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Subject: "Arcsin as different of two radicals"     Previous Topic | Next Topic
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jmolokach
Member since Aug-17-10
Nov-09-10, 01:20 PM (EST)
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"Arcsin as different of two radicals"
 
   In studying for upcoming topics for precalculus course, I was venturing into the ideas of the sum and difference formulas for sine, cosine, and tangent. I stumbled upon this graph:

f(x) = sqrt(1+x) - sqrt(1-x) and noticed it to be eerily similar to g(x) = arcsin(x)

I noticed the graphs to be a tad bit different so then I wondered if there might be some solution a for which:

sqrt(a+x) - sqrt(a-x) = arcsin(x)

I do not believe this to be possible but was wondering if someone might comment on why the graphs are so similar. A proof that the above equation is not solvable might find calculus useful I think (I am thinking concavity has something to do with it).

What about other roots... say cube root or 4th root or 1.5 root, etc...?

Is is preposterous to think that the arcsin function can be written as the difference of two radicals? Or did I just stumble upon a near coincidence?

molokach


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alexbadmin
Charter Member
2655 posts
Nov-09-10, 01:32 PM (EST)
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1. "RE: Arcsin as different of two radicals"
In response to message #0
 
   >f(x) = sqrt(1+x) - sqrt(1-x) and noticed it to be eerily
>similar to g(x) = arcsin(x)

The Maclaurin series of f is

x + x^3/8 + ...

That of g is

x + x^3/6 + ...

>I noticed the graphs to be a tad bit different so then I
>wondered if there might be some solution a for which:
>
>sqrt(a+x) - sqrt(a-x) = arcsin(x)

I'd guess, with a ≠1, the proximity of the two series would rather worsen because of the loss of symmetry. For a = 1, both functions arer odd.


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