OFFSET

1,3

COMMENTS

Nonnegative integers n satisfying sin(n)<sec(n). - Clark Kimberling, Aug 26 2014

This is proved by taking sec(n)=1/cos(n), therefore considering sin(n)*cos(n)<1 for cos(n)>0 and sin(n)*cos(n)>1 for cos(n)<0. Since sin(n)*cos(n)=sin(2n)/2, the first case becomes sin(2n)<2 for cos(n)>0 which is always correct, and the second case becomes sin(2n)>2 for cos(n)<0 which is never correct. - R. J. Mathar, Sep 07 2014

LINKS

Danny Rorabaugh, Table of n, a(n) for n = 1..10000

Eric Weisstein's World of Mathematics, Cosine

MATHEMATICA

Select[Range[150], Cos[#]>0&] (* Harvey P. Dale, Feb 02 2012 *)

PROG

(Sage) [i for i in range(200) if cos(i)>0] # Danny Rorabaugh, Mar 19 2015

(PARI) lista(nn) = for (n=0, nn, if (cos(n) > 0, print1(n, ", "))); \\ Michel Marcus, Mar 19 2015

CROSSREFS

KEYWORD

easy,nonn

AUTHOR

Stephen Casey (hexomino(AT)gmail.com), Aug 13 2007

EXTENSIONS

0 inserted by R. J. Mathar, Sep 06 2014

STATUS

approved