# Graphical Method For Summing Arctangents

Updated on August 14, 2016

In this hub I want to show you how to use Lill’s method to find the sum of 2 or more arctangents. I will illustrate this method of summation by applying it to one specific example.This method is ideal if the sum of the arctangents is less than 360 degrees.

## Obtaining a polynomial

In this hub I will show how this graphical method works by summing arctan(1) and arctan(2). arctan or tan-1 is the inverse of the tangent function, and the arctan function gives us an angle. Thus arctan(1)=45 degrees and arctan(2) is about 63.43 degrees. The sum adds to about 108.43 degrees.

Lill’s method is a graphical method of finding the roots of polynomial equations. I would recommend reading my hub about Lill’s method since I don’t want to make this hub too long. Since Lill’s method deals with polynomial equations we should create an equation first. We will create a quadratic equation that has the roots z1=1 and z2=2. Since we already know the roots, the equation can be derived like this: (z-1)(z-2)=z2-3z+2. Another method for deriving our polynomial equation is to use Vieta’s formulas (see this detailed Wikipedia page). If we have a general quadratic equation a2z2+a1 z+a0 =0 , using Vieta’s formulas we know that z1+z2=- a1/ a2 and z1z2= a0/ a2. To make things easy we can choose a2=1 and since we already know our 2 roots we get 1+2=3=- a1 and (1)(2)=2= a0. Thus a2=1, a1=-3, a0=2 and we get the same quadratic z2-3z+2=0.

## Lill’s method and the sum of arctangents

In Lill’s method the coefficients of the polynomial have corresponding segments linked together. Starting from a point of origin O we will draw some segments using the formula akei(n-k)π/2 where ak is the coefficient corresponding to the segment, i is the imaginary number and 0≤k≤n . n=2 since we are dealing with a quadratic equation. Thus the segment corresponding to a2 will have the value a2ei(2-2)π/2=1e0=1 (from O goes 1 unit to right at A). The segment corresponding to a1 has the value a1ei(2-1)π/2=-3eiπ/2=-3i (from A goes 3 units down to B). Finally the segment corresponding to a0 has the value a0ei(2-0)π/2=2e=2(-1)=-2 (from B goes 2 units to the left to C). Image 1 shows how our polynomial z2-3z+2=0 looks if we use Lill’s method. In image 1 you can also see that the angle (AOC) is about 108.43 degrees , and indeed AOC=arctan(1) + arctan (2). Those who looked at my hub that dealt with Lill’s method in more details can see that due to the convention that I chose, the value of angle AOC is actually -108.43 degrees. So it is more correct to say that the absolute value of angle AOC= arctan(1) + arctan (2).

In this specific example, the segment OA corresponds to the coefficient a2=1 and the point C is the terminus point of the segment BC that corresponds to the coefficient a0=2. The angle AOC is a visual representation of the angle given by arctan(1) + arctan (2). This method of obtaining the sum of arctangents is most suited when you have to add 2 arctangents since the sum will be between -180 dgrees and 180 degrees.

If you want to add 3 arctangents arctan(a)+arctan(b)+arctan(c), then you will have to create a cubic equation that has the roots z1=a, z2=b and z3=c. The algebraic equation can be obtained by the 2 methods that I showed above, and the Lill method representation of that cubic equation can have at maximum an additional segment CD. To obtain your angle you measure the angle between the segment AO corresponding to the coefficient a3=1 and the segment made by O and the terminus of the segment corresponding to a0. Most probably you will have an additional segment CD corresponding to a0, then angle AOD= arctan(a)+arctan(b)+arctan(c). The absolute value of 3 arcangents is always less than 270 since arctan(x) is always less than 90 (you cannot have a root x=infinity). If the sum of the 3 arctangents is over 180 you will probably see easily if you plot the graph using Lill’s method. The angle AOD can be measured clockwise and counterclockwise, with the segment OA being the zero degree reference. In the case where the sum is over 180, you simply choose the angle AOD made by the rotation that gives the largest angle (the clockwise AOD +anticlockwise AOD=360).

The last thing to remember is that due to the convention I chose, the angle obtained using this method is the negative of the true value. In Lill's method the angle AOC = -108.43, but arctan(1) + arctan (2)=108.43.

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