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6th Years and Leavers :: Chat Thread #2 :: Revenge of the Ape watch

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    (Original post by Ape Gone Insane)
    :sigh:

    Why weren't we taught that in Higher?
    :shrug:

    It's just another piece of notation though, as far as I'm aware you don't even need to know it at AH.
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    (Original post by ukdragon37)
    I just write "atan" in exams to same time :awesome:

    And it saves from confusion when you want to write "1/(tan x)"
    Haha, in fourth year (and a good part of fifth year:o: ) I always thought sin^{-1}x was equivalent to \frac{1}{sinx}
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    (Original post by Ape Gone Insane)
    :sigh:

    Why weren't we taught that in Higher?
    Because there are also people doing it who are not necessarily naturally good at maths and introducing different names for the same thing would just cause confusion.

    Same reasoning for why "haversine" and "covercosine" are not taught anymore.
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    (Original post by Ape Gone Insane)
    You like to keep me wondering :ninja:
    :p: Sorry, I started to reply along the lines of namedeprived, but left it for a couple minutes too long, so the answer had already been given.

    (Original post by namedeprived)
    Haha, in fourth year (and a good part of fifth year:o: ) I always thought sin^{-1}x was equivalent to \frac{1}{sinx}
    Well, it doesn't help that, normally, \left\begin{matrix}f^n = &\underbrace{f\circ f \circ \cdots \circ f}&\\&n\right\end{matrix}, and that the trig functions then (for convenience, since that definition's almost entirely useless for them) mix the two notations (of exponentiating the whole thing and of f^{-1} denoting the inverse).
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    (Original post by Ape Gone Insane)
    Any other hidden methods/names I don't know :ninja: ?
    Oh there are plenty, see the class of "Versines", "Inverse Trigonometric Functions" and if you really want to be smartarsed (or just enjoy saying "arc-hacoversin-ch" ) see "Hyperbolic functions"
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    (Original post by Ape Gone Insane)
    Haha, awesome.

    Time to learn and to put to use.

    :cool:
    Problem is, your teacher/examiner probably don't know what any of the "versines" are and probably can't remember what the definition of a hyperbolic function is. Use at your peril :p:
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    (Original post by Vitamin D)
    45.
    Same :fan:

    I've forgotten the exact values thing, again :sad:. Do we get this in the exam? Also, when we're working with the line crap is tan inverse? (I missed the whole straight line topic in college so it's kinda tits up in a way.)
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    (Original post by LuhLah)
    I've forgotten the exact values thing, again :sad:. Do we get this in the exam?
    Yes you'll get questions that expect you to know them, no you don't get given the exact values.

    Just remember that for the angles 30, 45 and 60, you have sin = a half, one over root 2, and 3 over root two. Cos is the same, but reversed (so cos 60 = 1/2). Tan is then just dividing the two above.
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    (Original post by LuhLah)
    Same :fan:

    I've forgotten the exact values thing, again :sad:. Do we get this in the exam?
    You don't. You either need to remember or use some tricks to remember. (I know two)
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    (Original post by LuhLah)
    Same :fan:

    I've forgotten the exact values thing, again :sad:. Do we get this in the exam?
    Yes.No. I thought you meant does it come up in the exam. :o:

    If you don't want to memorise them, you can construct right angled triangles that will tell you. Alternativley you can just remember the values of sin30, sin45 and sin60. Remember that the cosine values are just the above ones in reverse. so sin60=cos30 and sin30=cos60. For tan you can remember that tan30=\frac{1}{\sqrt3} , tan 45 is the numerator and tan60 is the denominator of this fraction. Or just use the fact that \frac{sinx}{cosx}=tanx
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    (Original post by TheUnbeliever)
    Yes you'll get questions that expect you to know them, no you don't get given the exact values.

    Just remember that for the angles 30, 45 and 60, you have sin = a half, one over root 2, and 3 over root two. Cos is the same, but reversed (so cos 60 = 1/2). Tan is then just dividing the two above.
    Thank you!!! and tan90 is undefined :holmes:
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    or draw the triangles i probably would have failed higher maths without those little triangles
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    (Original post by TheUnbeliever)
    Yes you'll get questions that expect you to know them, no you don't get given the exact values.

    Just remember that for the angles 30, 45 and 60, you have sin = a half, one over root 2, and 3 over root two. Cos is the same, but reversed (so cos 60 = 1/2). Tan is then just dividing the two above.
    sin60=\frac{\sqrt3}{2} :p:
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    Can the exact value things be applied to any topic that involves tan etc?

    THIS is why I prefer differentiation (even though I can't spell it )
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    Ok, what about this?

    A, B, C and D are the points (3,2), (2,3) (-2,-1) and (p,5) respectively. AB is parallel to CD. What is the value of p?

    I worked out that the gradient of both lines is -1, and thought maybe you had to substitute the -1 into the formula for Mcd, then re-arrange and find p. But I'm stuck :/ Is there some other way of doing it?
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    (Original post by LuhLah)
    ...
    You can also remember the values for sin by remembering that it's always 2 in the denominator and the numerators are root(1), root(2), and root(3) for angles 30 degrees, 45 degrees and 60 degrees. You then reverse the order of the numerators for cos. For tan put the numerator of sin on top and numerator of cos on the bottom for each angle. (Trick 1)

    Or you can draw triangles (Trick 2)
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    (Original post by namedeprived)
    sin60=\frac{\sqrt3}{2} :p:
    :p: And this, ladies and gentlemen, is why people who copied my answers usually dropped about half a dozen marks over how they would have done otherwise.

    (Original post by LuhLah)
    Can the exact value things be applied to any topic that involves tan etc?

    THIS is why I prefer differentiation (even though I can't spell it )
    Spelling's correct. Can you clarify what you mean by the question? Any time that those values come up, you can sub in those numbers, yes. But there aren't exact values for any argument, no.
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    (Original post by LuhLah)
    Can the exact value things be applied to any topic that involves tan etc?

    THIS is why I prefer differentiation (even though I can't spell it )
    It can come up in any topic that involves trigonometry, including definite differentiation and integration of trigonometric functions.
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    What's wrong with Vista and what makes 7 so godlike? :facepalm2:
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    (Original post by Vitamin D)
    Ok, what about this?

    A, B, C and D are the points (3,2), (2,3) (-2,-1) and (p,5) respectively. AB is parallel to CD. What is the value of p?

    I worked out that the gradient of both lines is -1, and thought maybe you had to substitute the -1 into the formula for Mcd, then re-arrange and find p. But I'm stuck :/ Is there some other way of doing it?
    You know that the gradient \displaystyle\frac{y_2 - y_1}{x_2 - x_1} = -1 for CD. Does that help? Hint: sub in.

    EDIT: Actually, is this what you're trying to do? I sort of glossed over the 'Mcd'.
 
 
 
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