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Simple but forgetful trig.

  1. Sep 9, 2007 #1
    1. The problem statement, all variables and given/known data
    Well I have a couple trig questions that are most likely very simple, but I can't really remember the theory behind them. The first is:
    sin(pi)/2.
    I asked my teacher how to find this and he told me to just remember it was sqrt(2)/2. The thing is, I'd like to understand why this is true. I have been searching the internet and have found some things but I just need a simple explanation of what the sin function actually does.
    The other problem is:
    1/(1-cos(2*pi). This one I'm really at a loss for how to solve for the same reason as the first one, I just don't understand the math.


    3. The attempt at a solution

    I know that sin(pi)/2=sqrt(2)/2, so I have the solution, I would like to understand it though.

    The second question, I don't even know where to start, so if anyone could explain the method I could give it a shot. Thanks for your help.
     
  2. jcsd
  3. Sep 9, 2007 #2

    bel

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    The sine function takes a radius in a unit circle and sweeps out angles (the arguments) fromt the x- axis, which is conventionally angle of zero degree or radian, in a counter-clockwise manner, and then gives the y co-ordinate of the point of the radius at the perimeter of the unit circle as the output. The cosine function does the same thing but gives the x- co-ordinate as the out put. That, in fact is why [tex]cos^2(x)+sin^2(x)=1 [/tex].
     
  4. Sep 9, 2007 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    ?? there is no reason why it's true- it's untrue. sin(pi/2)= 1, of course. Did you mean to say sin(pi/4)? How you determine sine (or cosine) of that depends upon what definition of sine and cosine you use.

    One simple definition of the sine of an angle is "opposite side over hypotenuse" when the angle is one angle of a right triangle.

    pi/4 radians is the same as 45 degrees, 1/2 of 90 degrees. Since the two acute angles of a right triangle must add to 90 degrees, if one angle is 45 degrees, so is the other- two angles the same= two sides the same. That's an isosceles right triangle. Suppose the two equal sides have length 1. Then, by the Pythagorean theorem, the hypotenuse is given by c2= 12+ 12= 2. That is, the length of the hypotenuse is [itex]\sqrt{2}[/itex]. The ratio is [itex]1/\sqrt{2}= \sqrt{2}/2[/itex].

    Another, more general definition of sine and cosine is the "circle definition". Draw a unit circle with center (0,0) on an xy- coordinate system. To find sin(x) or cos(x), measure a distance x around the circumference of the circle in a counter-clockwise direction. The point you end up at had coordinates (cos(x), sin(x). For x= [itex]\pi/4[/itex] in particular, measure from the point (1,0) counter-clockwise around the circle a distance [itex]\pi/4[/itex]. Since the unit circle has total circumference [itex]2\pi[/itex] that is 1/8 of the whole circle. The point is exactly where the circle, defined by the equation x2+ y[/sup]2[/sup]= 1 intesects the line y= x. Putting y= x in the equation of the circle, we get x2+ x2= 2x2= 1 so x2= 1/2 and [itex]x= \sqrt{1/2}= \sqrt{2}/2[/itex].


    Well, that one is "undefined". both sine and cosine are periodic with period 2pi. cos(2pi)= 1 so 1/(1- cos(2*pi))= 1/(1-1)= 1/0 which is "undefined"- you can't divide by 0.


    3. The attempt at a solution

    I know that sin(pi)/2=sqrt(2)/2, so I have the solution, I would like to understand it though.

    The second question, I don't even know where to start, so if anyone could explain the method I could give it a shot. Thanks for your help.[/QUOTE]
    You can't use the "right triangle" definition since a triangle can't have an angle of [itex]2\pi= 360[/itex] degrees. You can use the "circle" definition. If you start at the point (1, 0) and measure around the unit circle a distance [itex]2\pi[/itex] you will have gone completely around the circle and arrived back at the point (1, 0). [itex]cos(2\pi)= 1[/itex] and [itex]sin(2\pi)= 0[/itex].
     
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