+14 So for some reason, I decided to see if there was a nice relationship between the x value of a tangent to a sin curve and the number of times it intercepts that same curve. This made sense, as the closer to the peak of the curve (where x = pi/2, 3pi/2, etc...) the more solutions there should be; the gradient of the tangent gets shallower and shallower until it is 0, and at those points there will be an infinite number of solutions.
To this end I formulated an equation, with a variable "a" representing the difference in the x coordinates of the original point, where x = X, and an intercept, where x = X + a.
y - y1 = m(x - x1);
Points: {X, sin(X)} {X + a, sin(X + a)}
Gradient: gradient of tangent to sin(x) = cos(X)
sin(X) - sin(X + a) = cos(X)*(X - (X + a))
sin(X) - sin(X + a) + cos(X)*(a) = 0
sin(X) - sin(a)cos(X) - cos(a)sin(X) + cos(X)*(a) = 0
-----divide through by cos(X)-------
tan(X) - sin(a) - cos(a)tan(X) + a = 0
a - sin(a) + tan(X)*(1 - cos(a)) = 0
Now this seems to work. For example, we would expect a = 0 to be a solution for any X, as the tangent must pass through at least that point, and X + 0 = X. This works, as
0 - sin(0) + tan(X)*(1 - cos(0)) = 0
0 + tan(X)*(1 - 1) = 0
tan(X)*0 = 0
0 = 0
Picking a point X = pi/2, yields an equation which implies infinity, because of the nature of tan(pi/2).
Furhermore, choosing a different point [ X = (pi + 0.4)/2 ] generated four solutions when I plugged it into Wolfram Alpha. Using GeoGebra showed these "a" values corresponded to the intercepts of the tangent at the point.
So I guess my question is, can you write "a" in terms of X from the equation
a - sin(a) + tan(X)*(1 - cos(a)) = 0 and is there a way to determine the relationship between X and how many solutions there are for "a", not just "a" itself. I imagine it looks something like the tan graph.
