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Simple radians explanation

  1. Apr 29, 2010 #1
    Im currently in Calc II and am finding myself totally lost when it comes to solving things like sin(pi/3) and trig of that nature. I am very reliant on my calculator and am extremely fond of degrees, but I NEED to be able to find the definite integral of cos(x) from 0 to (pi/3) and problems that are similar. I was hoping that someone could help explain how this works. Its something I really should know how to do, but I dont know where else to go to learn.

    Thanks in advance guys.
  2. jcsd
  3. Apr 29, 2010 #2


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    In general you will get more useful answers if you ask a specific question.

    Can you narrow it down a bit? Just what is that you are having trouble with?
  4. Apr 29, 2010 #3
    Sorry about that. Basically I have absolutely no idea how to evaluate an expression like sin(pi/3) without using my calculator. Would someone be able to explain to me how to do this?

    I know that the answer is sqrt(3)/2 but I do not know how to do this without using my calculator.
  5. Apr 29, 2010 #4
  6. Apr 29, 2010 #5
    Thanks! For the unit circle, the points (x,y)

    does x correspond with cos and y correspond with sin? How does that part work?
  7. Apr 29, 2010 #6
    Yes, that's right. Wikipedia can probably explain it better than I can for now (it's getting late here)
  8. Apr 30, 2010 #7
    Yes. Basically, it's SOH CAH TOA, in which case H, the hypotenuse, is 1 for unit circles. This makes sin(theta) dependent on the y value, cos(theta) dependent on the x value, and etc. I'm a bit curious why you've haven't learned this before calc.
  9. Apr 30, 2010 #8
    There's very little you need to do when thinking about these things luckily.

    pi/3 is the same as 180°/3 = 60°.

    I find it immensely easier to think in this way, so sin(60°)= (√3/2)

    The reason I know this is from using SOH, CAH, TOA on the famous 30°, 60° & 90° triangles.


    [tex] \int_{0}^{ \frac{ \pi }{3}} cos(x)\,dx \ = \ sin( \frac{\pi}{3}) \ - \ sin(0) \ = \ \frac{ \sqrt{3} }{2} [/tex]

    Try doing it this way until you intuitively get that pi/3 = 60°, pi/4 = 45° etc... and don't be afraid to draw and redraw the 30°, 60°, 90° triangles in a margin so that you don't have to memorize everything, you can recall all of them with no hassle this way.
  10. May 1, 2010 #9


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    To find [itex]sin(30)= sin(\pi/6)[/itex] or [itex]sin(60)= sin(\pi/3)[/itex] and the cosines, think of an equilateral triangle. If you drop a perpendicular from one vertex to the opposite side, it bisects that opposite side. Now you have two right triangles with angles of 30 degrees= [itex]\pi/6[/itex] radians and 60 degrees= [itex]\pi/3[/itex] radians. If each side of the equilateral triangle, and the hypotenuse of each right triangle, was 1, the side opposite the 30 degree angle has length 1/2 and, by the Pythagorean theorem, the side opposite the 60 degree angle is [itex]\sqrt{3}/2[/itex]. Now, you can calculate all trig functions from the definitions.

    For 45 degrees= [itex]\pi/4[/itex], use an isosceles right triangle with legs of length 1. The hypotenuse has length [itex]\sqrt{2}[/itex].
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