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Radian measures with Trig Functions?

  1. Sep 17, 2004 #1
    Guys, I am in Calculus 2, but I still have trouble seeing the answer when u plug in a radian measure into a trig function. My teacher assumes we know the answer right away. I know Sin(0)= 0 and Cos(0)=1, but that's about it.

    I don't know things like Sin(pi/2) I have to input it into the calculator. How do I find this out, or memorize it. Is it from the Unit Circle? And if not what should I refer to to memorize these values. Also, how do u use the Unit Circle, what information does it give you?

    Thanks for your time.
  2. jcsd
  3. Sep 17, 2004 #2
    as a general rule you can think x is cosine and y is sine values in many aspects of math (NOT ALL) looking at the unit circle (1 = x^2 + y^2) you see that all the typical radian mesures (0, pi/4, pi/2, etc) correspond to points on the unit circle memorizing these points provides you with the answer you are looking for, suprisingly you typically do this in pre-calculus and trig. For your pi/2 example, you can do sin (pi/2) = 1 and cos(pi/2) = 0 because the point on the unit circle that is pi/2 or 90 degrees is (1,0), another example is sin(pi/4) and cos(pi/4) both equal sqrt(2)/2 because the corresponding point on the unit circle is (sqrt(2)/2, sqrt(2)/2)

    as for you calculator assuming your using a TI-8x or 9x make sure your mode is set to radians and you can enter in pi/2 and it will return radian answers.
  4. Sep 17, 2004 #3


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    You should memorize the 'easy' points to remember, but the radians are easy to remember.

    A radian is an arc along the circumference that is equal in length to the radius. The circumference of a circle is 2(pi) times the radius.

    Half of a circle? 1(pi) radians or 180 degrees.
    Quater of a circle? 1/2(pi) radians or 90 degrees.
    Eighth of a circle? 1/4(pi) radians or 45 degrees.
    One sixth of a circle? 1/3(pi) radians or 60 degrees.
    One twelfth? 1/6(pi) radians or 30 degrees.

    And, you should have the sine and cosine of the 'easy' angles memorized from trig class:

    (sin 0 rad) or (sin 0 deg) = (sqrt 0)/2 or, more sensibly, 0
    (sin 1/6 pi) or (sin 30 deg) = (sqrt 1)/2 or, more sensibly, 1/2
    (sin 1/4 pi) or (sin 45 deg) = (sqrt 2)/2
    (sin 1/3 pi) or (sin 60 deg) = (sqrt 3)/2
    (sin 1/2 pi) or (sin 90 deg) = (sqrt 4)/2 or, more sensibly 1

    Cosine just runs the opposite direction from 1 to 0.
  5. Sep 17, 2004 #4
    Wow thanks guys, im gona print this out.
  6. Sep 18, 2004 #5


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    I don't really remember all those points, I always think of a picture of the unit circle.
    Then I imagine a point on the circle whose line through that point and the origin makes an angle of [itex]\theta[/itex] with the x-axis.

    Then all you need to know is that [itex]\cos(\theta)[/itex] gives the x-coordinate and [itex]\sin(\theta)[/itex] gives the y-coordinate of that point.

    Since for [itex]\theta=0[/itex], the point lies on the x-axis. We have [itex]\cos(0)=1, \sin(0)=0[/itex] (Well..allright, I simply KNOW these ^_^)
    For [itex]\theta=\Pi[/itex] you are on the point (-1,0), you can just visualize this. You are halfway the circle, since for [itex]\theta=2\pi[/itex] you've made one complete revolution. So [itex]\sin(\pi)=0, \cos(\pi)=-1[/itex].
    For [itex]\theta=\pi/2[/itex] you are at the top: Coordinates (0,1).
    For [itex]\theta=3\pi/2[/itex] you are at the bottom: Coordinates (0,-1).
    For [itex]\theta=\pi/4[/itex] the angle is 45 degrees, so the cosine and the sine must have the same value. Using [itex]\cos^2(\theta)+\sin^2(\theta)=1[/itex] the coordinates must be:[itex]\frac{1}{2}\sqrt{2}[/itex]

    For the others it's more tricky, but you can see that for [itex]\theta=1/6\pi[/itex] the cosine of theta must be greater than the sine of theta.
    So it's probably: [itex]\cos(\frac{1}{6}\pi)=1/2, \sin(\frac{1}{6}\pi)=\frac{1}{2}\sqrt{3}[/itex], because I remember 1/2 and the half the squareroot of three in one of those expressions.

    Ok, that last bit may not be so advisable, but it works for me.
  7. Sep 18, 2004 #6
    Looks like you've already got real sound advice. I'd just like to add that you could reinforce your memorisation of the formulas by remembering how the sin(x) and cos(x) curves look like.
  8. Sep 18, 2004 #7
    thanks galileo and recon , great help.
  9. Sep 18, 2004 #8
    In my high school Calculus, we memorized a unit circle diagram, like this one:

    http://www.math.lsa.umich.edu/~zacht/teaching/unit_circle.html [Broken]

    The ordered pairs correspond to (cos(t), sin(t)).

    Basically, all I have to remember is that the cosine of 60 degrees is 1/2, and then I can reconstruct the rest of the chart if I need to.
    Last edited by a moderator: May 1, 2017
  10. Sep 18, 2004 #9

    THank you for the chart =)
    Last edited by a moderator: May 1, 2017
  11. Sep 19, 2004 #10
    All I do is just visualize the graph and where it lies on the points. For example, take sin(pi). If you just visualize the graph of sine you'd find that it interesects the x-axis at pi which makes it equal to 0.
  12. Apr 10, 2009 #11
    All i had to remember was: the Sine is op hy! then, cosine is adj/hy and tangent is op/adj. use the right triangle formed by dropping a vertical from the point on the circle intercepted by the angle's radius. so as op and hy become equal for the angle pi/2, sine=1, cos=0 and tan approaches 1/0...
    this may be old hat for all of you -- i was just refreshing my memory of radian measure.
  13. Apr 10, 2009 #12
    This is over 4 years old...
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