# Homework Help: Arcsin(sqrt 2)

1. Dec 6, 2013

### srfriggen

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

What is arcsin(√2)?

3. 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-eb-ai)/2 = -(√2)/i

2e-b+ai-2eb-ai = -2√2*(1/i)

...converting using Euler's Formula:

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

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

(2e-b-2eb)cos(a) + (2e-b+2eb)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-2eb)cos(a) = 2√2 ,

{2} (2e-b+2eb)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.

2. Dec 6, 2013

### vela

Staff Emeritus
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.

3. Dec 6, 2013

### 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?

4. Dec 6, 2013

### vela

Staff Emeritus
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.

5. Dec 6, 2013

### 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-2eb)cos(a) = 0 ,

{2} (2e-b+2eb)sin(a) = 2√2

Factoring out 2's yields:

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

{2'} (e-b+2eb)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.

6. Dec 6, 2013

### vela

Staff Emeritus
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.

7. Dec 6, 2013

### srfriggen

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

8. Dec 6, 2013

### vela

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

9. Dec 6, 2013

### srfriggen

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

10. Dec 6, 2013

### PeroK

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.

11. Dec 6, 2013

### vela

Staff Emeritus
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).

12. Dec 6, 2013

### 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?

13. Dec 6, 2013

### vela

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

14. Dec 6, 2013

### PeroK

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

15. Dec 6, 2013

### srfriggen

Now I end up with the last two formulas:

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

[2] (e-b+eb)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+eb=2√2.

Thoughts?

16. Dec 6, 2013

### PeroK

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

17. Dec 6, 2013

### vela

Staff Emeritus
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 $x$.

18. Dec 6, 2013

### srfriggen

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

Last edited: Dec 6, 2013
19. Dec 6, 2013

### haruspex

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

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

Last edited: Dec 6, 2013
20. Dec 6, 2013

### PeroK

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

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

21. Dec 6, 2013

### srfriggen

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

22. Dec 6, 2013

### 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.

23. Dec 6, 2013

### srfriggen

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?

24. Dec 6, 2013

### srfriggen

so solving for x I get x=2√2±1.

Which means eb=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.

25. Dec 6, 2013

### haruspex

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.