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.
$a_1=1,a_n=sin\left(a_{n-1}\right)$
$b_1=1,b_n=cos\left(b_{n-1}\right)$

Which of the following is true?

• A. $\left\{a_n\right\}$ is increasing, $\left\{b_n\right\}$ is decreasing
• B. $\left\{a_n\right\}$ is decreasing, $\left\{b_n\right\}$ is increasing
  
• C. $\left\{a_n\right\}$ is increasing, $\left\{b_n\right\}$ is non-monotonic
• D. $\left\{a_n\right\}$ is decreasing, $\left\{b_n\right\}$ is non-monotonic

Answer 1

The correct option is D $\left\{a_n\right\}$ is decreasing, $\left\{b_n\right\}$ is non-monotonic

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.
$a_1=1,a_n=sin(a_n–1)b_1=1,b_n=sin(b_n–1)$

$a_1=1,a_2=sin1\Rightarrow a_2<a_1{a}_3=sin\left(a_2\right)=sin\left(sin1\right)\Rightarrow a_3<a_2$
$\therefore \left\{a_n\right\}$ is decreasing
$b_1=1,b_2=cos1\Rightarrow b_2<b_1{b}_3=cos\left(cos1\right)>cos1\Rightarrow b_3>b_3$
similarly $b_4<b_3$
$\therefore \left\{b_n\right\}$ is non-monotonic