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    I have a feeling this is a silly question, but:

    If I know the arcsin of a degree, is there any "automatic" way of telling the arccos of the same angle, or is it simply a matter of knowing them by heart, or, for less well-known values:

    1: Looking up the angle for which the known figure is an arcsin, then:

    2: Just looking up the cos for that angle?

    Thank you.
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    \cos(\arcsin(x))=\sqrt{1-x^2} if that is what you are looking for.
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    That only works for -1 < x < 1. For an angle greater than that (radians obviously), you will just have to use pythagoras.

    EDIT: should be complex, not pythagoras.
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    (Original post by DeanK2)
    That only works for -1 < x < 1. For an angle greater than that (radians obviously), you will just have to use pythagoras.
    arcsin(x) isn't defined outside those parameters of x.
    (Unless you're using complex x, but I'm assuming not.)
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    Thanks, folks, but I am looking for a way to tell the ARCcos if I already know the ARCsin, and vice versa. For example, if the arcsin of an angle is 0.6, the arccos is 0.8. But is there a quick way of working out the 0.8 figure other than the method I put above?
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    I think OP meant getting arccos from arcsin, rather than cos(arcsin(x))

    Draw yourself a triangle, OP. Put the top number on the opposite side, and bottom on the hypotenuous.

    E.G.

    arcsin = 3/5

    than arccos = ((5^2 - 3^2)^1/2)/5



    EDIT 1: So, general formula would be

    arcsin = O/H

    arccos = ((H^2 - O^2)^1/2)/H

    arctan = O/((H^2 - O^2)^1/2)


    etc...



    EDIT 2:
    nevermind, just realised nota bene's way is better...
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    (Original post by Snagglepuss)
    Thanks, folks, but I am looking for a way to tell the ARCcos if I already know the ARCsin, and vice versa. For example, if the arcsin of an angle is 0.6, the arccos is 0.8. But is there a quick way of working out the 0.8 figure other than the method I put above?
    WTF? :confused:

    There is a formula above for that very question.

    Note that 0.8 = 8/10.
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    (Original post by p.stick)
    I think OP meant getting arccos from arcsin, rather than cos(arcsin(x))

    Draw yourself a triangle, OP. Put the top number on the opposite side, and bottom on the hypotenuous.

    E.G.

    arcsin = 3/5

    than arccos = ((5^2 - 3^2)^1/2)/5
    That would make more sense.
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    The inverse functions change an trig ratio into an angle. as such your question is confusing.
    As such you cannot arc cos an angle...
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    arccos x + arcsin x = π/2
    You can prove it if you want. Substitute a = arccos x and b = arcsin x
    Then see how you get on.
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    Thanks all. Yes, I wrote the question incorrectly. But you have effectively answered me in any case. One does have to go through a little formula, whereas I was wondering if there was some way of seeing it immediately that I hadn't twigged.
 
 
 
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