what is the best way of remebering the unit circle?
595637
if you dont have one already, make a nice big copy of the entire circle with everything labeled.  repetition and review is really just the best way to remember anything.  when i try to think of the unit circle values for 30-60-90 triangles, i just think of it in terms of the long side and the short side.  the short side is always 1/2 and the long side is always (sq. root of 3)/2.  if i need to remember the values for, say, a 120 degree angle, i always draw a quick little sketch- just enough to show which are the short and long sides.  the horizonal is short, so the cosine is 1/2.  the vertical is long, so the sine is (root 3)/2.  this method works for all angles that are multiples of 30 degrees.  45, 135, 225, and 315 degrees are easy because you only have to remember one value for all the sine and cosines: (root 2)/2\n\njust dont forget to look at your sketch to see which of the values is negative! (and if the picture method doesnt work for you, theres a trick to knowing those too . . .\n\nhere are the quadrant numbers: (imagine the axis in between)\n\n2 1\n3 4\n\nuse this acronym to remember what trig functions are positive in each quadrant:\n\nALL---------> ALL of them are positive\nStudents----> Sine is positive\nTake--------> Tangent is positive\nCalculus----> Cosine is positive
183842
Write it down and look at it every once in a while. The other thing I do to remember numbers is to associate it with something that I cannot forget. Analogies work greatly in memory!
68512
Ask Dr. Math is my favorite.  You can also try Mathworld, though I usually find that I leave it more confused than when I entered.
978661
Have fun:\nhttp://www.math.nyu.edu/~griffith/teaching/06numanalysis/labs/bisection/
156603
Draw a line.  Set the compass to the distance you want the point to be from the line so the circle passes through the point at the top of its arc.  Draw the circle.  Then draw the same size circle again at another point on the line and connect the tops of both circles.
837293
I haven't done one of these in quite a while. \n\nWhat would you differential element be?  How about r*(dtheta)*dr?\n\nthat sort of a ds*dr   where ds depends on the radius.  Try it! \n\nNote that without the sin theta you'd get the int rdthetadr with r 0 to R and theta 0 to 2pi  You'd get the correct area of a circle.\n\nTry the intergral, see if it makes sense.  Then sketch the figure and estimate it's area to compare.  My confidence is increasing as I write this.
767419
http://library.thinkquest.org/28807/data/circ2.htm
893599
You may want to be more specific on this one... are you trying to find the largest possible area? You can inscribe lots of different trapezoids in a circle. \n\nAnd which theta are we talking about? The angle that the base makes with the circle?\n\nTry that one again,\n\nUMRmathmajor
722230
http://www.dianahacker.com/resdoc/p04_c08_s2.html#27\n\nthis site will tell u exactly how to do it.
714881
First, you take a REALLY BIG tape measure and measure around the whole earth at the equator.  That gives you the circumference of the two dimensional circle defined by the equator.  Divide that number by pi.  Then divide the resulting quotient by 2.  Now you have the radius.\n\nIf you don't have a tape measure that big, and your mom won't give you the recipe for pi, you can try a different method.  Check out this method to get the circumference:  http://staff.imsa.edu/science/astro/astrometry/ground1.html\n\nThe Greek had of the library of Alexandria did it more than two thousand years ago.  All you have to do it get to Egypt at noon on the 21st of June for two years in a row.  And you have to find two wells.  \n\nOnce you have the curcumference, you can use my recipe for pi.  (3.160484) Once again, divide the circumference of the earth by pi, and then by 2.\n\nHope this helps.
70409
Go back to math class and figure it out!
396223
lets assume you are talking about a radio signal just for the example like a lost hiker\ntake a direction reading from each of your coordinates\nlike say from the first your signal comes in best from the west, from point two its south-east, and from point three its north\non your map mark your three coordinates and a line representing the direction of best radio reception\nwhere the three lines converge is the aproximate location of the radio the lost hiker was using\nhope this helps you you understand triangulation better
726789
What are the co-ordinates of the other point? just use the midppint formula for a segment. All the best!\n\nLife's Lovely! Love & Live Life!
663777
SOrry, I can't help you. If I had the formulas I could do it, but there is no way I remember that off the top of my head! Good luck
1147102
pi*(r^2)\n\nIt doesn't matter which unit you use, just make sure you're consistent.
355812
You can't.\nIt was initialized a long long time back.
900535
This type of problem will eventually seem EASY for you, so hang in there.  Let's begin by stressing that "a circle with radius 5" is simple to draw -- I recommend marking the 4 points that are "5 units" above, below, right and left of the center point.\n\nThat is, mark (-6,7) on your graph paper, then sketch four more points at (-1,7), (-11,7), (-6, 2) and (-6,12).  "Connect the dots" by drawing all 4 quarter-circle arcs between the points.\n\nThe equation for any circle -- in "rectangular coordinates" (which means (x,y)) -- is normally:  (x-a)^2 + (y-b)^2 = r^2, where (a,b) is the center point and r is the radius.\n\nBut what you're hoping to find is the equation of a circle in "polar coordinates" (which means (r,theta)).\n\nThe reason a sample triangle was drawn inside the given circle (I'm assuming you're looking at a textbook) is to illustrate the relationships among x, y, theta and r.  Let's inspect, using rules from trignometry:\n\nx = r cos (theta) + a\ny = r sin (theta) + b\nr = sqrt ((x-a)^2 + (y-b)^2)\ntheta = Arctan ((y-b)/(x-a))\n\nwhere (x,y) is any point on the circle, (a,b) is the circle's center, and r is the radius.\n\nThe formula for theta that I just provided may be precisely what you are looking for.  It works simply by first finding the "slope" of the line connection (x,y) and (a,b), then taking the inverse tangent of that slope, which 'converts' it to an "angle."\n\nHope this helps!