Determining Measure of Angle Without Tables or Calculators
Date: 03/05/98 at 00:18:18 From: Guy Vooys Subject: Unit Circles - calculating angles I'm 38 years old, and I am embarking on another calculus course. It seems I've forgotten everything, so I am back reviewing my old texts. My current problem is with the unit circle. I find I have no problems determining the x and y co-ordinates of P(x,y) or finding the values of the trig functions. My problem lies with determining what the angle theta is in either radians or degrees. If I knew one, it would be simple to find the other. I can't seem to find any formula or relation between P(x,y) and angle XOP - X being the x-axis, O being the origin. The only formula I have found is theta = s/r where s is the radians of the arc, and r is the radius of the arc. How do I find a relation between theta and P(x,y), or between s and P(x,y), without using tables? Any help would be greatly appreciated. I've been going crazy for 4 days over this. I can't find a triangle or anything to help me. I'm sure it's something easy, as the exercises list the questions of this type first, with the tougher ones (which I can do) following. Thank you, Guy Vooys
Date: 03/05/98 at 10:11:45 From: Doctor Rob Subject: Re: Unit Circles - calculating angles Let P(x,y) be a point on the unit circle with center at the origin O (0,0). Let theta be the angle between the radius OP and the positive x-axis. Probably the formula you are seeking is tan(theta) = y/x. Then theta = arctan(y/x) Usually one uses a calculator or a table of values of the trigonometric functions to find theta from y/x. One can also use the infinite series arctan(u) = u - u^3/3 + u^5/5 - u^7/7 + ..., where -1 < u < 1. Here, the answer comes out in radians, and between -Pi/2 and Pi/2. If u > 1 or u < -1, you can use arctan(u) = Pi/2 - arctan(1/u), and then use the above series, because -1 < 1/u < 1. You also know that tan(-u) = -tan(u), and tan(u+Pi) = tan(u), which allows you to reduce the angle in question to the range between 0 and Pi/2, and so the tangent to positive values. For the special angles of Pi/6, Pi/4, Pi/3, and Pi/2 degrees, you know that the tangent is sqrt(3)/3, 1, sqrt(3), and infinite, respectively. Example: (x,y) = (sqrt(2)/2,sqrt(2)/2), y/x = 1. theta = arctan(1) = Pi/4, a special angle. Example: (x,y) = (5/13,12/13), y/x = 12/5. theta = arctan(12/5), = Pi/2 - arctan(5/12), = Pi/2 - 5/12 + (5/12)^3/3 - (5/12)^5/5 + (5/12)^7/7 - ..., = 1.5708 - .4167 + .0241 - .0025 + .0003 - ..., which is roughly 1.1760 radians, or about 67.38 degrees. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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