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Sine & Cosine
( Math | Algebra | Trig | Sine&Cosine)

 

Trig Functions: Sine and Cosine Definition

Definition: An algebraic approach

From defining a few general properties of the sine and cosine functions, we can algebraically derive the sine and cosine functions themselves.

First, define the sine and cosine functions to have these properties:

[d/dx]sin x = cos x
[d/dx]cos x = - sin x
sin 0 = 0
cos 0 = 1
Now, write the sine function as an arbitrary power series, in that let sin x = a0x0 + a1x1 + a2x2 + a3x3 + a4x4 + ...
    sin 0 = 0 = a0
          a0 = 0

By differentiating the power series and equating it with the cosine by the original properties, [d/dx]sin x = cos x = 0 + (1)a1x0 + (2)a2x1 + (3)a3x2 + (4)a4x3 + ...
    cos 0 = 1 = (1)a1
            a1 = 1

Continuing, [d/dx]cos x = -sin x = -(0 + 0 + (2)(1)a2x0 + (3)(2)a3x1 + (4)(3)a4x2 + ...)
    -sin 0 = 0 = -(2)a2
             a2 = 0

[d/dx]-sin x = cos x = -(0 + 0 + 0 + (3)(2)(1)a3x0 + (4)(3)(2)a4x1 + ...)
     cos 0 = -1 = (3)(2)(1)a3
              a3 = -1/3!

[d/dx]cos x = sin x = 0 + 0 + 0 + 0 + (4)(3)(2)(1)a4x0 + ...
    sin 0 = 0 = (4)(3)(2)(1)a4
            a4 = 0

Continue on and you get values of all an
   if n is even then an = 0
   if n is odd then an = 1/n! alternating positive and negative.
Or stated as:
   a2n = 0
   a2n+1 = (-1)n/(2n+1)!


Plugging these values into the equation for sine:
    sin x = x1/1! - x3/3! + x5/5! - x7/7! + x9/9! - ... + (-1)nx2n+1/(2n+1)! + ...

          = [sum n=0..inf](-1)nx2n+1 / (2n+1)!

and for cosine:
    cos x = 1 - x2/2! + x4/4! - x6/6! + x8/8! - ... + (-1)nx2n/(2n)! + ...

          = [sum n=0..inf](-1)nx2n / (2n)!

These series converge for all x ÎÂ.

Source: Jeff Yates.
  
 
  

 
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