A Trigonometric Dilemma!

If we graph the line given by $y=\dfrac{x}{100}$ and the curve $y=\sin(x)$ , the number of points they have in common is what we require.

When $x$ is measured in degrees , it can be easily seen that the line intersects sine graph at $2$ distinct points and the origin , that is $3$ in total.Thus $A=3$ .

Now consider when $x$ is measured in radians . Since $y=\dfrac{x}{100}$ , as value of $y$ exceeds $1$ , the line goes away from the sine curve and doesnot intersect.Let $x=k\pi$ for some real $k$ such that $\dfrac{k\pi}{100}=1 \Rightarrow k=\dfrac{100}{\pi} \approx \dfrac{700}{22} \approx 31.8$ . Since $31\pi < 31.8\pi < 32\pi$ , the last point that the line intersects lies on the part of curve that terminates on $31\pi$ . Thus the line intersects the curve at $31$ distinct points on right side of $Y$ axis. By symmetry , it also intersects the curve at $31$ distinct points on left side of $Y$ axis. Origin is also common point to the line and curve. And hence $B=31+31+1=63$ .

Thus $A+B=3+63=66$ .

Consider the graph sin(x)=f(x) and x/100=g(x) separately.

-1<f(x)<1 inclusive of the values, as its limit.

g(x) has +- infinity as its limit.

The idea is to find the number of cycles g(x) passes through f(x), and later rationalize the number of intersections.

If x=radian,

then for g(x)=1, x=100 rad.

This means, number of cycles=100 rad/(2pi)=15.9 cycles

At f(0.9x2pi)=-0.5...

Hence, from x= 0 to 100 rad. , we know there must be 16x2=32 points of intersection, since g(x) is a continuous linear function, and for each cycle g(x) must pass through two points at f(x)'s positive crest. Note that at the first cycle g(x) passes through the origin.

From x=0 to -infinity inclusive, f(x) and g(x) is a reflection of f(x) and g(x) in its region in the positive values of x, about the x-axis. So points of intersection= 32 points, including the point at the origin.

Hence, total points of intersection = 64-1=63

If x=degree, and it is also assumed that the 100 in g(x) has a unit of degrees,

the same method is repeated and it will yield a total point of intersection of 3.

Thus, A+B= 63+3=66.