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Question Number 17233 by sushmitak last updated on 02/Jul/17
How to find out if  cos (cos x)>sin (sin x)  cos (sin x)>sin (cos x)  cos (sin (cos x))>sin (cos (sin x))  cos (cos (cos x))>sin (sin (sin x))  exam questions.  calculators not allowed.
$$\mathrm{How}\:\mathrm{to}\:\mathrm{find}\:\mathrm{out}\:\mathrm{if} \\ $$$$\mathrm{cos}\:\left(\mathrm{cos}\:{x}\right)>\mathrm{sin}\:\left(\mathrm{sin}\:{x}\right) \\ $$$$\mathrm{cos}\:\left(\mathrm{sin}\:{x}\right)>\mathrm{sin}\:\left(\mathrm{cos}\:{x}\right) \\ $$$$\mathrm{cos}\:\left(\mathrm{sin}\:\left(\mathrm{cos}\:{x}\right)\right)>\mathrm{sin}\:\left(\mathrm{cos}\:\left(\mathrm{sin}\:{x}\right)\right) \\ $$$$\mathrm{cos}\:\left(\mathrm{cos}\:\left(\mathrm{cos}\:{x}\right)\right)>\mathrm{sin}\:\left(\mathrm{sin}\:\left(\mathrm{sin}\:{x}\right)\right) \\ $$$$\mathrm{exam}\:\mathrm{questions}. \\ $$$$\mathrm{calculators}\:\mathrm{not}\:\mathrm{allowed}. \\ $$
Commented by prakash jain last updated on 03/Jul/17
One option is to manually check  for x from 0 to 2π, in the intervals  for (π/8)  0,(π/8),((2π)/8),((3π)/8),((4π)/8),...,..((16π)/8)  I am not 100% sure.  but Tinkutara has also been woking  on many such problem he may  have some suggestion.  mrW1, ajfour any comments?
$$\mathrm{One}\:\mathrm{option}\:\mathrm{is}\:\mathrm{to}\:\mathrm{manually}\:\mathrm{check} \\ $$$$\mathrm{for}\:{x}\:\mathrm{from}\:\mathrm{0}\:\mathrm{to}\:\mathrm{2}\pi,\:\mathrm{in}\:\mathrm{the}\:\mathrm{intervals} \\ $$$$\mathrm{for}\:\frac{\pi}{\mathrm{8}} \\ $$$$\mathrm{0},\frac{\pi}{\mathrm{8}},\frac{\mathrm{2}\pi}{\mathrm{8}},\frac{\mathrm{3}\pi}{\mathrm{8}},\frac{\mathrm{4}\pi}{\mathrm{8}},…,..\frac{\mathrm{16}\pi}{\mathrm{8}} \\ $$$$\mathrm{I}\:\mathrm{am}\:\mathrm{not}\:\mathrm{100\%}\:\mathrm{sure}. \\ $$$$\mathrm{but}\:\mathrm{Tinkutara}\:\mathrm{has}\:\mathrm{also}\:\mathrm{been}\:\mathrm{woking} \\ $$$$\mathrm{on}\:\mathrm{many}\:\mathrm{such}\:\mathrm{problem}\:\mathrm{he}\:\mathrm{may} \\ $$$$\mathrm{have}\:\mathrm{some}\:\mathrm{suggestion}. \\ $$$$\mathrm{mrW1},\:\mathrm{ajfour}\:\mathrm{any}\:\mathrm{comments}? \\ $$
Commented by 1234Hello last updated on 03/Jul/17
Maybe there is an idea: Range of  cos (cos x) is [cos 1, 1] and that of  sin (sin x) is [−sin 1, sin 1] and second  is the graphical method.
$$\mathrm{Maybe}\:\mathrm{there}\:\mathrm{is}\:\mathrm{an}\:\mathrm{idea}:\:\mathrm{Range}\:\mathrm{of} \\ $$$$\mathrm{cos}\:\left(\mathrm{cos}\:{x}\right)\:\mathrm{is}\:\left[\mathrm{cos}\:\mathrm{1},\:\mathrm{1}\right]\:\mathrm{and}\:\mathrm{that}\:\mathrm{of} \\ $$$$\mathrm{sin}\:\left(\mathrm{sin}\:{x}\right)\:\mathrm{is}\:\left[−\mathrm{sin}\:\mathrm{1},\:\mathrm{sin}\:\mathrm{1}\right]\:\mathrm{and}\:\mathrm{second} \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{graphical}\:\mathrm{method}. \\ $$

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