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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.


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Are you working in the reals or complex numbers? – copper.hat Aug 28 '12 at 2:25
I don't think they've covered imaginary numbers. So I would assume we're sticking with real numbers. – Cody Aug 28 '12 at 2:26
Then there is no solution, as the answers below attest :-(. – copper.hat Aug 28 '12 at 2:35
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... – copper.hat Aug 28 '12 at 2:38
It's definitely arcsin(2), but good note. – Cody Aug 28 '12 at 2:54
up vote 4 down vote accepted

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? – Cody Aug 28 '12 at 2:27
@DoctorOreo Correct. Examine the function plotted:… – Argon Aug 28 '12 at 2:27
in your context, yes "no solution". – James S. Cook Aug 28 '12 at 2:39

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.

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


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.


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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:

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


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