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Hard time with trigonomic functions.

  1. Nov 24, 2006 #1
    I am currently a student in a Engineering based calculus class and well things are about average in terms of my performance, Just there is one thing that I just don't get and for that matter have never gotten. That is trig functions, I just don't understand them. Ive tried memorizing and reading old text books, i just cant learn them. I need a way to learn these things before they kill me. Any suggestions
     
  2. jcsd
  3. Nov 25, 2006 #2
    Well you can think of them in a lot of different ways. I'm sure some really great suggestions will come about from this thread, so you will have to suffice with some mediocre ones for the time being (from moi).

    Trig functions have really interesting properties. For example, remember this:

    "SOHCAHTOA"

    Where:
    Sin(theta) = O/H
    Cos(theta) = A/H
    Tan(theta) = O/A

    So these functions take the value of the angle for a ratio of a triangle. That's pretty neat.

    Or how about the fact that if you paramterize a curve as, x = cos(theta) and y = sin(theta) you can feed x(theta), y(theta) a value from 0 to 2pi and draw a a circle.

    Or better yet you can get into the very interesting property of representing "any" (as far as my knowledge is concerned) function as a representation of sin and cos functions via a fourier series. That means that yes you can turn that straight line into a bunch of sin and cos functions.

    With cos and sin, you can stop using time as the domain and instead use the frequency domain which makes things very nice in some situations. Phasors? Have yoiu heard of those... very nice.

    Anyways, all I can say is just play around with them. The more practice you get, the better you will be. What don't you understand about them? Ask away!!!
     
  4. Nov 25, 2006 #3
    I could neither understand nor memorise most of the trig functions/formulae initially, but surprisingly, I got a good grasp after working with a lot of problems in calculus involving trig. functions.
     
  5. Nov 25, 2006 #4

    mathwonk

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    a basic problem is to measure arclength on a circle. this equivalent to measuring angles, since the angle cut out by the arc of circle is proprotional to the length of the arc.

    the basic trig function "sin" is just the name given to one of these arc length functions, or rather its inverse.

    i.e. think of the unit circle, and an arc beginning at (1,0) AND RUNNING UP COUNTERCLOCKWISE and ending at a point (x,y). Think of the arc length of that arc as a function of y.

    that arclength function is called arcsin, ND ITS INVERSE IS SIN.

    i.e. the arclength of the qarc ending at the point (x,y) is arcsin(y) = t, and if you know the arclength t of this arc, then sin(t) = y is the ehight of the endpoint.

    all thi makes good sense at least for points in the first quadrant. for other points, the arcsin is not single valued, but the sin is, so we use sina s a primary function.

    then cos is just sqrt(1-sin^2), and tan is sin/cos, and sec = 1/cos, and thats about all you usually need.


    another very useful way to view these functiions, that is a klittle more sophisticated, is as the real and imaginary parts of the complex exponential function. i.e. e^(it) = cos(t) + isin(t).

    this is actually the best way to understand the complicated addition formuals for sin and cos. since e^i(s+t) = e^is e^it, by multiplying out we get:

    cos(s+t) + isin(s+t) = [cos(s)cos(t)-sin(s)sin(t)] + i [cos(s)sin(t)+sin(s)cos(t)],

    and setting real and imag parts equal, gives the laws:

    cos(s+t) = [cos(s)cos(t)-sin(s)sin(t)] and
    sin(s+t) = [cos(s)sin(t)+sin(s)cos(t)].

    these are hard to remember otherwise.


    also cos and sin are important as basic solutions to the differential equation

    f'' + f = 0. try e^it in there also and see why these functions are related.
     
  6. Nov 25, 2006 #5

    mathwonk

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    i myself missed trig in high school and did not know any when i got to college (and no calculus), where fortunately for me they were presented ab initio using power series.
     
  7. Nov 25, 2006 #6
    I love when high level math individuals go through the basics! It's very interesting as well as helpful to see how they think about certain topics. Thanks! mathwonk for your post, I especially enjoyed what I quoted above from you. It's interesting to think of them as arc length functions.
     
  8. Nov 25, 2006 #7
    If you start with the standard basics,

    The sine of an angle is a function which assigns to each angle the ratio of the opposite side to the hypotenuse as if the angle was in a right triangle.

    The cosine of an angle is is a function which assigns to each angle the ratio of the adjacent side to the hypotenuse as if the angle was in a right triangle.

    The tangent of an angle is a function which assigns to each angle the ratio of the opposite side to the adjacent side as if the angle was in a right triangle. Thus it is also equivalent to the ratio of sine to cosine.

    [tex]
    \frac{O/H}{A/H} = \frac{O}{A}
    [/tex]

    However, these only work for acute angles. For obtuse angles, sin and cosine are defined in respect to the complements of these angles.

    Now, many of the properties of the trigonometric functions arise from these simple definitions.

    For instance: the area of any triangle as (1/2) AB sin theta

    is a direct consequence of the definition of area as (1/2)(Base times height). This is because, if B is the base of the triangle, A times sin theta is the height.

    Also, the pythagorean identity is a direct consequence of the pythagorean theorem (hence the name).

    EDIT: Of course, to give definite values to the trig functions, one needs power series (except in the simple cases such as 0, 30, 45, 60, and 90 degrees).
     
    Last edited: Nov 25, 2006
  9. Dec 1, 2006 #8
    im a little confused on what you guys mean by a ratio. so is Sin a ration of Opposite:Hypotenuse and if so what does that mean.


    sorry for the delay I have been real busy.
     
  10. Apr 14, 2009 #9
    the length of the side opposite side from the angle divided by the length of the side of the longest side (the hypotenuse) is the sine of the angle in a right angle. Try that and verify it is correct.
     
  11. Apr 15, 2009 #10
    Sine and cosine are easy.

    Step 1) Take a circle with of radius 1 centered at the origin.

    Step 2) Start at the coordinate (1, 0) -- (this would be 3 o'clock on a clock).

    Step 3) Walk around the circle in a counter-clockwise direction.

    Step 4) After you have traveled for a distance of x units, you are at the new coordinate (cos x, sin x). (The parameter x is in radians, not degrees).

    And that's one way to define cosine and sine.

    Some examples of using this definition.

    * If you stay in place, you have traveled 0. You started at (1, 0) and you didn't move anywhere, so cos 0 = 1 and sin 0 = 0.

    * If you walk all the way around the circle, you have traveled the entire perimeter, and again, you end where you started. The perimeter has a length of 2 * pi. So cos(2 * pi) = 1, sin(2 * pi) = 0.

    * If you walk half way around the circle, you are going from (1, 0) to (-1, 0). Half the perimeter is pi, so cos pi = -1 and sin pi = 0.

    * If you walk a quarter of the way around the circle, you are going from (1, 0) to (0, 1). A quarter of the perimeter is pi / 2, so cos(pi / 2) = 0 and sin(pi / 2) = 1.

    I hope that helps a bit.
     
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