60567

Funkcje Cyklometryczne

1.    Wyznaczyć dziedzinę funkcji:

(a) / (x) = aresin (4x — 2),

(d)    / (x) = y/4 aresin 3x — ;r,

(e)    / (x) = aresin (aretg x),

2.    Naszkicować wykres funkcji:

(a)	/(*)	= 3 aresin (|x| — 2),
(d)	f(x)	= aretg (|x - 1|) + 7T,
(e)	/(*)	— [arccos (x+ 1) - ^
(g)	/(*)	= aresin (sin x),
(i)	/(*)	= arccos (cosx),
(k)	/(*)	= aretg (tgx),
(1)	/(*)	= arcctg (ctgx).

(b)    / (x) = arccos (2^2r+l — 3),

(c)    / (x) = log.j [2 arccos (1 — 2x) —

(f)    / (i) = \/3arcctg (1 “ 2x) — 2TT.

(b)    f(x) = arccos (f - l) - § _f

(c)    / (z) = 2arcctg (x + 2) - tt, (f) / (x) = 3 arccos (3x — 0) — tr. (h) / (x) = sin (aresin x),

(j) / (x) = cos (arccosx),

(1) / (x) = tg (arctgx)

(m) f{x) = ctg (arcctgx).

3. Wykazać tożsamości:

	aresin (—x) = — aresin x,
[-i.i]	aresin x + arccos x =
Ki]	arccos (— x) — i: — arccos x,
[-i.i]	aresin x = aretg
[-i.i]	arccos x = arcctg j,
IR ari	"tg ( x) = —arctgx,
IR arcctg (—x) = tt — arcctgx,
R arctgx + arcctgx =
[0,1]	aresin x = arccos \/l — X^2,
[0,1]	aresin x = arcctg'
[0,1]	arccos x = aresin \/l - x*,
[0.1]	arccos x = aretg
(0,+oc)	arctgx = arcctg j,
(0,+oo)	arcctgx = aretg j,
(0,+oo)	arctgx = arccos
(0,+oc)	arcctgx = aresin

(a)

(b)

(c) W)

(e)

(0

(g)

(h)

(i)

(j) 00 (1)

(m)

(n)

(o) (P)

Vx g VxG Vx G Vx G Vx G Vx G Vx G Vx G Vx G Vx G VxG Vx G Vx G Vx G Vx G Vx G

4. Obliczyć wartość: aresin —fi + arccos fi + aretg (tg-^p¹) + arcctg (-\/3).

1