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Data publicarii: 16 Octombrie, 2008

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Identitati remarcabile:

${\sin ^2}{a}+{\cos^2}{a}=1,\forall{a}\in{\mathbb{R}}.$ {\sin ^2}{a}+{\cos^2}{a}=1,\forall{a}\in{\mathbb{R}}.
$\sin(-x)=-\sin{x} ,\forall{x}\in{\mathbb{R}}.$ \sin(-x)=-\sin{x} ,\forall{x}\in{\mathbb{R}}.
$\cos{(-x)}=\cos{x} ,\forall{x}\in{\mathbb{R}}.$ \cos{(-x)}=\cos{x} ,\forall{x}\in{\mathbb{R}}.
${tgx} = \frac{\sin{x}}{\cos{x}},\forall{x}\in{\mathbb{R}}\setminus{\begin{Bmatrix} (2k+1)\frac{\pi}{2}|k\in{\mathbb{Z}}\end{Bmatrix}}.$ {tgx} = \frac{\sin{x}}{\cos{x}},\forall{x}\in{\mathbb{R}}\setminus{\begin{Bmatrix} (2k+1)\frac{\pi}{2}|k\in{\mathbb{Z}}\end{Bmatrix}}.

${ctgx} = \frac{\cos{x}}{\sin{x}},\forall{x}\in{\mathbb{R}}\setminus\{{k\pi}|k\in{\mathbb{Z}}\}.$ {ctgx} = \frac{\cos{x}}{\sin{x}},\forall{x}\in{\mathbb{R}}\setminus\{{k\pi}|k\in{\mathbb{Z}}\}.

${tg(-x)} = {- tgx}, \forall{x}\in{\mathbb{R}}\setminus{\begin{Bmatrix}(2k+1)\frac{\pi}{2}|k\in{\mathbb{Z}}\end{Bmatrix}}.$ {tg(-x)} = {- tgx}, \forall{x}\in{\mathbb{R}}\setminus{\begin{Bmatrix}(2k+1)\frac{\pi}{2}|k\in{\mathbb{Z}}\end{Bmatrix}}.

${ctg(-x)}={- ctgx},\forall{x}\in{\mathbb{R}}\setminus{\begin{Bmatrix}k\pi|k\in{\mathbb{Z}}\end{Bmatrix}}.$ {ctg(-x)}={- ctgx},\forall{x}\in{\mathbb{R}}\setminus{\begin{Bmatrix}k\pi|k\in{\mathbb{Z}}\end{Bmatrix}}.

${secx}=\frac{1}{cosx},\;\forall{x}\in{\mathbb{R}}\setminus{\begin{Bmatrix}(2k+1)\frac{\pi}{2}|k\in{\mathbb{Z}}\end{Bmatrix}}.$ {secx}=\frac{1}{cosx},\;\forall{x}\in{\mathbb{R}}\setminus{\begin{Bmatrix}(2k+1)\frac{\pi}{2}|k\in{\mathbb{Z}}\end{Bmatrix}}.

${cosecx}=\frac{1}{sinx},\;\forall{x}\in{\mathbb{R}}\setminus{\begin{Bmatrix}k\pi|k\in{\mathbb{Z}}\end{Bmatrix}}.$ {cosecx}=\frac{1}{sinx},\;\forall{x}\in{\mathbb{R}}\setminus{\begin{Bmatrix}k\pi|k\in{\mathbb{Z}}\end{Bmatrix}}.

$\cos{(a+b)}=\cos{a}\cos{b}-\sin{a}\sin{b}, \forall{a,b}\in{\mathbb{R}}.$ \cos{(a+b)}=\cos{a}\cos{b}-\sin{a}\sin{b}, \forall{a,b}\in{\mathbb{R}}.

$\cos{2a}={{\cos}^2}{a}-{{\sin}^2}{a} = 2{{\cos}^2}{a} - 1 = 1 - 2{{\sin}^2}{a},\forall{a}\in{\mathbb{R}}.$ \cos{2a}={{\cos}^2}{a}-{{\sin}^2}{a} = 2{{\cos}^2}{a} - 1 = 1 - 2{{\sin}^2}{a},\forall{a}\in{\mathbb{R}}.

$\cos{3a} =\ cos{a}(4{\cos^2}{a}-3),\forall{a}\in{\mathbb{R}}.$ \cos{3a} =\ cos{a}(4{\cos^2}{a}-3),\forall{a}\in{\mathbb{R}}.

$\cos{(a-b)}=\cos{a}\cos{b}+\sin{a}\sin{b}, \forall{a,b}\in{\mathbb{R}}.$ \cos{(a-b)}=\cos{a}\cos{b}+\sin{a}\sin{b}, \forall{a,b}\in{\mathbb{R}}.

$\cos{(\frac{\pi}{2}-x)}=\sin{x},\forall{x}\in{\mathbb{R}}.$ \cos{(\frac{\pi}{2}-x)}=\sin{x},\forall{x}\in{\mathbb{R}}.

$\cos{(\pi-x)}= -\ cos{x}, \forall{x}\in{\mathbb{R}}.$ \cos{(\pi-x)}= -\ cos{x}, \forall{x}\in{\mathbb{R}}.

$\cos{(x+2k\pi)} =\ cos{x},\forall{x}\in{\mathbb{R}},\forall{k}\in{\mathbb{Z}}.$ \cos{(x+2k\pi)} =\ cos{x},\forall{x}\in{\mathbb{R}},\forall{k}\in{\mathbb{Z}}.

$\sin{(a+b)} =\ sin{a}\cos{b} +\ cos{a}\sin{b},\forall{a,b}\in{\mathbb{R}}.$ \sin{(a+b)} =\ sin{a}\cos{b} +\ cos{a}\sin{b},\forall{a,b}\in{\mathbb{R}}.

$\sin{2a} = 2\sin{a}\cos{a},\forall{a}\in{\mathbb{R}}.$ \sin{2a} = 2\sin{a}\cos{a},\forall{a}\in{\mathbb{R}}.

$\sin{3a} =\ sin{a}(3 - 4{\sin^2}{a}), \forall{a}\in{\mathbb{R}}.$ \sin{3a} =\ sin{a}(3 - 4{\sin^2}{a}), \forall{a}\in{\mathbb{R}}.
$\sin{(a-b)}= \sin{a}\cos{b} - \cos{a}\sin{b}, \forall{a,b}\in{\mathbb{R}}.$ \sin{(a-b)}= \sin{a}\cos{b} - \cos{a}\sin{b}, \forall{a,b}\in{\mathbb{R}}.
$\sin{(\frac{\pi}{2}-x)} = \cos{x},\forall{x}\in{\mathbb{R}}.$ \sin{(\frac{\pi}{2}-x)} = \cos{x},\forall{x}\in{\mathbb{R}}.
$\sin{(\pi-x)} =\ sin{x},\forall{x}\in{\mathbb{R}}.$ \sin{(\pi-x)} =\ sin{x},\forall{x}\in{\mathbb{R}}.
$\sin{(x + 2k\pi)} =\ sin{x}, \forall{x}\in{\mathbb{R}},\forall{k}\in{\mathbb{Z}}.$ \sin{(x + 2k\pi)} =\ sin{x}, \forall{x}\in{\mathbb{R}},\forall{k}\in{\mathbb{Z}}.
${tg(x + k\pi)} = {tgx},\forall{x}\in{\mathbb{R}}\setminus{\begin{Bmatrix}(2k+1){\frac{\pi}{2}},\forall{k}\in{\mathbb{Z}}\end{Bmatrix}}.$ {tg(x + k\pi)} = {tgx},\forall{x}\in{\mathbb{R}}\setminus{\begin{Bmatrix}(2k+1){\frac{\pi}{2}},\forall{k}\in{\mathbb{Z}}\end{Bmatrix}}.
${ctg(x + k\pi)} = {ctgx},\forall{x}\in{\mathbb{R}}\setminus{\begin{Bmatrix}k\pi,\forall{k}\in{\mathbb{Z}}\end{Bmatrix}}.$ {ctg(x + k\pi)} = {ctgx},\forall{x}\in{\mathbb{R}}\setminus{\begin{Bmatrix}k\pi,\forall{k}\in{\mathbb{Z}}\end{Bmatrix}}.
${tg({a}\pm{b})} = \frac{{tga}\pm{tgb}}{1\mp{tga}{tgb}},\forall{a,b,{a}\pm{b}\in{\mathbb{R}}}\setminus{\begin{Bmatrix}(2k+1)\frac{\pi}{2},{k}\in{\mathbb{Z}}\end{Bmatrix}}.$ {tg({a}\pm{b})} = \frac{{tga}\pm{tgb}}{1\mp{tga}{tgb}},\forall{a,b,{a}\pm{b}\in{\mathbb{R}}}\setminus{\begin{Bmatrix}(2k+1)\frac{\pi}{2},{k}\in{\mathbb{Z}}\end{Bmatrix}}.
${tg2a} =\frac{2tga}{1-{{tg}^{2}}{a}}, \forall{a}\in{\mathbb{R}}\setminus{\begin{Bmatrix}(2k+1)\frac{\pi}{2}, k\in{\mathbb{Z}}\end{Bmatrix}}\cup\begin{Bmatrix}(2k+1)\frac{\pi}{4}, k\in{\mathbb{Z}}\end{Bmatrix}.$ {tg2a} =\frac{2tga}{1-{{tg}^{2}}{a}}, \forall{a}\in{\mathbb{R}}\setminus{\begin{Bmatrix}(2k+1)\frac{\pi}{2}, k\in{\mathbb{Z}}\end{Bmatrix}}\cup\begin{Bmatrix}(2k+1)\frac{\pi}{4}, k\in{\mathbb{Z}}\end{Bmatrix}.
$\sin{a}=\frac{2tg{\frac{a}{2}}}{1+{tg}^{2}{\frac{a}{2}}}, \forall{a}\in{\mathbb{R}} \setminus{\begin{Bmatrix}(2k+1)\pi,k\in{\mathbb{Z}}\end{Bmatrix}}.$ \sin{a}=\frac{2tg{\frac{a}{2}}}{1+{tg}^{2}{\frac{a}{2}}}, \forall{a}\in{\mathbb{R}} \setminus{\begin{Bmatrix}(2k+1)\pi,k\in{\mathbb{Z}}\end{Bmatrix}}.
$\cos{a} =\frac{1-{tg}^{2}{\frac{a}{2}}}{1+{tg}^{2}{\frac{a}{2}}}, \forall{a}\in{\mathbb{R}}\setminus{\begin{Bmatrix}(2k+1)\pi,k\in{\mathbb{Z}}\end{Bmatrix}}.$ \cos{a} =\frac{1-{tg}^{2}{\frac{a}{2}}}{1+{tg}^{2}{\frac{a}{2}}}, \forall{a}\in{\mathbb{R}}\setminus{\begin{Bmatrix}(2k+1)\pi,k\in{\mathbb{Z}}\end{Bmatrix}}.
${tga} =\frac{2tg\frac{a}{2}}{1-{{tg}^{2}}{\frac{a}{2}}}, \forall{a}\in{\mathbb{R}}\setminus{\begin{Bmatrix}(2k+1)\frac{\pi}{2}, k\in{\mathbb{Z}}\end{Bmatrix}}\cup\begin{Bmatrix}(2k+1)\pi, k\in{\mathbb{Z}}\end{Bmatrix}.$ {tga} =\frac{2tg\frac{a}{2}}{1-{{tg}^{2}}{\frac{a}{2}}}, \forall{a}\in{\mathbb{R}}\setminus{\begin{Bmatrix}(2k+1)\frac{\pi}{2}, k\in{\mathbb{Z}}\end{Bmatrix}}\cup\begin{Bmatrix}(2k+1)\pi, k\in{\mathbb{Z}}\end{Bmatrix}.
$\sin{a} +\ sin{b} = 2\sin{\frac{a+b}{2}}\cos{\frac{a-b}{2}}, \forall{a,b}\in{\mathbb{R}}.$ \sin{a} +\ sin{b} = 2\sin{\frac{a+b}{2}}\cos{\frac{a-b}{2}}, \forall{a,b}\in{\mathbb{R}}.
$\sin{a}-\ sin{b} = 2\sin{\frac{a-b}{2}}\cos{\frac{a+b}{2}}, \forall{a,b}\in{\mathbb{R}}.$ \sin{a}-\ sin{b} = 2\sin{\frac{a-b}{2}}\cos{\frac{a+b}{2}}, \forall{a,b}\in{\mathbb{R}}.
$\cos{a} +\ cos{b} = 2\cos{\frac{a+b}{2}}\cos{\frac{a-b}{2}}, \forall{a,b}\in{\mathbb{R}}.$ \cos{a} +\ cos{b} = 2\cos{\frac{a+b}{2}}\cos{\frac{a-b}{2}}, \forall{a,b}\in{\mathbb{R}}.
$\cos{a} -\ cos{b} = -2\sin{\frac{a+b}{2}}\sin{\frac{a-b}{2}}, \forall{a,b}\in{\mathbb{R}}.$ \cos{a} -\ cos{b} = -2\sin{\frac{a+b}{2}}\sin{\frac{a-b}{2}}, \forall{a,b}\in{\mathbb{R}}.
${tga}\pm{tgb}=\frac{\sin({a}\pm{b})}{{cosa}{cosb}}, \forall{a,b}\in{\mathbb{R}}\setminus{\begin{Bmatrix}(2k+1)\frac{\pi}{2}, k\in{\mathbb{Z}}\end{Bmatrix}}.$ {tga}\pm{tgb}=\frac{\sin({a}\pm{b})}{{cosa}{cosb}}, \forall{a,b}\in{\mathbb{R}}\setminus{\begin{Bmatrix}(2k+1)\frac{\pi}{2}, k\in{\mathbb{Z}}\end{Bmatrix}}.
${1 + cos{2a}} = 2{{\cos}^2}{a}\Leftrightarrow{|\cos{a}|} =\sqrt{\frac{1+\cos{2a}}{2}},\forall{a}\in{\mathbb{R}}.$ {1 + cos{2a}} = 2{{\cos}^2}{a}\Leftrightarrow{|\cos{a}|} =\sqrt{\frac{1+\cos{2a}}{2}},\forall{a}\in{\mathbb{R}}.
${1 - cos{2a}} = 2{{\sin}^2}{a}\Leftrightarrow{|\sin{a}|} =\sqrt{\frac{1-\cos{2a}}{2}},\forall{a}\in{\mathbb{R}}.$ {1 - cos{2a}} = 2{{\sin}^2}{a}\Leftrightarrow{|\sin{a}|} =\sqrt{\frac{1-\cos{2a}}{2}},\forall{a}\in{\mathbb{R}}.
${\sin{a}\cos{b}}=\frac{1}{2}\cdot[\sin{(a-b)}+\sin{(a+b)}],\forall{a,b}\in{\mathbb{R}}.$ {\sin{a}\cos{b}}=\frac{1}{2}\cdot[\sin{(a-b)}+\sin{(a+b)}],\forall{a,b}\in{\mathbb{R}}.
${\cos{a}\cos{b}}=\frac{1}{2}\cdot[\cos{(a-b)}+\cos{(a+b)}],\forall{a,b}\in{\mathbb{R}}.$ {\cos{a}\cos{b}}=\frac{1}{2}\cdot[\cos{(a-b)}+\cos{(a+b)}],\forall{a,b}\in{\mathbb{R}}.
${\sin{a}\sin{b}}=\frac{1}{2}\cdot[\cos{(a-b)}-\cos{(a+b)}],\forall{a,b}\in{\mathbb{R}}.$ {\sin{a}\sin{b}}=\frac{1}{2}\cdot[\cos{(a-b)}-\cos{(a+b)}],\forall{a,b}\in{\mathbb{R}}.
${\arcsin{x} +\ arccos{x}} =\frac{\pi}{2},\forall{x}\in{[-1,1]}.$ {\arcsin{x} +\ arccos{x}} =\frac{\pi}{2},\forall{x}\in{[-1,1]}.
${arctgx + arcctgx} =\frac{\pi}{2},\forall{x}\in{\mathbb{R}}.$ {arctgx + arcctgx} =\frac{\pi}{2},\forall{x}\in{\mathbb{R}}.
${\arcsin{(-x)}} = - {\arcsin}{x},\forall{x}\in{[-1,1]}.$ {\arcsin{(-x)}} = - {\arcsin}{x},\forall{x}\in{[-1,1]}.
${\arccos{(-x)}} = \pi -{\arccos}{x},\forall{x}\in{[-1,1]}.$ {\arccos{(-x)}} = \pi -{\arccos}{x},\forall{x}\in{[-1,1]}.
${arctg(-x)}=-{arctgx}, \forall{x}\in{\mathbb{R}}.$ {arctg(-x)}=-{arctgx}, \forall{x}\in{\mathbb{R}}.
${arcctg(-x)}=\pi -{arcctgx},\forall{x}\in{\mathbb{R}}.$ {arcctg(-x)}=\pi -{arcctgx},\forall{x}\in{\mathbb{R}}.
${{\sin}^2}{x} = \frac{{tg}^{2}{x}}{1+{tg}^{2}{x}},\forall{x}\in{\mathbb{R}}\setminus{\begin{Bmatrix}(2k+1)\frac{\pi}{2},k\in{\mathbb{Z}}\end{Bmatrix}}.$ {{\sin}^2}{x} = \frac{{tg}^{2}{x}}{1+{tg}^{2}{x}},\forall{x}\in{\mathbb{R}}\setminus{\begin{Bmatrix}(2k+1)\frac{\pi}{2},k\in{\mathbb{Z}}\end{Bmatrix}}.
${{\cos}^2}{x} = \frac{1}{1+{tg}^{2}{x}},\forall{x}\in{\mathbb{R}}\setminus{\begin{Bmatrix}(2k+1)\frac{\pi}{2},k\in{\mathbb{Z}}\end{Bmatrix}}.$ {{\cos}^2}{x} = \frac{1}{1+{tg}^{2}{x}},\forall{x}\in{\mathbb{R}}\setminus{\begin{Bmatrix}(2k+1)\frac{\pi}{2},k\in{\mathbb{Z}}\end{Bmatrix}}.

Valori remarcabile ale functiilor trigonometrice:

	0°	30°	45°	60°	90°	180°	270°	360°
sin	0	$\frac{1}{2}$ \frac{1}{2}	$\frac{\sqrt{2}}{2}$ \frac{\sqrt{2}}{2}	$\frac{\sqrt{3}}{2}$ \frac{\sqrt{3}}{2}	1	0	-1	0
cos	1	$\frac{\sqrt{3}}{2}$ \frac{\sqrt{3}}{2}	$\frac{\sqrt{2}}{2}$ \frac{\sqrt{2}}{2}	$\frac{1}{2}$ \frac{1}{2}	0	-1	0	1
tg	0	$\frac{\sqrt{3}}{3}$ \frac{\sqrt{3}}{3}	1	$\sqrt{3}$ \sqrt{3}	/	0	/	0
ctg	/	$\sqrt{3}$ \sqrt{3}	1	$\frac{\sqrt{3}}{3}$ \frac{\sqrt{3}}{3}	0	/	0	/

Ecuatii trigonometrice fundamentale:

$1)\,sinx={a}\in{[-1,+1]}\Leftrightarrow{x_k}={(-1)}^{k}arcsina+k\pi,{k}\in{\mathbb{Z}}.$ 1)\,sinx={a}\in{[-1,+1]}\Leftrightarrow{x_k}={(-1)}^{k}arcsina+k\pi,{k}\in{\mathbb{Z}}.

$2)\,cosx={a}\in{[-1,+1]}\Leftrightarrow{x_k}=\pm{arccosa}+2k\pi,{k}\in{\mathbb{Z}}.$ 2)\,cosx={a}\in{[-1,+1]}\Leftrightarrow{x_k}=\pm{arccosa}+2k\pi,{k}\in{\mathbb{Z}}.

$3)\,tgx={a}\in{\mathbb{R}}\Leftrightarrow{x_k}=arctga+k\pi,{k}\in{\mathbb{Z}}.$ 3)\,tgx={a}\in{\mathbb{R}}\Leftrightarrow{x_k}=arctga+k\pi,{k}\in{\mathbb{Z}}.

$4)\,ctgx={a}\in{\mathbb{R}}\Leftrightarrow{x_k}=arcctga+k\pi,{k}\in{\mathbb{Z}}.$ 4)\,ctgx={a}\in{\mathbb{R}}\Leftrightarrow{x_k}=arcctga+k\pi,{k}\in{\mathbb{Z}}.

Ecuatii trigonometrice reductibile la ecuatii algebrice:

Sunt ecuatiile de forma P(trig(mx)) = 0, unde P este un polinom de gradul n > 1, cu

coeficenti reali; "trig" simbolizeaza o functie trigonometrica oarecare, iar m este numar

real.

Notand trig(mx) = y, obtinem ecuatia algebrica P(y) = 0, de gradul n, cu radacinile

reale

y1, y2, ... ,yk, k in {1, 2, 3, ... ,n}; in continuare se rezolva ecuatiile

trig(mx) = yi, i = 1, 2, 3, ... , k.

Ecuatii omogene in sinus si cosinus:

Sunt ecuatiile de forma P(sinx, cosx) = 0, unde P(u,v) este un polinom omogen, cu

doua variabile, ai carui termeni sunt monoame de acelasi grad k.

Exemplu:

sin²x - 3sinxcosx + 2cos²x = 0,

unde P(u,v) = u² - 3uv + 2v², k = 2.

Impartind aceasta ecuatie prin cos²x, (evident, cosx este nenul, caci daca admitem

cosx = 0, rezulta din ecuatie si sinx = 0, ceea ce este imposibil), se ajunge la

ecuatia tg²x - 2tgx + 2 = 0 etc.

In mod similar se procedeaza in cazul altor valori ale lui k.

Observatie:

Daca ecuatia are forma P(u,v) = m, unde m este numar real nenul, iar k = 2k ', atunci

ecuatia se omogenizeaza, scriind-o sub forma:

$P(sinx,cosx)={m}\cdot{({{\sin}^{2}}{x}+{{\cos}^{2}}{x})}^{k$ P(sinx,cosx)={m}\cdot{({{\sin}^{2}}{x}+{{\cos}^{2}}{x})}^{k'}.

Ecuatii liniare in sinus si cosinus:

Sunt ecuatiile de forma asinx + bcosx + c = 0, unde a, b, c sunt numere reale, astfel

incat a·b·c este numar real nenul (alte cazuri conduc la ecuatii usor de analizat).

Distingem urmatoarele metode de rezolvare:

a) Metoda unghiului auxiliar:

Se imparte prin a si se obtine sinx + (b/a)cosx = c/a; se noteaza b/a = tgα, deci

α = arctg(b/a), α € (- π/2, π/2).

Dupa cateva calcule se ajunge la ecuatia elementara sin(x + α) = (c/a)·cosα etc.

b) Metoda substitutiei:

Cu ajutorul formulelor

$sinx=\frac{2tg{\frac{x}{2}}}{1+{tg}^{2}{\frac{x}{2}}}$ sinx=\frac{2tg{\frac{x}{2}}}{1+{tg}^{2}{\frac{x}{2}}}

$cosx=\frac{1-{tg}^{2}{\frac{x}{2}}}{1+{tg}^{2}{\frac{x}{2}}},$ cosx=\frac{1-{tg}^{2}{\frac{x}{2}}}{1+{tg}^{2}{\frac{x}{2}}},

obtinem o ecuatie de gradul al doilea cu necunoscuta tg(x/2) etc.

Observatie:

Intrucat numarul tg(x/2) nu exista daca x = (2k + 1)π, k € Z, rezulta ca eventualele

solutii de aceasta forma se pot pierde; prin urmare, in final, trebuie verificate in

ecuatia initiala si numerele respective.

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Răspunsuri şi comentarii

Emil Dumitrescu

Emil Dumitrescu, 30.12.2008 00:00

Critica a fost constructiva!;) Mi-am dat silinta si am reusit sa pacalesc LaTeX-ul! Acum nu mai apare "tan","ctan"...ci "tg", "ctg"...!!! Acum astept ...felicitarile!;))

Mihai

Mihai, 22.10.2008 00:00

le ai scris in franceza stimabile domn :) caci la noi nu exista ctan ci ctg la fel si ptr tan - tg si cele cu arc ( cotangenta si tagenta) dar sunt folositoare ! multumesc

Emil Dumitrescu

Emil Dumitrescu, 22.10.2008 00:00

Notatiile "tan" , "ctan" si altele sunt impuse de limbajul pe care-l folosesc in tehnoredactarea textului matematic...;n-am de ales!

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