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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. – Doctor Oreo 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. – Doctor Oreo Aug 28 '12 at 2:54

3 Answers 3

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? – Doctor Oreo 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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