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?
 A: Note
$$ \sin(2\theta)=2\sin\theta\cos\theta,\cos(2\theta)=2\cos^2\theta-1.$$
Let $\theta=2\arcsin(\frac35)$. It is easy to get
$$ \sin\theta=\frac{24}{25},\cos\theta=\frac{7}{25} $$
and hence
$$\sin(2\theta)=2\cdot\frac{24}{25}\cdot\frac{7}{25}=\frac{336}{625},\cos(2\theta)=2(\frac7{25})^2-1=\frac{527}{625} $$
and
$$\sin(4\theta)=2\sin(2\theta)\cos(2\theta)=\frac{354144}{390625},\cos(4\theta)=2\cos^2(2\theta)-1=\frac{164833}{390625}. $$
Thus
\begin{eqnarray}
&&\sin(10\arcsin(\frac35))\\
&=&\sin(5\theta)\\
&=&\sin(4\theta)\cos(\theta)+\cos(4\theta)\sin(\theta)\\
&=&\frac{354144}{390625}\cdot\frac7{25}+\frac{164833}{390625}\cdot\frac{24}{25}\\
&=&\frac{10296}{15625}.
\end{eqnarray}
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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$\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}

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

$$
f_{10} = 2^{10}\
\underbrace{\cos\pars{{\pi \over 2} - 2\arctan\pars{3 \over 4}}}
_{\ds{=\ {24 \over 25}}} = \bbx{24576 \over 25} = 983.04
$$
A: Simply writing it out by hand quickly yields
$$f_1=\frac{6}{5},\qquad f_2=\frac{96}{25},\qquad f_3=\frac{936}{125},\qquad f_4=\frac{384}{25},$$
and the rest should only take a few minutes. Here's a few more, it's not that hard:
$$f_5=\frac{3744}{125},\qquad f_6=\frac{1536}{25},\qquad f_7=\frac{14976}{125},\qquad f_8=\frac{6144}{25}.$$
I must say that $f_8$ and $f_{10}$ took slightly longer, but all in all this is no more than ten minutes work:
$$f_9=\frac{59904}{125},\qquad f_{10}=\frac{24576}{25}.$$
A: Let $g_n = \sqrt{4^n - f_n^2}$
You have noticed that the sequence seem well-behaved at first, and in particular, $f_{n+1}$ and $g_{n+1}$ should look like linear combinations of $f_n$ and $g_n$ except for the issue of the sign of the square root.
$f_{n+1} = \frac 15 (8 f_n + 6 g_n)$ is easy to get, and 
$g_{n+1}^2 = 4^{n+1}-f_{n+1}^2 = 4(g_n^2+f_n^2)-\frac 1{25}(64f_n^2+96f_ng_n+36g_n^2) = \frac 1{25}(36f_n^2-96f_ng_n+64g_n^2) = (\frac 15(6f_n-8g_n))^2$
And since $g_n$ is positive, we have $g_{n+1} = \frac 15|6f_n - 8g_n|$ (this absolute value is a giant trap and is also why your general trigonometric formula is wrong)
Now I only did that so that it was easier to compute the sequence (no need to take any square root !)
You start with $(f_0,g_0) = (0,1)$ and you apply those recurrence relations 10 times and you will get
$f_{10} = 983.04 = \frac{24576}{25}$

Despite the square root sign apparently screwing things up, it's still possible to say some more :
Let $z_n = f_n + ig_n$. The recurrence relation say that $z_{n+1} = \frac 15(8-6i)z_n$ or its conjugate (whichever has positive imaginary part)
So letting $\theta_n$ be an argument of $z_n$, we can pick $\theta_0 = \pi/2$, then $\theta_{n+1} = \theta_n - \arcsin(3/5)$ if it's positive, and $\theta_{n+1} = - \theta_n + \arcsin(3/5)$ otherwise.
If the second case happens, $\theta_{n+1} - \arcsin(3/5) = - \theta_n < 0$, and so $\theta_{n+2} = -\theta_{n+1}+\arcsin(3/5) = \theta_n$ : we enter an infinite $2$-cycle.
Which means that $z_{n+2} = 4z_n$ forall $n$ from that point on.
It turns out that $\theta_2 = \pi/2 - 2\arcsin(3/5) < \arcsin(3/5)$ and so $z_{10} = 4^4 z_2$
Thus $f_{10} = 256 f_2$ (which is why the denominator of $f_{10}$ isn't $5^{10}$ but $5^2$).
A: Squaring both sides
$$
25f_{n+1}^2-80f_{n}f_{n+1}+100f_n^2 = 36 \times 4^n
$$
or
$$
(5 f_{n+1}-\lambda_1f_n)(5f_{n+1}+\lambda_2f_n) = 36\times 4^n
$$
with $\lambda =  8\pm i 6$  This can be handled as
$$
\left\{
\begin{array}{}
5 f_{n+1}-\lambda_1f_n & = & a\\
5 f_{n+1}-\lambda_2f_n & = & b
\end{array}
\right.
$$
with $a \times b = 4^n$
then
$$
f_{n+1} = \frac{ ((3 + 4 i) a + (3 - 4 i) b)}{30}\\
f_{n} = \frac{i(a-b)}{12}
$$
etc.
For real solutions $a=b \Rightarrow f_{n+1} = \frac{2^{n+1}}{10}$
A: De Moivre's theorem says
$(\cosθ+i\sinθ)^{10}=\cos{10θ}+i\sin{10θ}$
$$\sin{10θ}=10\sinθ\cos^9θ-_{10}C_3\sin^3θ\cos^7θ+_{10}C_5\cos^5θ\sin^5θ
-_{10}C_7\cos^3θ\sin^7θ+10\cosθ\sin^9θ$$
$=\dfrac{10*3*4^9-10*3*4*3^3*4^7+9*2*7*3^5*4^5-10*3*4*3^7*4^3+10*4*3^9}{5^{10}}$
$$=\dfrac{1476984}{9765625}.$$ The result of wolfram alpha give same value.
