Find $f_{10}$ of the sequence defined by $f_{n+1}=\frac{8f_n}{5}+\frac{6 \sqrt{4^n-f_n^2}}{5}$ Find $f_{10}$ of the sequence defined by $$f_{n+1}=\frac{8f_n}{5}+\frac{6 \sqrt{4^n-f_n^2}}{5}$$ given $f_0=0$
My Approach:
Letting $f_n=2^n b_n$ we get
$$b_{n+1}=\frac{4b_n}{5}+\frac{3 \sqrt{1-b_n^2}}{5}$$
Now letting $b_n=\cos(x_n)$ we get
$$\cos(x_{n+1})=\cos(x_n-\theta)$$ where $\cos(\theta)=\frac{4}{5}$
Now Since $f_0=0$ we have $b_0=0$ and $x_0=\frac{\pi}{2}$
Now we have
$$x_{n+1}=x_n-\theta$$
Putting $n=0,1,2,3 \cdots 10$ and adding all we get
$$x_{10}=\frac{\pi}{2}-10\theta$$
Hence
$$b_{10}=\cos\left(\frac{\pi}{2}-10\theta\right)=\sin(10\theta)=\sin\left(10\arcsin\left(\frac{3}{5}\right)\right)$$
How to proceed further?

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{f_{n} =
{8f_{n - 1} \over 5} + {6\root{4^{n - 1} - f_{n - 1}^{2}} \over 5}:\ {\large ?}\,,\qquad f_{0} = 0}$.

Lets $\ds{f_{n} = 2^{n}\cos\pars{x_{n}}}$:
\begin{align}
2^{n}\cos\pars{x_{n}} & =
{8 \over 5}\,2^{n - 1}\cos\pars{x_{n - 1}} +
{6\root{2^{2n - 2} - 2^{2n - 2}\cos^{2}\pars{x_{n - 1}}} \over 5}
\\[5mm] & =
{8 \over 5}\,2^{n - 1}\bracks{\cos\pars{x_{n - 1}} +
{3 \over 4}\verts{\sin\pars{x_{n - 1}}}}\quad
\pars{\begin{array}{l}
\mbox{Note that the correct term is}
\\
\ds{\color{red}{\verts{\sin\pars{x_{n - 1}}}}}\,\,\, \mbox{instead of}
\\
\ds{\sin\pars{x_{n - 1}}}\ \mbox{since}
\\
\ds{\left.\root{a^{2}}\right\vert_{\ a\ \in\ \mathbb{R}} = \verts{a}}.
\end{array}}
\\[5mm] & =
{8 \over 5}\,2^{n - 1}\bracks{\cos\pars{x_{n - 1}} +
\tan\pars{\theta}\verts{\sin\pars{x_{n - 1}}}}
\end{align}

where $\ds{\theta = \arctan\pars{3 \over 4}}$ and $\ds{x_{0} = \pi/2}$.

\begin{align}
\cos\pars{x_{n}} & =
{4 \over 5}\,\sec\pars{\theta}\cos\pars{x_{n - 1} -
\mrm{sign}\pars{\sin\pars{x_{n - 1}}}\theta}
\\[5mm] & =
{4 \over 5}\,\root{\pars{3/4}^{2} + 1}
\cos\pars{x_{n - 1} - \mrm{sign}\pars{\sin\pars{x_{n - 1}}}\theta}
\\[5mm] \implies &
\bbx{\cos\pars{x_{n}} = \cos\pars{x_{n - 1} -
\mrm{sign}\pars{\sin\pars{x_{n - 1}}}\theta}}
\end{align}

Moreover,
\begin{align}
x_{n} & = x_{n - 1} -
\mrm{sign}\pars{\sin\pars{x_{n - 1}}}\theta
\\[5mm]
\sum_{k = 1}^{n}x_{k} & = \sum_{k = 1}^{n}x_{k - 1} -
\theta\sum_{k = 1}^{n}\mrm{sign}\pars{\sin\pars{x_{k - 1}}}
\\[5mm]
-\,{\pi \over 2} + \sum_{k = 0}^{n}x_{k} & = \sum_{k = 0}^{n - 1}x_{k} -
\theta\sum_{k = 1}^{n}\mrm{sign}\pars{\sin\pars{x_{k - 1}}} =
\sum_{k = 0}^{n - 1}x_{k} -
\theta\sum_{k = 0}^{n - 1}\mrm{sign}\pars{\sin\pars{x_{k}}}
\\[5mm]
\implies &
\bbx{x_{n} =
{\pi \over 2} -
\bracks{\sum_{k = 0}^{n - 1}\mrm{sign}\pars{\sin\pars{x_{k}}}}\theta\,,
\qquad x_{0} = {\pi \over 2}\,,\quad\theta = \arctan\pars{3 \over 4}}
\end{align}

$$
\mbox{Then,}\quad x_{0} = {\pi \over 2}\,,\ x_{1} = {\pi \over 2} - \theta\quad \mbox{and}\quad
\left\{\begin{array}{rcl}
\ds{x_{2}} & \ds{=} & \ds{{\pi \over 2} - 2\theta}
\\[1mm]
\ds{x_{3}} & \ds{=} & \ds{{\pi \over 2} - 3\theta}
\\[1mm]
\ds{x_{4}} & \ds{=} & \ds{{\pi \over 2} - 2\theta}
\\[1mm]
\ds{x_{5}} & \ds{=} & \ds{{\pi \over 2} - 3\theta}
\\[1mm]
\ds{\vdots} & \ds{\vdots} & \ds{\phantom{AA}\vdots}
\\[1mm]
\ds{x_{10}} & \ds{=} & \ds{{\pi \over 2} - 2\theta}
\end{array}\right.
$$

$$
f_{10} = 2^{10}\
\underbrace{\cos\pars{{\pi \over 2} - 2\arctan\pars{3 \over 4}}}
_{\ds{=\ {24 \over 25}}} = \bbx{24576 \over 25} = 983.04
$$
