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Basic trig functions

  1. Jun 3, 2014 #1
    I just wanted to clear a couple of things up in terms of strict mathematical definition...

    Is the correct definition of the trigonometric ratios:

    cos[itex]\varphi[/itex]=[itex]\frac{|x|}{r}[/itex], sin[itex]\varphi[/itex]=[itex]\frac{|y|}{r}[/itex]

    as opposed to:

    cos[itex]\varphi[/itex]=[itex]\frac{x}{r}[/itex], sin[itex]\varphi[/itex]=[itex]\frac{y}{r}[/itex]

    (note the lack of absolute value definition for the axis projections)

    Also, is it correct to define a circle with the equation:

    r = rcos[itex]\varphi[/itex]+rsin[itex]\varphi[/itex]?

  2. jcsd
  3. Jun 3, 2014 #2


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    What do you mean by "strict mathematical definition"?

    is sin(x)=|sin(x)|?
    No, sine and cosine can be negative (and complex) in general
    Sometimes we are interested in a limited domain where this is true though.
    0<=x<=90 degrees
    0<=x<=pi/2 radians

    The equation of a circle in polar coordinates is
    where r is the distance from the center and R is the radius
    you can however write it using the Pythagorean theorem
    which reduces to
    Last edited: Jun 3, 2014
  4. Jun 3, 2014 #3


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    I think a comma would make your meaning clearer. :wink:

  5. Jun 3, 2014 #4
    By R do you mean |r|? I think this is where I am getting confused... so the circle equation would be:

    r = |r|cos[itex]\varphi[/itex]+|r|sin[itex]\varphi[/itex]


    r = |r|(cos[itex]\varphi[/itex]+sin[itex]\varphi[/itex])

    Is this right?
  6. Jun 3, 2014 #5


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    The equation in polar form of a circle of radius = a centered at the origin is simply r = a.

    It's not clear what your equation represents.
  7. Jun 3, 2014 #6
    One thing that doesn't get sufficient coverage, in my opinion, is the fact that there are actually (at least) three different versions of the trigonometric functions;

    There are right-triangle trig functions that are defined in terms of ratios of side lengths of right triangles. The arguments (domain) of these functions are angles; specifically angles between 0 and 90 degrees (whether or not we include 0 and 90 depends on how you want to define "right triangle"). As ratios of side lengths, the outputs (range) of the right-triangle trig functions are positive real numbers.

    Then there are the "unit-circle" trig functions which are defined in terms of coordinates of points on the unit circle. The arguments of these functions are technically arc lengths, and the outputs are, again, real numbers; e.g. for the unit-circle version of ##\sin##, ##\sin a## is the ##y##-coordinate of the point on the unit circle that is ##a## counter-clockwise units around the circle.

    Finally, there are the analytic trig functions, which have power series definitions. The domains and ranges of these functions are real (or complex) numbers. These are the functions of primary interest in a calculus course.

    Now all of the versions of the trig functions can be understood in terms of the others - e.g. it is common for students to use "reference angles" and right-triangle trig functions to aid in their understanding of the unit-circle trig functions - so they are often considered to be the same. But I believe there is value in realizing that they're fundamentally different kinds of functions that happen to be comparable.

    *Remark: It's reasonable to consider a fourth class of trig functions, which I would call the "rotational" trig functions, but one could argue that this class is the same as the "unit-circle" class.
  8. Jun 5, 2014 #7
    OK so for a non unit circle, a radius vector with magnitude r has component vectors x and y

    by the Pythagorean theorem:

    r2 = x2 + y2

    In parametric form:

    x = |r|cos[itex]\varphi[/itex]
    y = |r|sin[itex]\varphi[/itex]

    therefore the equation in polar form would be:

    r2 = (|r|cos[itex]\varphi[/itex])2 + (|r|sin[itex]\varphi[/itex])2

    Is this correct?
  9. Jun 5, 2014 #8


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    Looks fine to me. I will just mention that taking the absolute value of ##r## is redundant, since magnitude is by definition nonnegative. Also, ##\varphi## is not defined if ##x=y=0##, but of course in that case ##r = 0##.
  10. Jun 5, 2014 #9
    So can you take the square root of both sides to get:

    r = rcos[itex]\varphi[/itex] + rsin[itex]\varphi[/itex]

    r = r(cos[itex]\varphi[/itex] + sin[itex]\varphi[/itex])
  11. Jun 5, 2014 #10


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    No, ##\sqrt{\cos^2 \theta + \sin^2 \theta} \neq \cos\theta + \sin\theta##. In fact, the left hand side is 1 for all ##\theta##, whereas ##\cos \theta + \sin \theta = \sqrt{2}\sin(\theta + \pi/4)##.
  12. Jun 5, 2014 #11
  13. Jun 5, 2014 #12
    This only makes sense as a function in polar form if ##r## is a variable and ##|r|## is a constant. This can be rewritten as ##r^2=|r|^2(\cos^2\theta+\sin^2\theta)=|r|^2## which is like saying ##y=y##. It sure is true, but it is not the graph that you are looking for. If you are writing ##r(\theta)##, and you want a circle with radius ##a## (which is a constant), then the functional notation that you are looking for is

    ##r^2=(|a|^2\cos^2\theta+|a|^2\sin^2\theta)=|a|^2(\cos^2\theta+\sin^2 \theta)=|a|^2##

    This clearly becomes ##r(\theta)=a## since ##a## is already positive as we chose it to be the radius of the circle. I think this is what you were originally asking, but maybe I misunderstood.
  14. Jun 6, 2014 #13
    Thanks, I wasn't making the connection between the basic trigonometric identity [itex]\sin^2 + \cos^2 = 1[/itex]
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