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?

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.
