Question

sắp xếp theo thứ tự tằng dần

a, sin 40, cos28, sin65, cos88, cos20

b, tan32 độ 48, cot28 độ 36, tan 56độ 32, cot 67 độ 18

Answer 1

a) sin 40, cos 28, sin 65, cos88, cos20

ta có: \(cos28^0=sin62^0\)

\(cos88^0=sin2^0\)

\(cos20^0=sin70^0\)

vì \(sin2^0< sin40^0< sin62^0< sin65^0< sin70^0\)

nên \(cos88^0< sin40^0< cos28^0< sin65^0< cos20^0\)

b) \(tan32^048',cot28^036',tan56^032',cot67^018'\)

ta co: \(cot28^036'=tan62^036'\approx tan63^0\)

\(cot67^018'=tan23^018'\approx tan23^0\)

\(tan32^048'\approx tan33^0\)

\(tan56^032'\approx tan57^0\)

vi \(tan23^0< tan33^0< tan57^0< tan63^0\)

nen \(cot67^018'< tan32^048'< tan56^032'< cot28^036'\)