This article covers the maths of some algorithms to calculate sinus-tables. No code is given. You should be able to implement it byself. Following methods are covered:

• Taylor series
• Extrapolation using trigonometrical addition theorems
• The harmonical oscilator approach.
• Another Algebraic approach
• Fix by nao

Here we go:

The more elements you add, the more accurate the approximation of sin(x) is getting. x^7 or x^9 is already quite accurate.

Please note, that k!=1*2*3...k (faculty). 5!=1*2*3*4*5=120 for example.

So just use this and your sinustable-generator will work fine:

Trigonometric addition theorems:

1) cos(a+c) =cos(a)*cos(c) - sin(a)*sin(b)
2) cos(a-c) =cos(a)*cos(c) + sin(a)*sin(b)

2) forms into:

3) sin(a)*sin(b) =cos(a-c) - cos(a)*cos(c)

inserting 3) into 1) gives the final equation:

cos(a+c) =2*cos(a)*cos(c) - cos(a-c)

This equation makes recursive sinustable calculation possible. how ? easy:

f(x)=2*const*f(x-1) - f(x-2)

So everything we need now are the first two values of the sinus (cosinus) list. f(0)=cos(0)=1 and f(1)=cos(c). Also the constant const is needed. It is const=cos(c)=f(1)."c" is the stepwidth of the sinus-list.

With the usage of this algorithm really short sinustable generators are possible. But you have to take care of the accuracy. (14bit integer is NOT enough for a 1024 entry table)

Here is the derivation:

Assumption:

f(x+d)=f(x)+d*f'(x) with d>0

This isnt mathematically 100% correct, it is only an approximation. But still accurate enough for our purposes - and quite handy.

Since we want to calculate sinus we set:f(x)=sin(x) d is the stepwidth of the sinus table we want to generate.

sin(x+d)=sin(x)+d*cos(x)

we derive the whole equation and multiply with d:

d*cos(x+d)=d*cos(x)-d*d*sin(x)

1) sin(x+d) = sin(x) + d*cos(x)
2) d*cos(x+d) =d*cos(x) - d*d*sin(x)

now we have got two equations, which complete each other to a recursive function to calculate sinus.

a' = a + b
b' = b - a*d^2

Fixed version: (read below)

a' = a + b
b' = b - d^2*a'

The starting value of a is: sin(0)=0
The starting value of b is: cos(0)*d=1*d
Well and then we also need the constant d^2

This algorithm seems to be a little bit easier to handle than the trigonometric one when it comes to integer implementation. Well - the more accurate the calculations are , the later the function will degenerate.
btw. this calculation can also be applied to a 2d-array with some additional blurring - you will get nice water waves out of it :)

Assumption:

f(x+d)=f(x)+d*f'(x)
f'(x+d)=f'(x)+d*f''(x) with d>0

Now we set: f(x)=sin(x) - thus f'(x)=cos(x) and f''(x)=-sin(x). Using this facts we can change the equations a bit.

f(x+d)=f(x)+d*f'(x)
f'(x+d)=f'(x)-d*f(x)

This gives us a nice recursive formula calculating both sin(x)=f(x) and cos(x)=f'(x):

a' = a + d * b
b' = b - d *a

Fixed version: (read below)

a' = a + d * b
b' = b - d *a'

The starting value of a is: sin(0)=0
The starting value of b is: cos(0)=1
Well and then we also need the constant d, which is the stepwidth.

Lets start :

a' = a + b*d
b' = b - d*a

associated matrix (A) is:

| 1 -d | = A
| d 1 |

Det(A)= 1 + d^2

now...we expected an orthogonal matrix, that have determinant equal to 1 or -1.

Let's say, in physic words, that this matrix don't keep 'energy' constant. In fact, area covered by rotation, increase, and we obtain a spiral instead a circle.

Now, i rewrote A, with a little correction:

| 1 d | = A
| -d (1-d^2) |

Det(A) = 1 - d^2 +d^2 = 1

OK..now the matrix is (real) correct, but what about new formulas?

a' = a + b*d
b' = -d*a + b - b*d^2

AHHH..new ones are quite odd!

But, let try to substitute a' instead a in OLD b' formula:

b' = -d*a + b - b*d^2

WOW! exactly like new b' calculus, then we have:

a' = a + b*d
b' = b - a'*d