Associated Topics || Dr. Math Home || Search Dr. Math

### 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
got

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])

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/
```
Associated Topics:
High School Trigonometry

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search