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The Exact Value of the Sine of 1 Degree

Date: 05/24/2000 at 02:57:53
From: Lance Ward
Subject: The exact sine of 1 degree

Has anyone ever figured out the exact value for the sine of 1 degree 
or pi/180 radians? If so, can you tell me what it is?

Date: 05/25/2000 at 15:30:41
From: Doctor Rob
Subject: Re: The exact sine of 1 degree

Thanks for writing to Ask Dr. Math, Lance.

1. Probably, but I don't know who.

2. I can tell you how to figure it out for yourself. Start with

     sin(30 degrees) = 1/2
     cos(30 degrees) = sqrt(3)/2
     sin(36 degrees) = sqrt(10-2 sqrt[5])/4
     cos(36 degrees) = (1+sqrt[5])/4

Then use the identity

     cos(A-B) = cos(A)*cos(B) + sin(A)*sin(B)

with A = 36 degrees, B = 30 degrees to figure out cos(6 degrees). Then 
use the identity

     sin(A/2) = sqrt([1-cos(A)]/2)

to figure out sin(3 degrees). (The positive square root here is right 
because A/2 is in the first or second quadrant.) When I did this, I 

     sin(3 degrees) = sqrt(8-sqrt[3]-sqrt[15]-sqrt[10-2*sqrt(5)])/4
     cos(3 degrees) = sqrt(8+sqrt[3]+sqrt[15]+sqrt[10-2*sqrt(5)])/4

but you should do the calculation yourself to make sure that I did 
this correctly.

The last step is to use the identity

     sin(3*A) = 3*sin(A) - 4*sin^3(A)

with A = 1 degree, to create a cubic equation of which sin(A) is a 
root, and then to solve it. If s = sin(3 degrees), then sin(1 degree) 
is a root of the cubic equation

     4*x^3 - 3*x + s = 0

If we multiply by 2, and put y = 2*x, we get an equivalent equation

     y^3 - 3*y + 2*s = 0

Now the roots of this cubic are

     y = cbrt(-s + sqrt[s^2-1]) + cbrt(-s - sqrt[s^2-1])
     y = w*cbrt(-s + sqrt[s^2-1]) + w^2*cbrt(-s - sqrt[s^2-1])
     y = w^2*cbrt(-s + sqrt[s^2-1]) + w*cbrt(-s - sqrt[s^2-1])


     w = (-1+sqrt[-3])/2  and  w^2 = (-1-sqrt[-3])/2

are the complex cube roots of 1. Despite the appearance of complex 
numbers, the three roots will all be real, two positive and one 
negative. The smallest positive root is the one you want.


     sin(1 degree) = [cbrt(sqrt[s^2-1]-s) - cbrt(sqrt[s^2-1]+s)]/2

provided you take the cube roots of the complex numbers to have the 
smallest imaginary part.

- Doctor Rob, The Math Forum   
Associated Topics:
High School Trigonometry

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