# ARCTAN

Also found in: Dictionary, Thesaurus, Encyclopedia, Wikipedia.
AcronymDefinition
ARCTANArcus Tangent (trigonometry)
References in periodicals archive ?
[f.sub.ap(e,o)3] = [2[f.sub.a0]/[pi]] arctan [square root of (-[B.sub.(e,o)] + [square root of ([B.sup.2.sub.(e,o)] - 4[A.sub.(e,o)][C.sub.(e,o)])]/2[A.sub.(e,o)])] (4c)
[[alpha].sub.1] = arctan [h/z] [right arrow] [[alpha].sub.2] [approximately equal to] [h/z] (3)
[DELTA][[phi].sub.mismatch] < arctan (0.0069/2) = 0.2[degrees] (E5)
It is interesting to observe that in case of small injection amplitudes, the arctan function can be approximated by its argument and, hence, (12) is reduced to (9).
[W.sub.cp] = [gamma]/2[pi] (arctan x + b/z - arctan x - b/z) + i [gamma]/4[pi] ln [z.sup.2] + [(x + b).sup.2]/[z.sup.2] + [(x - b).sup.2] (3)
[[theta].sub.1] = (arctan (y/x) - arccos ([a.sup.2]-[b.sup.2]+[x.sup.2]+[y.sup.2]/2a[square root of ([x.sup.2] + [y.sup.2])]) (14)
arctan [k.sub.d] [[omega].sup.[lambda].sub.c] sin (([pi]/2)[lambda])/ [k.sub.p] + [k.sub.d] [[omega].sup.[lambda].sub.c] cos(([pi]/2)[lambda])
[theta] = arctan [[P.sub.T]/[P.sub.z]] = arctan [[square root of -2 ln [R.sub.1]]/[a.sub.z][square root of -2 ln [R.sub.2]] cos (2[pi][R.sub.3]) + [b.sub.z]].
[phi]' = arctan (r cos [[theta].sub.n,m]sin [[theta].sub.n,m]/ r cos [[theta].sub.n,m] cos [[phi].sub.n,m] + (n- 1)[DELTA]).
[[theta].sub.2] = arctan [y.sub.c]/[square root of ([x.sup.2.sub.c] + [y.sup.2.sub.c]) - [(S/2).sup.2]], (1)
Moreover, it was showed that all paraxial shape-invariant beams have a universal Gouy phase of arctan form .
[[beta].sub.A, i] = [[theta].sub.s, i] arctan ([d.sub.i] - [R.sub.cell]/[H.sub.HAPS]) for i = 2, ...
Site: Follow: Share:
Open / Close