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?

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 $5^2$).
