<s>[INST] 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?
 [/INST] 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 $a=b \Rightarrow f_{n+1} = \frac{2^{n+1}}{10}$
</s>