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Is-there-any-n-N-such-that-sin-n-Q-




Question Number 209519 by depressiveshrek last updated on 12/Jul/24
Is there any n∈N such that  sin(n)∈Q ?
$$\mathrm{Is}\:\mathrm{there}\:\mathrm{any}\:{n}\in\mathbb{N}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{sin}\left({n}\right)\in\mathbb{Q}\:? \\ $$
Commented by Rasheed.Sindhi last updated on 12/Jul/24
n in degrees?
$${n}\:{in}\:{degrees}? \\ $$
Commented by depressiveshrek last updated on 12/Jul/24
No that would be too easy
$$\mathrm{No}\:\mathrm{that}\:\mathrm{would}\:\mathrm{be}\:\mathrm{too}\:\mathrm{easy} \\ $$
Commented by mr W last updated on 12/Jul/24
no! there is even no rational number  n, except zero, such that sin (n) is  rational.  actually the only rational values of  sin (x) are: 0, ±(1/2), ±1.  ⇒see also Niven′s theorem
$${no}!\:{there}\:{is}\:{even}\:{no}\:{rational}\:{number} \\ $$$$\boldsymbol{{n}},\:{except}\:{zero},\:{such}\:{that}\:\mathrm{sin}\:\left(\boldsymbol{{n}}\right)\:{is} \\ $$$${rational}. \\ $$$${actually}\:{the}\:{only}\:{rational}\:{values}\:{of} \\ $$$$\mathrm{sin}\:\left({x}\right)\:{are}:\:\mathrm{0},\:\pm\frac{\mathrm{1}}{\mathrm{2}},\:\pm\mathrm{1}. \\ $$$$\Rightarrow{see}\:{also}\:{Niven}'{s}\:{theorem} \\ $$

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