Since high school math in trigonometry, it has been a fascination of mine to figure the exact value for sin (1°) and cos (1°).  I am not a mathematician, please excuse the messy nature of my calculations.  Thanks to WolframAlpha, we can do so with some basic algebra. 

We start with Sin (5x) given by Wolfram. This was deduced using Euler's equation and binomial algebra.  In a later edit, I will attempt to show the steps needed for this. 

Given:

\[sin (5x) = 16 sin^5(x) - 20 sin^3(x) + 5 sin(x)\]

\[sin (3x) = 3 sin(2x) - 4sin^3(2x)\]

\[sin (2x) = 2 sin(x) cos(x)\]

 

Sin (5x) = 5 Sin (x) - 20 Sin3 (x) + 16 Sin5 (x)

Substitute:

Sin (x) = k

Sin (5x) = 5k - 20k2 + 16k5

Factor:

Sin (5x) = k(5 - 20k2 + 16k4)

Substitute:

U = k2

Sin (5x) = k(5 - 20U + 16U2)

Here we can use the quadratic formula to find when our substitution is zero.

-b ± b2 - 4ac2a

U + -20±202-(4)(16)(5)(2)(16) = 0

U + -20 ±400-32032 = 0

U - 2032 ± 8032 = 0

U ± 58 - 58 = 0

Put it together and re-substitute:

Sin(5x) = k(k2+58-58)(k2-58-58)

Quadratic formula to factor k, twice:

-0±02-(4)(1)(58-58)(2)(1) and -0±02-(4)(1)(-58-58)(2)(1)

±10-52 and