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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 byprakash 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 by1234Hello 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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