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Trig & Physics: How does Cos(90+θ)= -sin(θ) relate to Vectors

  1. Jan 9, 2012 #1
    1. The problem statement, all variables and given/known data


    [PLAIN]http://i1094.photobucket.com/albums/i451/LearningMath/vectorconfusion.png[/PLAIN]


    Question 1: Is this mainly saying that when the angle is greater than 90 degrees use the reference angle to evaluate a trig function?

    Question 2: Is this then saying that if you choose to evaluate a trig function without using a reference angle, you have to go through a lot of work, such as using these trig identities:

    sin(180 + θ) = -sin(θ) or cos(90 + θ)= -sin(θ)

    and doing so allows you to see that this following equation is true:

    sin(180 + θ) + cos(270 - θ) = 0 ?



    2. Relevant equations

    If I have the basic idea of what this text is saying in general, then i can move on to say that I have no idea what this text is saying in specifics.

    I understand the first part. I use a reference angle by convention because I know of no other way to evaluate a trig function with an angle greater than 90. Reference angles allows me to see the angle as right triangle and thus is possible to evaluate.

    However, when the text goes on to say that if you choose not to use the reference angle, you'll have to evaluate using the given trig identities. I have two questions about that:

    1. I don't understand how this equation is possible: cos(90 + θ)= -sin(θ)

    For example: If I have angle 2∏/3 and I find that sin(2∏/3) = √3/2
    and then I put angle 2∏/3 into cos(∏/2+2∏/3) and find that cos(7∏/6) = (-√3/2)

    Then if I divide both sides of cos(7∏/6) = (-√3/2) by -1, I get -cos(7∏/6) = √3/2

    Then I could say -cos(7∏/6) = sin(2∏/3), then if i divide that by -1, I get

    cos(7∏/6) = -sin(2∏/3) or cos(∏/2+2∏/3) = -sin(2∏/3)

    So is this logic/evaluating correct? If so, then this is my next question:


    2) How does that have anything to do with finding the components of a vector?



    3. The attempt at a solution

    Please help, I feel lost on this.
     
  2. jcsd
  3. Jan 9, 2012 #2

    SammyS

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    Your last question,
    2) How does that have anything to do with finding the components of a vector?​
    is related to my first question.

    What are the expressions alluded to in the first sentence of the image you provided?

    Other answers in no particular order:
    One of the things it's referring to is reference angle. It may also be referring to SOHCAHTAH .

    Another way to evaluate a trig. function with an angle greater than 90° is to use the unit circle definition of the trig. functions.

    cos(90° + θ) = cos(90°)cos(θ) - sin(90°)sin(θ) = 0 - sin(θ) = -sin(θ)​

    By The Way: Where is this image from? Who wrote that?
     
  4. Jan 10, 2012 #3
    It's an online hw program called MasteringPhysics that my school and many other colleges and universities force on us students. I just pressed print screen, copied to microsoft paint, cropped it a bit, uploaded to photobucket, then inserted the image here. Most people seem dissatisfied with this MasteringPhysics hw program because of it not being very user friendly in terms of typing in your answers and it provides very few examples or explanations. I guess it's better than nothing though. That green image was the most detailed explanation of a concept in MasteringPhysics that I've come across so far, but it still doesnt provide any image or example of what its explaining. You can go to the site and get access to the program without having to enroll in a class by clicking self study if you want to check it out for yourself.

    Here is my own image of what I understand so far:

    vectorconfusiona.jpg

    For the first circle I got the x and y components, which I guess if you consider the radius a vector, the x and y would be vector components.

    However, for the second circle, I don't see how I would draw any vector components to make it fit into the equation Cos(90+θ)= -sin(θ)

    It just feels like comparing apples and oranges to me. I would guess there's only one way to create vector components for a given vector, and that one way would be to create a right triangle.

    So if I were to take a guess in answering your first question, I would say that the first sentence and paragraph are saying that if I have a vector of 1 and an angle 2∏/3, I should be able to find its x and y components by using the given identities.

    And then I guess from there I can say if I take the angle 2∏/3 and add 90 degrees to it, then the cosine of that new angle will automatically be my x component?

    So if I go back and continue trying to find the vector components of angle 2∏/3 in the unit circle sketch:

    vectorconfusionb-1.jpg

    But the x component and y component of the second method does not match the x component and y component of the first method. Is my math wrong or is my entire approach incorrect?
     
    Last edited: Jan 10, 2012
  5. Jan 11, 2012 #4
    SammyS, it still isn't clear to me what I'm supposed to be grasping here. How am I supposed to find the x and y components of an angle greater than 90 degrees by using those trig identities mentioned in my first post of the thread? What am I not understanding here?
     
  6. Jan 12, 2012 #5

    SammyS

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    LearninDaMath, I thank you for your patience.

    I have not been trying to ignore your questions. I've been busy, and it looked like it was going to take some time to understand what that screen shot was saying, and precisely what was bothering you about it.

    The first paragraph, plus the first sentence of the second paragraph, lead me to believe that there is some method of finding vector components which MasteringPhysics is trying to avoid.
    Is trigonometry not a prerequisite for you physics course? If not, then I take it that they are trying to help anyone who is new to trig how to work with vectors that terminate in a Quadrant other than Q1 . Otherwise, that situation should be no problem for anyone with a trig back ground. ​
    You seem to understand 'reference angles', so I wouldn't be too concerned with the method they're presenting.

    Of course, if you use the reference angle idea, you need to know the signs of the three main trig functions (sine, cosine, and tangent) in Quadrants I, II, III, and IV .

    I'll try to get back to this later.
     
  7. Jan 13, 2012 #6
    Thanks SammyS, I appreciate it. I am comfortable with finding the components of vectors in any direction using basic trig functions (sin cos tan), finding the magnitudes of vectors using pythagorean theorum, and finding the angle, if needed, using arctan. So, I understand the convential methods used for evaluating basic 2 dimensional vectors, but this other method the image is referring to is what I am completely lost on.
     
  8. Jan 16, 2012 #7
    any ideas?? If not it's all good. The most pressing problem i'm having at the moment is the thread entitled "geometric vs component addition." And regarding the problem in this thread, I'll just ask my professor at school about it and if he can provide some insight on it, i'll relay that here on physicsforums. He's probably been asked this question before, seeing that it's from the homework assigned by him.
     
    Last edited: Jan 16, 2012
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