# A Fifth Way to Calculate Sine Without Trig

In a previous post, I showed Four Ways to Calculate Sine Without Trig. While reading up on rational bezier curves, I came across a 5th way!

A rational bezier curve post will be coming in the (hopefully near) future, so I won’t go into too many details of that, but the important thing to know is that with a rational 2d quadratic bezier curve, you can EXACTLY represent any conic section, including circle arcs. With a regular (integral) bezier curve, you can’t.

You can in fact see that in action in an interactive HTML5 demo i made here: Rational Quadratic Bezier Curve. Note that on my main page, i have links to other 1d/2d bezier/trig rational/integral interactive curve demos. go check them out if you are interested in seeing more: http://demofox.og.

After I made the 2d quadratic bezier demo I realized something… The x and y positions of that curve are calculated independently of each other, and if you wanted to draw a circle in a program, you’d calculate the x and y positions independently, using sine for the y axis value and cosine for the x axis value. That means that if rational curves can – as people say – perfectly represent conic sections, that the two 1d curves that control each axis of that circle arc curve must be exact representations of sine and cosine, at least for the 90 degrees of that arc.

It turns out that this is true enough. I am seeing some variation, but have an open question on math stack exchange to try to get to the bottom of that.

Let’s do some math to come up with our curve based sine equation!

## Math Derivation – Skip If You Want To

A rational Bezier equation is defined like this, which is basically just one Bezier curve divided by another: $\bf{B}(t) = \frac{\sum\limits_{i=0}^n\binom {n} {i}(1-t)^{n-i}t^iW_iP_i}{\sum\limits_{i=0}^n\binom {n} {i}(1-t)^{n-i}t^iW_i}$

So, here is the equation for a rational quadratic bezier curve (n = 2): $\bf{B}(t) = \frac{(A*W_1*(1-t)^2 + B*W_2*2t(1-t) + C*W_3*t^2)}{(W_1*(1-t)^2 + W_2*2t(1-t) + W_3*t^2)}$ $A,B,C$ are the control points and $W_1,W_2,W_3$ are the weightings associated with those control points.

For the first 90 degrees of sine, $A=0, B=1, C = 1, W_1 = 1, W_2 = cos(arcAngle/2), W_3 = 1$.

arcAngle is 90 degrees in our case, so $W_2=cos(45)$ or $W_2=1/sqrt(2)$ or $W_2=0.70710678118$.

If we plug those values in and simplify, and treat it as an explicit (1d) equation of $y = f(x)$, instead of $P=f(t)$, we come up with the equation in the next section.

## Final Equation $y = \frac{(0.70710678118*2x(1-x) + x^2)}{((1-x)^2 + 0.70710678118*2x(1-x) + x^2)}$

or, so you can copy paste it:
y=(0.70710678118*2x(1-x) + x^2) / ((1-x)^2 + 0.70710678118*2x(1-x) + x^2)

That gives us the first quadrant (first 90 degrees) of sine. To get the second quadrant, we just horizontally flip the first quadrant. To get the third quadrant, we vertically flip the first quadrant. To get the fourth quadrant, we vertically and horizontally flip the first quadrant.

## Equation In Action

Here’s a screenshot from a shadertoy where I implemented this. red = true sine value, green = the value made with the curve, yellow = where they overlap and are equal. Any place you see red or green peaking out means they aren’t quite equal. I sort of expected it to be exactly dead on correct, but like I said, I have an open math stack exchange question to find out why it isn’t and will update this with my findings if they are relevant or interesting. It still is pretty darn good though.

Also, this formula may not be the most efficient way to calculate sine without trig that I’ve shown, but it is interesting, and that’s why I wanted to show it (:

You can see this in action at Shadertoy: Sin Without Trig IV 