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Math Help - Compute inverse trig functions

  1. #1
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    Compute inverse trig functions

    This is in the review section of my calculus books so I think it should be solvable without calc but I'm not sure. I dont even know how to start this problem.

    Using only arctangent and other commonly available functions compute the arcsine and the arcosine.

    Any and all help is greatly appreciated!
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Alice96 View Post
    This is in the review section of my calculus books so I think it should be solvable without calc but I'm not sure. I dont even know how to start this problem.

    Using only arctangent and other commonly available functions compute the arcsine and the arcosine.

    Any and all help is greatly appreciated!
    You want to draw out the triangle - it will help.

    \arcsin(x) is the angle \theta such that \sin(\theta)=\frac{x}{1}. So if the opposite side of \theta is x and the hypotenuse is 1, what is the adjacent side to \theta?

    Well, using the Pythagorean Theorem, we know that it equals \sqrt{1-x^2}. Now that we know the opposite and adjacent sides to \theta, we can find \tan(\theta), which equals \frac{x}{\sqrt{1-x^2}}.

    So we know that:

    \arcsin(x)=\theta
    \tan(\theta)=\frac{x}{\sqrt{1-x^2}}\implies\theta=\arctan\left(\frac{x}{\sqrt{1-x^2}}\right)

    Thus \arcsin(x)=\arctan\left(\frac{x}{\sqrt{1-x^2}}\right)

    Using an analogous method to this one, see if you can compute \arccos(x).
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  3. #3
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    I got arccos(x) = arcsin(sqrt(1-x^2)).
    I'm pretty sure this is right as I've checked it on my calculator.
    Thanks for your help!
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Alice96 View Post
    I got arccos(x) = arcsin(sqrt(1-x^2)).
    I'm pretty sure this is right as I've checked it on my calculator.
    Thanks for your help!
    Yes it is, though if your professor doesn't want it in terms of \arcsin then it would be:

    \arccos(x)=\arctan\left(\frac{\sqrt{1-x^2}}{x}\right)
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