Here we have sin((x+1)*(pi/2)):
Now let's say we have a point at (x=2, y=1)...
And we draw a line to the origin (x=0, y=0)...
If we create a point where the line intersects the sine function, we get a new point. Is there a way to find out the (x,y) of this new point?
The $x$ coordinate of the point of intersection is the unique positive number such that $\sin(\pi(x+1)/2) = x/2$. This equation cannot be solved in closed form but Newton's method shows the solution is approximately $0.7539340187865513$.
To apply Newton's method, define $f(x)=\sin(\pi(x+1)/2) - x/2$. Then $$N(x)=x-\frac{f(x)}{f'(x)} = x-\frac{\sin \left(\frac{1}{2} \pi (x+1)\right)-\frac{x}{2}}{\frac{1}{2} \pi \cos \left(\frac{1}{2} \pi (x+1)\right)-\frac{1}{2}}.$$
Iterating $N$ from the starting point $x_0 = 1.0$, we generate the sequence
$$ \begin{array}{l} 1.0000000000000000 \\ 0.7585469929947761 \\ 0.7539390100743306 \\ 0.753934018792478 \\ 0.7539340187865515 \\ 0.7539340187865514 \\ 0.7539340187865513 \\ 0.7539340187865514 \\ 0.7539340187865513 \\ \end{array} $$
Thus, it appears the solution is between $0.7539340187865513$ and $0.7539340187865513$.
The line through $\;(2,1)\;$ and through the origin is $\;y=\frac12x\;$ , so you need to solve
$$\sin\left(\frac\pi2(x+1)\right)=\frac12x$$
This yields a transcendental equation which would hardly have a simple, nice-looking zero, but:
$$h(x)=\sin\left(\frac\pi2(x+1)\right)-\frac12x\implies\begin{cases}h(0)=1\\{}\\h(1)=-\frac12\end{cases}$$
Now apply the Mean Value Theorem (=Lagrange's Theorem) to $\;h\;$ and deduce the existence of a point like the one you're interested in. If you want the actual value of the point's coordinates you may need to repeat the above several times, or use some other approximation methods (say, Newton's)
