For all θ in [0, π/2] show that cos(sinθ)≥sin(cosθ).


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Question Number 60849 by Kunal12588 last updated on 26/May/19

$F o r a l l θ i n [0, π / 2] s h o w t h a t c o s (s i n θ) ⩾ s i n (c o s θ) .$
Commented by Prithwish sen last updated on 26/May/19

$when θ \in [0, \frac{π}{2}]$ $\sin θ and \cos θ lies between 0 to 1$ $∴ For α \in [0, 1]$ $Cos α > Sin α for \forall α \in [0, 1]$ $i . e Cos (Sin θ) > Sin (Cos θ)$
Commented by Kunal12588 last updated on 26/May/19

$t h a n k s s i r$
Commented by Kunal12588 last updated on 26/May/19

$b u t i {d o n}^{'} t g e t i t s i r h o w$ $c o s α > s i n α \forall α \in [0, 1]$
Commented by Prithwish sen last updated on 26/May/19

$θ - \to 0 - - \to \frac{π}{2}$ $\cos (\sin θ) \to 1 - - - \to 0.5$ $\sin (\cos θ) \to 0.8 - - - \to 0$ $is this ok ?$
Commented by Kunal12588 last updated on 26/May/19

$g r e a t s i r y o u a r e a m a z i n g T h a n k s a l o t .$
	Answered by tanmay last updated on 26/May/19

	$w h e n θ \in [0, \frac{π}{2}] s i n θ \in [0, 1] s o c o s (s i n θ) \in [c o s 1, 1]$ $w h e n θ \in [0, \frac{π}{2}] c o s θ \in [0, 1] s o s i n (c o s θ) \in [0, s i n 1]$ $c o s 1 > 0$ $1 > s i n 1$ $s o c o s (s i n θ) > s i n (c o s θ)$ $a t t a c h i n g g r a p h . . .$
	Commented by tanmay last updated on 26/May/19

	Commented by Kunal12588 last updated on 26/May/19

	$t h a n k y o u s i r$
	Commented by Kunal12588 last updated on 26/May/19

	$s i r h o w u r f i r s t f o u r l i n e s i m p l y$ $c o s (s i n θ) < s i n (c o s θ)$ $p l s e x p l a i n$
	Commented by tanmay last updated on 26/May/19

	$p l s r e f e r t h e g r a p h . . . a l l p r o b l e m s c a n n o t b e s o l v e d s i m p l y$ $b y m a t h e m a t i c s . . . i t i s b e t t e r t o s e e t h e g r a p h s . .$
	Commented by Kunal12588 last updated on 26/May/19

	$t h a n k s s i r g r a p h s c l a r i f i e s t h e t h i n g .$
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