Can anybody solve the following using binomial theorem?

Question

If $x,y \in\mathbb{R}$ such that $x^2 + y^2 - 6x + 8y + 24 = 0$ then what is the greatest value of $$\frac{16\cos^2(\sqrt{x^2+y^2})}{5} - \frac{24\sin(\sqrt{x^2+y^2})}{5}?$$

Answer 1

Hint. Note that $x^2 + y^2 - 6x + 8y + 24 = 0$ is equivalent to $$(x-3)^2 + (y-4)^2 = 1$$ that is the circle $C$ centered in $(3,4)$ of radius $1$. Hence for any point $(x,y)$ on this circle its distance from the origin, that is $\sqrt{x^2+y^2}$, attains by continuity all the values in the interval $[\sqrt{3^2+4^2}-1,\sqrt{3^2+4^2}+1]=[4,6]$. Therefore \begin{align*} M:&=\max_{(x,y)\in C} \left(\frac{16\cos^2(\sqrt{x^2+y^2})}{5}-\frac{24\sin(\sqrt{x^2+y^2})}{5}\right)\\ &=\frac{1}{5}\max_{t\in [4,6]} \left(16\cos^2(t)-24\sin(t)\right)\\ &=\frac{1}{5}\max_{t\in [4,6]} \left(16(1-\sin^2(t))-24\sin(t)\right)\\ &=\frac{8}{5}\max_{t\in [4,6]} \left((\sin(t)+2)(1-2\sin(t))\right). \end{align*} Now the quadratic polynomial $(s+2)(1-2s)$ attains its maximum value at the midpoint of its roots that is $\frac{-2+1/2}{2}=-3/4$.

Is there a value of $t\in [4,6]$ such that $\sin(t)=-3/4$? What is $M$?

P.S. I am not sure you need the binomial theorem for this problem.