Sequence involving trig

Given that

$\begin{aligned} f(a_{n+1}) & = \sqrt{f'(a_n)} \\ \tan a_{n+1} & = \sec a_n \\ \tan^2 a_{n+1} & = \sec^2 a_n = \tan^2 a_n + 1 \\ \implies \tan^2 a_2 & = \tan^2 a_1 + 1 = \tan^2 \frac \pi 3 + 1 = 4 \\ \tan^2 a_3 & = \tan^2 a_2 + 1 = 5 \\ \tan^2 a_4 & = 6 \\ \implies \tan^2 a_n & = n + 2 \\ \tan a_n & = \sqrt{n+2} \\ \implies \sin a_n & = \sqrt{\frac {n+2}{n+3}} \end{aligned}$

Then we have:

$\begin{aligned} \prod_{i=1}^k \sin a_i & = \prod_{i=1}^k \sqrt{\frac {i+2}{i+3}} = \sqrt{\frac 3{k+3}} \\ \prod_{i=1}^k \sin a_i & < \frac 1{10} \\ \implies \sqrt{\frac 3{k+3}} & < \frac 1{10} \\ k & > 297 \end{aligned}$

Therefore the minimum $k$ is $\boxed{298}$ .

It's easy to derive from $\tan (a_{n+1})=\sec (a_n), a_1=\dfrac{π}{3}$ that

$\sin (a_k)=\sqrt {\dfrac{k+2}{k+3}}$ ,

and the given product reduces to $\sqrt {\dfrac{3}{k+3}}$ . So

$\sqrt {\dfrac{3}{k+3}}<\dfrac{1}{10}\implies k>297\implies k_{min}=\boxed {298}$ .

Let $b_n=\tan^2{a_n}$ , then

$b_1=\tan^2{\frac{\pi}{3}}=3\\\tan^2{a_{n+1}}=\tan^2{a_n}+1\\b_{n+1}=b_{n}+1$

Using arithmetic progression, then $b_n=n+2\\\tan^2{a_n}=b_n=n+2\\\sin{a_n}=\sqrt{1-\cos^2{a_n}}\\=\sqrt{1-\frac{1}{\sec^2{a_n}}}\\=\sqrt{1-\frac{1}{1+tan^2{a_n}}}\\=\sqrt{1-\frac{1}{1+n+2}}\\=\sqrt{\frac{n+2}{n+3}}$

$\prod_{n=1}^k\sin{a_n}=\prod_{n=1}^k\sqrt{\frac{n+2}{n+3}}\\=\sqrt{\frac34}\sqrt{\frac45}\sqrt{\frac56}\hdots\sqrt{\frac{k+2}{k+3}}\\=\sqrt{\frac{3}{k+3}}\\\\\sqrt{\frac{3}{k+3}}<\frac1{10}\\k>297$

Hence, the minimum k is 298