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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 

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
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Associated Topics:
High School Trigonometry

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