh{b} = 0$ [eq 1]

$\cos{a}\sinh{b} = 2$ [eq 2]

From 1, you should be able to see that $\cosh b$ can never equal zero. So $\sin a = 0$, yielding $a = n\pi$.

Put that into 2:

$\cos{n\pi}\sinh{b} = 2$

Consider separately $n$ odd and even, yielding:

$\sinh{b} = 2$ for even $n$

and $\sinh{b} = -2$ for odd $n$.

At this point, you can replace the $\sinh$ function with the exponential notation and solve the quadratic, etc. to get the value for b. Alternatively, just use the logarithmic formulation for the inverse hyperbolic sine directly (if you're allowed to) to get $arcsinh{2} = \ln(2 + \sqrt{5})$.

Either way, you should get the general solution quite easily (you can use $n = 2k$ and $n = 2k+1$ to express the even and odd cases more succinctly).

 

[srfriggen]

I'll assume your {1} and {2} are correct. In solving, you flipped a sign. It should be $e^{-b}-e^b = \pm 4$. The exponentials are subtracted.

I'm not sure what you mean by flipped a sign.

Do you mean when I was solving the quadratic? Or before that step?

 

[srfriggen]

Perhaps I could suggest that solution I was thinking of (and which I find easier). Don't introduce exponentials (yet). Leave things in hyperbolic form.

Following the method in my post 4595007, you will end up with:

$\sin{a}\cosh{b} = 0$ [eq 1]

$\cos{a}\sinh{b} = 2$ [eq 2]

From 1, you should be able to see that $\cosh b$ can never equal zero. So $\sin a = 0$, yielding $a = n\pi$.

Put that into 2:

$\cos{n\pi}\sinh{b} = 2$

Consider separately $n$ odd and even, yielding:

$\sinh{b} = 2$ for even $n$

and $\sinh{b} = -2$ for odd $n$.

At this point, you can replace the $\sinh$ function with the exponential notation and solve the quadratic, etc. to get the value for b. Alternatively, just use the logarithmic formulation for the inverse hyperbolic sine directly (if you're allowed to) to get $arcsinh{2} = \ln(2 + \sqrt{5})$.

Either way, you should get the general solution quite easily (you can use $n = 2k$ and $n = 2k+1$ to express the even and odd cases more succinctly).

I do see what you mean, and really appreciate this insight, but at this point I've written out the problem to be handed in very neatly and I'd rather just correct whatever mistake I have then re-work it.

 

[Curious3141]

I do see what you mean, and really appreciate this insight, but at this point I've written out the problem to be handed in very neatly and I'd rather just correct whatever mistake I have then re-work it.

No worries. It comes to pretty much the same idea as yours, but I just find it neater (and simpler) to leave the computation of $b$ to the end.

Basically, you've defined $b$ to be real. When you solve the quadratic for the respective cases, just take the solution that gives you a real logarithm (and hence a real $b$). Reject the other solution. It's that simple.

 

[srfriggen]

I'm not sure what you mean by flipped a sign.

Do you mean when I was solving the quadratic? Or before that step?

when solving the quadratic for a=0 this is what I get:

4±\sqrt{(-4^2)-4(1)(-1)/2}

= 4±\sqrt{20}/2

=2±√5 = e^b


b=ln(2±√5)

 

[Curious3141]

when solving the quadratic for a=0 this is what I get:

4±\sqrt{(-4^2)-4(1)(-1)/2}

= 4±\sqrt{20}/2

=2±√5 = e^b


b=ln(2±√5)

This is correct, but as I said in the edit in my last post, only $b = \ln (2 + \sqrt{5})$ is admissible, because $b$ is by definition real.

You should explicitly write that line into your derivation at the top, i.e. "Let the solution be $z = a + bi$, where $a$ and $b$ are real." EVERY complex number can be represented in this form, so there is no loss of generality.

 

[srfriggen]

This is correct, but as I said in the edit in my last post, only $b = \ln (2 + \sqrt{5})$ is admissible, because $b$ is by definition real.

You should explicitly write that line into your derivation at the top, i.e. "Let the solution be $z = a + bi$, where $a$ and $b$ are real." EVERY complex number can be represented in this form, so there is no loss of generality.

cool, I understand now. I just wasn't sure if I was "allowed" to dismiss the ln(2-√5).

but now I have values for a and b, namely, a=2pi+2npi and b=ln(2+√5).

Thanks!

 

[srfriggen]

cool, I understand now. I just wasn't sure if I was "allowed" to dismiss the ln(2-√5).

but now I have values for a and b, namely, a=2pi+2npi and b=ln(2+√5).

Thanks!

actually wait one second... if I choose a=∏ , then my b=ln(-2±√5), which is a real solution for ln(-2+√5).

So do I have two different a and b combination solutions? i.e. if a=pi, then b=n(2+√5), and if a=0, b=(2+√5)?