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Homework Help: Circular functions

  1. Sep 8, 2010 #1
    I'm stuck at the very beginning of circular functions because I don't understand what I'm doing. If someone could explain these questions to me I might be able to move on the the relevant stuff finally :redface:

    1. What does it mean to evaluate something like tan(-2.5)? as in the number in the brackets

    2. cos(theta) degrees = -0.5, what does -0.5 represent?

    3. tan(pi/2 - theta) given tan(theta) = 0.7... the answer in the book is 10/7 and cos(3pi/2 -x) the solution is cos(-pi/2 -x) = cos(pi/2 +x) = -sin x... what happened there?

    on the same note, apparently sin(pi/2 - theta) = cos(theta)... does it make a difference which way you solve?

    4. what's a quick way to solve cos(68pi/3)? the solution is -1/2... or sin(67pi/4)= 1/sq(2)

    5. What's going on in this solution?

    What the heck are they saying? is there an easier way to solve that equation? and one like this-

    for pi/2<a<pi with sin(a)=cos(b) where 0<b<pi/2, find a in terms of b
     
  2. jcsd
  3. Sep 8, 2010 #2
    1. The angle -2.5 degrees is measured clockwise and is therefore below the x axis.

    2. -0.5 is the factor you can multiply the radius by to get the x value. It is the x value when r = 1.

    3 and following:
    C = Pi*d = Pi * 2r
    C = 360 degrees
    sp Pi * 2r = 360
    For r = 1 (which is used in Trig to set up the value table)
    Pi = 180 degrees.
    That said, to get trig functions over 360 degrees (2 Pi), take out the extra trips around the circle, so 67 Pi is the same as Pi and is 180 degrees. 67Pi/2 would be 33.5 Pi which is the same as 1.5 Pi or 270 degrees. So 67Pi/4 is ___.

    OK?
     
    Last edited: Sep 8, 2010
  4. Sep 8, 2010 #3

    HallsofIvy

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    Starting at the point (1, 0) measure around the circumference of the unit circle, clockwise (because it is negative), a distance 2.5 (using the units your coordinate system is marked in- the same units in which the radius of the "unit" circle is 1.). Taking the coordinates of the point at which you end to be (x, y), the tan(-2.5) is y/x.
    If you were to draw a perpendicular from (x, y) to the x-axis and the line from (x,y) to (0, 0) you would get a right triangle in which the angle is 2.5 radians and y/x is "opposite side over near side" as in "trigonometric" functions.
    Notice that there are no units on the "-2.5" here. Strictly speaking, we are measuring a distance, not an angle. But since engineers tend to think of sine and cosine in terms of angles, by using radian measure, the angle is the same as the arc length on the unit circle.

    On the unit circle, "cos(theta)" is the y coordinate of the point (x, y) on the unit circle where the line from (0,0) to (x, y) makes angle theta with the x-axis. If you convert [itex]\theta[/itex] degrees to [itex](\theta/180)\pi[/itex] radians, that is also the distance from (1, 0) to (x, y) around the unit circle. If you draw the horizontal line y= -0.5, it will cross the unit circle at two points. They both correspond to "theta" such that cos(theta)= -0.5.

    cos(pi/2- theta)= sin(theta) and sin(pi/2- theta)= cos(theta)- you can see those by seeing that going from "theta" to "pi/2- theta" in angle (or measured around the circumference of the unit circle. So tan(pi/2- theta)= sin(pi/2- theta)/cos(pi/2- theta)= cos(theta)/sin(theta)= 1/tan(theta) and 1/0.7= 1/(7/10)= 10/7.

    If you are referring to that same problem, you need both cos(pi/2- theta)= sin(theta) and sin(pi/2- theta)= cos(theta).

    68/3= 22+ 2/3 the same as cos(11(2pi)+ (2pi/3)). The unit circle has circumference 2pi so going to 68pi/3 is just going around the circle 11 times and then an additional distance of 2pi/3- the same as if you had just gone directly to 2pi/3: cos(68pi/3)= cos(2pi/3). Now, if you go around the unit circle as distance 2pi/3 from (1, 0) you wind up a (-1/2, sqrt(3)/2). And "cosine" is the x coordinate.

    Same thing. 67/4= 8+ 3/4 so 67pi/4= 8pi+ 3pi/4= 4(2pi)+ 3pi/4. You have gone around the unit circle 4 times, arriving at the same point as if you had only gone 3pi/4.
    That point has coordinates (-1/sqrt(2), 1/sqrt(2)) and sine is the y coordinate.





    What? That makes the problem harder! Where would you go from "sin(a)= sin(b)"? They are using the fact that is sin(x)= sin(y) (and both are in the first quadrant) then x= y. To use that they change from "cos" to "sin" using just what I mentioned before: cos(x)= sin(pi/2- x). cos(x)= -sin(\pi/6) becomes sin(pi/2- x)= -sin(pi/6) and we can get rid of that "-" by using the identity sin(-t)= - sin(t) (moving "+t" is moving counter clockwise while moving "-t" is moving clockwise around the unit circle. Since we are starting from (1, 0), that just swap "+" and"-" on the y coordinate- which is sin(t) at every point) we get, finally,
    sin(pi/2- x)= sin(-pi/6) and so can say pi/2- x= -pi/6. Subtracting pi/2 from both sides, -x= -pi/6- pi/2= -pi/6- 3pi/6= -4pi/6= -2pi/3. Then multiplying both sides by -1, x= 2pi/3.
     
  5. Sep 14, 2010 #4
    Thanks HallsofIvy! great and informative reply. I realized that the book was trying to get me to recognize the symmetrical properties of the circular function which as you said made things harder then they needed to be... so I decided to skip to the relevant graphing stuff.

    I'm wondering, what would be your method of sketching a graph like -3cos(1/2(theta- 3pi/2) + 2 finding the relevant intercepts and coordinates of turning points and such? domain being equal to [-pi, pi]? and say if it were different obscure domain restrictions like [-11pi/3 , 4pi] what's the fastest way to adjust to that?
     
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