Question

We know that
(i) sin x < x, for any x > 0
(ii) sin x > x, for any x < 0
Thus for any natural number n, sin(n) < n.
But as $-1<sin\left(n\right)<1and(-1,1)\subset [-\frac{\pi }{2},\frac{\pi }{2}]$
So, we can conclude that :
sin (sin(n)) < sin(n) of sin (n) > 0.
and sin (sin(n)) > sin(n) of sin (n) < 0.
Let us define two recursion sequences {an} and {bn} as follows.
$a_1=1,a_n=sin\left(a_{n-1}\right)$
$b_1=1,b_n=cos\left(b_{n-1}\right)$
${lim}_{n\to \infty }{b}_n$

• A. does not exist
• B. lies between 1 and 2
• C. lies between ‘0’ and $\frac{1}{2}$
• D. lies between $\frac{1}{2}$ and 1

Answer 1

The correct option is D lies between $\frac{1}{2}$ and 1



$1ϵ(0,\frac{\pi }{2})$
We have cos 1<1
$\therefore cos\left(cos1\right)>cos1$
and $cos\left(cos\right(cos1\left)\right)<cos\left(cos1\right)$
$y=cosx$ is decreasing in $(0,\frac{\pi }{2})$
$\Rightarrow$ smaller angle $cos\left(cos1\right)of(0,\frac{\pi }{2})$ will have bigger cosine.
$\Rightarrow$ the numbers $b_n$ will oscillate between cos 1 and 1 and will ultimately converge.
But ${60}^{\circ }>1\Rightarrow cos{60}^{\circ }<cos1\Rightarrow \frac{1}{2}<cos1$
$\therefore$ the limit must between $\frac{1}{2}$ and 1