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[Post New]15 Apr 2008 11:52:32 IST
    Subject: Re:Trigonometry question
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hsbhatt (4470)

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Olaaa!! Perrrfect answer. 842  [972 rates]
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\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)
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