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How would I find $\arcsin 2$? I'm helping my little sister with her calculus "pre-test" before classes begin, and I don't remember how to do it in order to explain to her.

Help?

  • 1
    Are you working in the reals or complex numbers?2012-08-28
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    I don't think they've covered imaginary numbers. So I would assume we're sticking with real numbers.2012-08-28
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    Then there is no solution, as the answers below attest :-(.2012-08-28
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    Perhaps she meant $\arcsin \frac{1}{2}$? That would be a fairly usual number that can be easily worked out using an equilateral triangle (all angles 60°) of side 1 and bisecting one of the angles. Then use the definition of $\sin$ to get $\arcsin \frac{1}{2} = 30°$. Just a thought...2012-08-28
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    It's definitely arcsin(2), but good note.2012-08-28
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    I vaguely seem to recall that one of my answers here from a few months ago explained how to find a complex number whose sine is $2$. But I might not call that $\arcsin 2$. It can't be done with _real_ numbers.2012-08-28
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    $\sin (\frac{\pi}{2} + i \mathbb{arccosh} 2) = 2$.2012-08-28

4 Answers 4

1

In addition to the answers you received above, you might want to plot functions when not sure and see if that provides insights.

See: http://www.wolframalpha.com/input/?i=plot%20arcsin(x)&t=crmtb01

You may also want to look into a CAS like SAGE or Maxima.

Professional variants are Mathematica or Maple.

These are very helpful for students to learn as they are helpful for exploratory mathematics.

HTH

11

This is an answer that gives imaginary solutions. Note that the OP never specified that this couldn't be used.

You are trying to find:

$$\sin x=2$$

This obviously has no real solutions, so we will look for imaginary solutions. Rewrite $x$ as $a+bi$ where $a,b\in\Bbb{R}$. That is crucial.

$$\sin (a+bi)=2$$

Using the arguement sum property (not sure of the name) of $\sin$, you can isolate the $a$ and $bi$ into bite-sized parts. Also, multiply $-i$ and $i$

$$\sin a \cos bi+\cos a \sin bi=2$$

Now, we know some properties of hyperbolic trig allowing us to alter some of the stuff along the following rules. Those rules would be here. Now we are getting somewhere.

$$\sin a \cosh b+(\cos a) *(i\sinh b)=2$$ which we can rewrite as

$$\sin a \cosh b+i\cos a\sinh b=2$$

Because $a,b$ have to be real, and that $\sinh b\not=0$, you are forced to assume that $a=\pi/2+2\pi z$. The reason? Well, the term $i\cos a \sinh b$ is imaginary and the term (2) you are trying to get is strictly real. It is imaginary because both $\cos$ and $\sinh $ are functions of $\Bbb{R}\to\Bbb{R}$

So, we now know that $a=\pi/2+2\pi z$ from logical sense. (where $z\in\Bbb{Z}$)

$$\sin(\pi/2+2\pi z)\cosh b=2\implies \cosh b=2$$

Now since $\cosh$ is an even function:

$$b=\pm \text{arcosh}(2)$$

Plugging that back into the almost original equation, the one with $\sin(a+bi)$, we get:

$$\sin(\frac{\pi}{2}+2\pi z\pm i\text{arccosh}(2) )=2$$

Viola!

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    This is actually the correct answer. Sine function can extend over 1 in imaginary plane2017-07-16
5

You don't. Arcsine is the inverse function of sine. The domain of arcsine is the range of sine which is $[-1,1]$ and $2$ is not in there.

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    So, the answer is either complex or "no solution", correct?2012-08-28
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    @DoctorOreo Correct. Examine the function plotted: http://upload.wikimedia.org/wikipedia/commons/thumb/6/6c/Arcsine.svg/280px-Arcsine.svg.png2012-08-28
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    in your context, yes "no solution".2012-08-28
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The arcsin(2) is not a real number. Recall that $$x = \arcsin(2)$$ is equivalent to the equation $$\sin(x) = 2.$$ Since the range of $\sin(x)$ as a real valued function is $[-1,1]$, the original equation has no real solution. I doubt a calculus exam wants a complex valued solution.