Figure out the numerical values of sines and cosines?

[StephenPrivitera]

How do mathematicians figure out the numerical values of sines and cosines? I can figure out how to evaluate sin(pi/12), sin(pi/24), sin(pi/48), etc, using sin(pi/6) and half angle formulas. How would I find sin(pi/5), for example? Is there any way other than infinite sums to express the value of a sine?

Edit: stupid mistake

 

[Hurkyl]

Use the 1/n-th angle formula!


Well, it's not quite that simple... but one way to get the value is to do something like the following:

cos 3θ = cos (2θ + θ)
= cos 2θ cos θ - sin 2θ sin θ
= (2 (cos θ)^2 - 1) cos θ - 2 (sin θ)^2 cos θ
= 2 (cos θ)^3 - cos θ - 2 cos θ + 2 (cos θ)^3
= 4 (cos θ)^3 - 3 cos θ

If you plug in π / 9 for θ, you have a polynomial in cos θ that you can solve.

In general, though, one cannot write the value of sin (π p/q) in terms of +, -, *, /, and roots.


When I derive the value for sin (π/5), I use geometry. I draw a regular pentagon and all of its diagonals, and through some magic with similar triangles, I can get a simple equation I can solve.

 

[HallsofIvy]

Actually most mathematicians use calculators to figure out values of trig functions!

For most people, the simplest way to calculate (by "hand") the approximate values of is to use the "Taylor series".

For any x, sin(x)= x- (1/6)x³+ 1/(5!) x⁵- 1/(7!) x⁷+ ... + (-1)²ⁿ⁺¹/(2n+1)! x²ⁿ⁺¹ and
cos(y)= 1- 1/2 x²+ 1/4! x⁴- 1/6! x⁶+ ...+ (-1)²ⁿ/(2n)! x²ⁿ.
If x is reasonably small, you don't have to take n very large at all.
(x is in radians, of course.)

I used to think that that was how calculators and computers did trig functions but I have been told that they actually use a much more sophisticated set of algorithms. Unfortunately, I've forgotten what they are called!

 

[chroot]

Halls,

He asked for ways to calculate WITHOUT power series.

And most calculators use a very sophisticated method called "table lookup."

- Warren

 

[StephenPrivitera]

Originally posted by Hurkyl

In general, though, one cannot write the value of sin (π p/q) in terms of +, -, *, /, and roots.

That disappoints me. I guess I already knew that. It just seems so unlikely that something as simple as a circle can lead so directly to such complicated results.

 

[HallsofIvy]

originally posted by Hurkyl In general, though, one cannot write the value of sin (ð p/q) in terms of +, -, *, /, and roots.

In other words, sine and cosine are not "algebraic functions".

Actually, most functions are not algebraic (they are "transcendental functions").

 

[selfAdjoint]

And π indeed is a transcendental number. It cannot be the solution of any polynomial equation with rational (or algebraic) coefficients.