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Tuesday, August 26, 2014

2014/068) which is greater \cos(\sin\, x) or \sin(\cos\, x)

we have
\cos(\sin\, x)-\sin(\cos\, x)
= \cos(\sin\, x)-\cos(\dfrac{\pi}{2}- \cos\, x)
= \cos(\sin\, x)+\cos(\dfrac{\pi}{2}+ \cos\, x)
2 \cos \dfrac{\sin\, x+\dfrac{\pi}{2}+ \cos\, x}{2} \cos \dfrac{-\sin\, x+\dfrac{\pi}{2}+ \cos\, x}{2}
2 \cos\ t \ \cos \ s
where t= \dfrac{\sin\, x+\dfrac{\pi}{2}+ \cos\, x}{2}
and s=  \dfrac{-\sin\, x+\dfrac{\pi}{2}+ \cos\, x}{2}

is product of 2 terms and we can show that both t and s are between 0 and \frac{\pi}{2}

we have \sin\, x\pm \cos\, x= \sqrt{2}\sin ( x\pm \dfrac{\pi}{4}) so between - \sqrt{2} and \sqrt{2}

hence s and t both are positive and less than \frac{\pi}{2} so the difference \cos(\sin\, x)-\sin(\cos\, x)\gt \ 0 and hence
\cos(\sin\, x) \gt \sin(\cos\, x)

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