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Math Help - arccos(3/5)+arccot(1/7)

  1. #1
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    arccos(3/5)+arccot(1/7)

    how do I solve arccos(3/5)+arccot(1/7) ? am i supposed to do cos(arccos(3/5)+arccot(1/7)) making it (3/5)+cos(arccot(1/7)) ? but if so how do i continue?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by weasley74 View Post
    how do I solve arccos(3/5)+arccot(1/7) ? am i supposed to do cos(arccos(3/5)+arccot(1/7)) making it (3/5)+cos(arccot(1/7)) ? but if so how do i continue?
    when you say "solve" what do you mean. because there is no nice way to "simplify" this without a calculator, which is what you seem to be asking us to do. (what you did was totally wrong by the way. cosine is not a linear operator, so you can't apply it as you did. \cos (A + B) = \cos A \cos B - \sin A \sin B, secondly, if you are to simplify, you cannot do it by applying a trig function, you'd be changing the value!)
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  3. #3
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    but we're suppose to solve this, in some wierd way.. it's supposed to be 3pi/4.. but no one seems to know how..
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  4. #4
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    actually.. i think i got it..

    cos(arccos(3/5))*cos(arccot(1/7))-sin(arccos(3/5))*sin(arccot(1/7))=

    = (3/5)*(1/sqrt(50))-(4/5)*(7/sqrt(50))=

    = -5/sqrt(50)

    and -5/sqrt(50) = cos 3pi/4


    right?
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by weasley74 View Post
    actually.. i think i got it..

    cos(arccos(3/5))*cos(arccot(1/7))-sin(arccos(3/5))*sin(arccot(1/7))=

    = (3/5)*(1/sqrt(50))-(4/5)*(7/sqrt(50))=

    = -5/sqrt(50)

    and -5/sqrt(50) = cos 3pi/4


    right?
    yes, that's nice!
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    Talking

    yay me! ^^
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