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Math Help - Help my life is in game

  1. #1
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    Help my life is in game

    I need help with this (i know its easy but i dont know how to resolve this problem )
    cosalfa = 5/13 sin alfa is - , cos alfa is + , tg alfa is - and ctg alfa is -

    I need to find a) sin(60`-alfa) b) cos(alfa +45`)
    i found out that sin alfa is 12/13


    ps `is degree

    please help
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  2. #2
    Member mfetch22's Avatar
    Joined
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    Columbus, Ohio, USA
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    Quote Originally Posted by TheGame View Post
    I need help with this (i know its easy but i dont know how to resolve this problem )
    cosalfa = 5/13 sin alfa is - , cos alfa is + , tg alfa is - and ctg alfa is -

    I need to find a) sin(60`-alfa) b) cos(alfa +45`)
    i found out that sin alfa is 12/13


    ps `is degree

    please help
    [Note that in the below I substituted x for your "alfa", so simplify things]


    Your question is hard to understand, maybe try writing the problem in latex? Its not hard, its actually kinda cool when you get it to work. I kinda see your equations, but I dont understand the question because I am not certain what the equations are? Does "is" mean "=" ? I reccomend trying latex. Heres how you do it. You write these symbols (without the quotation marks, but including the "[" and "]" outside of the quotation marks) around any of your math stuff:

    ["math] equations go here [/math"]

    Note that the last slash is a forward slash not a back slash, that messes me up all the time. Now, if you type in (without the quotation marks):

    ["math]cos(x) = 5/13 [/math"]

    then you'll get:

    cos(x) = 5/13

    to make it allite more compact you can use a fraction, which is made by putting "\frac{top of fraction here}{bottom of fraction here}" into the "" symbols. So if you typed (without the quotations):

    ["math] cos(x) = \frac{5}{13} [/math"]

    you'll get a nice pretty fraction like this:

     cos(x) = \frac{5}{13}

    Standard notation uses "tan(x)" to denote the tangent of x, not 'tg x'. It will just confuse people, just so you know for the furture to make things simpler on yourself. Same with 'ctg x'. Most people will not know that you meant "cot(x)". As for typing these into the math symbol magic maker, do the same thing, like this (without the quotations of course):

    ["math] sin(x) < 0 [/math"]

    ["math] cos(x) > 0 [/math"]

    ["math] tan(x) < 0 [/math"]

    ["math] cot(x) < 0 [/math"]

    These give you:

     sin(x) < 0

     cos(x) > 0

     tan(x) < 0

     cot(x) < 0

    I'm starting to understand what your question is. It took me awhile to realize what you were asking. Just so you know also, the best way to say "the tangent of x is negitive" is to say that "tan(x) < 0", as seen above. Now, if you don't mind, try rewriting your problem with these magic math symbols that make it look all fancy, and add any more details you may have left out, and also please provide the exact wording of the question in your textbook or worksheet or wherever, that is if you have the text book or worksheet or whatever the question was written on. Post that all back here, and I'll be happy to help you out with this problem. You might also want to take a quick look at this thread:

    http://www.mathhelpforum.com/math-he...ns-153821.html

    The problems in there sorta relate to what I think your problem is.
    Last edited by mfetch22; August 18th 2010 at 09:49 AM.
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  3. #3
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    Hello, TheGame!

    \cos\alpha = \frac{5}{13},\;\sin\alpha < 0

    Find:
    . . (a)\;\sin(60^o - \alpha)
    . . (b)\;\cos(\alpha +45^o)

    Cosine is positive; \alpha is in quadrant 1 or 4.
    Sine is negative: \alpha is in quadrant 3 or 4.

    . . Hence: \alpha is in quadrant 4.
    . . And: . \sin\alpha \:=\:\text{-}\frac{12}{13}


    (a)\;\sin(60^o-\alpha) \;=\;\sin60^o\cos\alpha - \cos60^o\sin\alpha

    . . . . . . . . . . . . . =\;\left(\dfrac{\sqrt{3}}{2}\right)\left(\dfrac{5}  {13}\right) - \left(\dfrac{1}{2}\right)\left(\text{-}\dfrac{12}{13}\right)

    . . . . . . . . . . . . . =\;\dfrac{5\sqrt{3} +12}{26}



    (b)\;\cos(\alpha + 45^o) \;=\;\cos\alpha\cos45^o - \sin\alpha\sin45^o

    . . . . . . . . . . . . . =\; \left(\dfrac{5}{13}\right)\left(\dfrac{1}{\sqrt{2}  }\right) - \left(\text{-}\dfrac{12}{13}\right)\left(\dfrac{1}{\sqrt{2}}\ri  ght)

    . . . . . . . . . . . . . =\;\dfrac{5+12}{13\sqrt{2}} \;=\;\dfrac{17}{13\sqrt{2}}

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