Page 3 of 4 FirstFirst 1234 LastLast
Results 31 to 45 of 53

Math Help - work out missing angle?

  1. #31
    Super Member
    Joined
    Oct 2008
    From
    Bristol, England
    Posts
    897
    would you do inverse sin0.47= 28.03??
    Follow Math Help Forum on Facebook and Google+

  2. #32
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Why are you doing an inverse sin?
    Follow Math Help Forum on Facebook and Google+

  3. #33
    Super Member
    Joined
    Oct 2008
    From
    Bristol, England
    Posts
    897
    i wasn't really sure what to do. would this work:
    cos/ 0.47= -2???
    Last edited by andyboy179; December 15th 2010 at 11:21 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #34
    Super Member
    Joined
    Oct 2008
    From
    Bristol, England
    Posts
    897
    would it be cos/-0.47?
    because you posted an earlier exapmle 4=6x, which is, 4/6
    Follow Math Help Forum on Facebook and Google+

  5. #35
    Super Member
    Joined
    Oct 2008
    From
    Bristol, England
    Posts
    897
    could i answer the question by doing this:
    sin?/18 = sin/10
    sin?=18xsin11/10=0.34 inverse sin= 20.09???
    Follow Math Help Forum on Facebook and Google+

  6. #36
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    No, no, no. Cosine is a function. The notation cos(x) most emphatically does NOT mean cosine times x. It means "cosine of x". You are applying the cosine function to the number x.

    Here's the more general case: suppose you have the equation f(x) = y, and you want to solve for x. How would you go about it? Well, what you would want to do, if you were able, would be to invert function f thus:

    f^{-1}(f(x))=f^{-1}(y).

    Here, all I've done is apply the inverse function of f, denoted f^{-1}, to both sides. By the definition of function inverse, you now have

    f^{-1}(f(x))=x=f^{-1}(y).

    And there's your solution. Does that make sense? If so, how would you apply it to the equation

    \cos(x)=-0.47?

    Let's use the variable x instead of ? from here on out. Less confusing. So you want to solve for x now. How would you do that?
    Follow Math Help Forum on Facebook and Google+

  7. #37
    Super Member
    Joined
    Oct 2008
    From
    Bristol, England
    Posts
    897
    i don't understand that at all!
    Follow Math Help Forum on Facebook and Google+

  8. #38
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Ok, let's back up a minute. What is a function?
    Follow Math Help Forum on Facebook and Google+

  9. #39
    Super Member
    Joined
    Oct 2008
    From
    Bristol, England
    Posts
    897
    one quantity determins another quantity
    Follow Math Help Forum on Facebook and Google+

  10. #40
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    I suppose that's a perfectly acceptable "rough-and-ready" definition. You should probably add in there something like this: "one quantity uniquely determines another quantity".

    There are three aspects to a function: the domain, the range, and the rule of association. What are those, exactly?
    Follow Math Help Forum on Facebook and Google+

  11. #41
    Super Member
    Joined
    Oct 2008
    From
    Bristol, England
    Posts
    897
    i don't know
    Follow Math Help Forum on Facebook and Google+

  12. #42
    Super Member
    Joined
    Oct 2008
    From
    Bristol, England
    Posts
    897
    can i work it out like this:
    just flip the cos over to the other side and make it inverse so it looks like this, ?=cos^-1(-0.47)??
    Last edited by andyboy179; December 15th 2010 at 12:20 PM. Reason: typo
    Follow Math Help Forum on Facebook and Google+

  13. #43
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Ok. Here's the terminology as I learned it.

    A function is like a machine that converts objects of one kind into objects of another kind. The domain is all the allowed inputs to the function. The rule of association is the rule that tells you how the function converts the inputs into the outputs. The range is all the outputs that come, via the rule of association, from objects in the domain.

    Example:

    y=f(x)=x^{3}.

    The domain is all real numbers. The rule of association is the equation f(x)=x^{3}.

    The range is all real numbers.

    So the function inverse, if it exists (not all functions have inverses), gets you back to where you started. The domains and the ranges are flipped, the rule of association is inverted, etc. For the example above, the domain of the inverse is the range of the original function, and vice versa. The rule of association is

    f^{-1}(x)=\sqrt[3]{x}.

    In general, the way to find the inverse rule of association is to swap the symbols for x and y, and then solve for y.

    Example: y=f(x)=3x-4. The domain and the range are both the set of real numbers. Now I want to invert the function. First, swap x and y thus:

    x=3y-4. Now solve for y:

    x+4=3y

    y=\dfrac{x+4}{3}=f^{-1}(x).

    Does that make sense?

    The whole goal of this idea is to get back to where you started. That is, you want to be able to do this:

    f(x)=y

    f^{-1}(f(x))=f^{-1}(y)

    x=f^{-1}(y).

    Thus, you can solve for various variables.

    In your latest post, I think you have got it. You use the inverse cosine function, also known as the arccosine function, to solve for your question mark. What do you get?
    Follow Math Help Forum on Facebook and Google+

  14. #44
    Super Member
    Joined
    Oct 2008
    From
    Bristol, England
    Posts
    897
    I don't really understand that. For the ? I got, ?=cos^-1(-0.47) =118.03
    Follow Math Help Forum on Facebook and Google+

  15. #45
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    And you should double-check whether that answer is in degrees or radians, depending on which mode your calculator is set in.
    Follow Math Help Forum on Facebook and Google+

Page 3 of 4 FirstFirst 1234 LastLast

Similar Math Help Forum Discussions

  1. missing angle
    Posted in the Trigonometry Forum
    Replies: 9
    Last Post: February 28th 2011, 07:06 PM
  2. missing angle
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: January 31st 2011, 06:59 AM
  3. find missing angle?
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: January 17th 2011, 10:48 AM
  4. integration; work; missing 2
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 25th 2010, 04:27 AM
  5. missing angle
    Posted in the Geometry Forum
    Replies: 1
    Last Post: April 12th 2008, 05:34 PM

Search Tags


/mathhelpforum @mathhelpforum