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)/4 cos(36 degrees) = (1+sqrt)/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 got sin(3 degrees) = sqrt(8-sqrt-sqrt-sqrt[10-2*sqrt(5)])/4 cos(3 degrees) = sqrt(8+sqrt+sqrt+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]) where 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. Then 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 http://mathforum.org/dr.math/
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