I'm trying to find a general solution for the intersection of two trigonometric functions:
$$a(x)=500\sin \left( \frac{\pi x}{2} \right)+150$$ $$a(x)=200\cos \left( \frac{\pi }{2}\left( x+\frac{\pi }{2} \right) \right)+250$$
When graphing the equations together, it can be seen that they periodically intersect. However, I am trying to algebraically prove their intersection point by obtaining a general solution for the x-coordinates of their intersection, without using a graph.
Here's how I'm trying to solve it: $$500\sin \left( \frac{\pi x}{2} \right)+150=200\cos \left( \frac{\pi }{2}\left( x+\frac{\pi }{2} \right) \right)+250$$ $$\frac{500\sin \left( \frac{\pi x}{2} \right)}{200}-\frac{100}{200}=\cos \left( \frac{\pi }{2}\left( x+\frac{\pi }{2} \right) \right)$$ $$\left( \frac{500\sin \left( \frac{\pi x}{2} \right)}{200}-\frac{100}{200} \right)^{2}=1-\sin ^{2}\left( \frac{\pi }{2}\left( x+\frac{\pi }{2} \right) \right)$$ so that $$6.25\sin ^{2}\left( \frac{\pi x}{2} \right)-2.5\sin \left( \frac{\pi x}{2} \right)+\sin ^{2}\left( \frac{\pi x}{2}+\frac{\pi ^{2}}{4} \right)-0.75=0$$ But here's where I'm stuck. How can I eliminate the $\frac{\pi ^{2}}{4}$ term from the $\sin ^{2}\left( \frac{\pi x}{2}+\frac{\pi ^{2}}{4} \right)$ term so that I can have 2 like terms and be able to add the coefficients and solve the quadratic? Then I'd be able to find the reference angle for the general solution and provide an algebraic solution.
Any help appreciated.
Hint: let $a=\pi/4$.
Step (1) expand $\sin(2a(x+2a))$:
$$\sin(2a(x+2a))=\sin(2ax)\cos(4a^2)+\cos(2ax)\sin(4a^2)......(1)$$
Step (2) Change variable: $$\sin(2ax)=\frac{2t}{1+t^2}, \cos(2ax)=\frac{1-t^2}{1+t^2},t=\tan(ax)$$
Step (3) solve the resulting equation for $t$.
Step (4) solve $t=\tan(ax)$ for $x$.