IIT JEE , AIEEE Forum - Mathematics , Trignometry

PyareMohan

For all

in interval 0,

/2, show thwt cos(sin

) > sin(cos

).

LAMPARD

sinx is always between -1 and 1.So cos(sinx) is very close to 1.cosx is also between 1 and -1.Thus,sin(cosx) is almost 0.And thus,cos(sinx)>sin(cosx)

edison

Use these identites to solve such problems

Inverse-forward identities are

Forward-inverse identities are

Inverse sum identities include

anandghegde

but sir how does your info help to solve this problem??

hsbhatt

$\sin \theta + \cos \theta < \frac{\pi}{2}$

Hence, $\sin \theta < \frac{\pi}{2} - \cos \theta$

Also, since $\theta \in [0, \frac{\pi}{2}]$ , $\sin \theta , cos \theta \in [0, \frac{\pi}{2}]$

where cosx is a decreasing function.

Hence $\cos (\sin \theta) > \cos (\frac{\pi}{2} - \cos \theta) = \sin (\cos \theta)$

coolriya

still cant get it ! !