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Resenja iz geometrije za zajednicki

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Puzzler	Feb 10 2008, 05:51 PM Post #1
Group: Članovi Joined: 2-July 06 Member No.: 74 Status: Van MGa Škola/Razred: MG/IIb (proud of it)	Skenirano se toliko užasno vidi, da ću morati ovako... You owe me big time.. Papir "16" 3. ${tg^2}{\frac{\pi}{8}}={\frac{1-cos{\frac{\pi}{4}}}{1+cos{\frac{\pi}{4}}}}={\frac{1-{\frac{sqrt{2}}{2}}}{1+{\frac{sqrt{2}}{2}}}}={\frac{2-sqrt{2}}{2+sqrt{2}}}=$ $={\frac{{(2-sqrt{2})}^2}{2}}$ $tg{\frac{\pi}{8}}=sqrt{2}-1 {\Rightarrow} ctg{\frac{\pi}{8}}=sqrt{2}+1$ $ctg{\frac{\pi}{8}}-tg{\frac{\pi}{8}}=2$ $tg{\frac{\pi}{12}}=tg({{\frac{\pi}{3}}-{\frac{\pi}{4}}})={\frac{tg{\frac{\pi}{3}}-tg{\frac{\pi}{4}}}{1+tg{\frac{\pi}{3}}tg{\frac{\pi}{4}}}}=...=2-sqrt{3}$ $tg{\frac{\pi}{12}}=2-sqrt{3} {\Rightarrow} ctg{\frac{\pi}{12}}=2+sqrt{3}$ $tg{\frac{\pi}{12}}+ctg{\frac{\pi}{12}}=4$ ${sin^4}({\frac{\pi}{16}})+{sin^4}({\frac{3{\pi}}{16}})+{sin^4}({\frac{5{\pi}}{16}})+{sin^4}({\frac{7{\pi}}{16}})={sin^4}({\frac{\pi}{16}})+{sin^4}({\frac{3{\pi}}{16}})+{cos^4}({\frac{3{\pi}}{16}})+{cos^2}({\frac{\pi}{16}})=$ $=({sin^2}({\frac{\pi}{16}})+{cos^2}({\frac{\pi}{16}}))^2-2{sin^2}({\frac{\pi}{16}}){cos^2}({\frac{\pi}{16}})+({sin^2}({\frac{3{\pi}}{16}})+{cos^2}({\frac{3{\pi}}{16}}))^2-2{sin^2}({\frac{3{\pi}}{16}}){cos^2}({\frac{3{\pi}}{16}})=$ $1-{\frac{{sin^2}({\frac{\pi}{8}})}{2}}+1-{\frac{{sin^2}({\frac{3{\pi}}{8}})}{2}}=2-{\frac{1-cos{\frac{\pi}{4}}}{4}}-{\frac{1-cos{\frac{3{\pi}}{4}}}{4}}=$ $={\frac{3}{2}}+{\frac{cos{\frac{\pi}{4}}+cos{\frac{3{\pi}}{4}}}{4}}={\frac{3}{2}}$ Zasad samo ovaj, sorry... Ne pitaj koliko mi je trebalo. --------------------

NeverMore21	Feb 10 2008, 06:10 PM Post #2
Group: Članovi Joined: 21-September 06 From: 21. блок Member No.: 115 Status: Učenik MGa Ime i prezime: Bojan Zukic Škola/Razred: Matematicka gimnazija IVb	Au Nenade, ovo ne da je prosto, a ni to ne znam ... --------------------

Puzzler	Feb 10 2008, 06:11 PM Post #3
Group: Članovi Joined: 2-July 06 Member No.: 74 Status: Van MGa Škola/Razred: MG/IIb (proud of it)	Papir "16" 4. ${sin^3}{\alpha}(1+ctg{\alpha})+{cos^3}{\alpha}(1+tg{\alpha})={sin^3}{\alpha}+{sin^2}{\alpha}cos{\alpha}+{cos^3}{\alpha}+sin{\alpha}{cos^2}{\alpha}=$ ${sin^3}{\alpha}+{cos^3}{\alpha}+sin{\alpha}cos{\alpha}(sin{\alpha}+cos{\alpha})=$ $(sin{\alpha}+cos{\alpha})({sin^2}{\alpha}-sin{\alpha}cos{\alpha}+{cos^2}{\alpha})+sin{\alpha}cos{\alpha}(sin{\alpha}+cos{\alpha})=$ $(sin{\alpha}+cos{\alpha})(sin^2{\alpha}+cos^2{\alpha})=sin{\alpha}+cos{\alpha}=sqrt{2}sin({\alpha}+{\frac{\pi}{4}})=$ $cos({\frac{\pi}{2}}-{\alpha}-{\frac{\pi}{4}})=cos({\frac{\pi}{4}}-{\alpha})$ $cos({\frac{3{\pi}}{10}}-{\alpha})-cos({\frac{{\pi}}{10}}-{\alpha})-cos({\frac{3{\pi}}{10}}+{\alpha})+cos({\frac{{\pi}}{10}}+{\alpha})=$ $(cos({\frac{3{\pi}}{10}}-{\alpha})+cos({\frac{{\pi}}{10}}+{\alpha}))-(cos({\frac{{\pi}}{10}}-{\alpha})+cos({\frac{3{\pi}}{10}}+{\alpha}))=$ $=2cos{\frac{\pi}{5}}cos({\alpha}-{\frac{\pi}{10}})-2cos{\frac{\pi}{5}}cos({\alpha}+{\frac{\pi}{10}})=2cos{\frac{\pi}{5}}(cos({\alpha}-{\frac{\pi}{10}})+cos({\alpha}+{\frac{\pi}{10}}))=$ $=4cos{\frac{\pi}{5}}sin{\alpha}sin{\frac{\pi}{10}}=sin{\alpha} {\cdot} 2 {\cdot} {\frac{2sin{\frac{\pi}{10}}cos{\frac{\pi}{10}}cos{\frac{\pi}{5}}}{cos{\frac{\pi}{10}}}}=sin{\alpha} {\cdot} {\frac{2sin{\frac{\pi}{5}}cos{\frac{\pi}{5}}}{cos{\frac{\pi}{10}}}}=sin{\alpha} {\cdot} {\frac{sin{\frac{2{\pi}}{5}}}{cos{\frac{\pi}{10}}}}=sin{\alpha}$ --------------------

Puzzler	Feb 10 2008, 06:24 PM Post #4
Group: Članovi Joined: 2-July 06 Member No.: 74 Status: Van MGa Škola/Razred: MG/IIb (proud of it)	Papir "16" 5. $f(x)=sin^5x=(sin^2x)^2 {\cdot} sin x=({\frac{1-cos 2x}{2}})^2 {\cdot} sin x={\frac{1}{4}}(1-2 cos 2x+cos^2x) {\cdot} sin x=$ ${\frac{1}{4}}(sin x - 2sin xcos 2x+ sinx {\cdot}{\frac{1+cos 4x}{2}})=$ $={\frac{1}{8}}(2sin x-2(sin x-sin 3x)+sin x +sin xcos 4x)=$ ${\frac{1}{8}}(2sinx-2sin3x+2sinx+sinx+{\frac{1}{2}}(sin 5x-sin3x))=$ $={\frac{1}{16}}(10sinx-4sin3x+sin5x-sin3x)={\frac{1}{16}}(10sinx-5sin3x+sin5x)$ $T_1=2{\pi}; T_2={\frac{2}{3}}{\pi}; T_3={\frac{2}{5}}{\pi}$ $T=[T_1,T_2,T_3]=2{\pi}$ I drugi skroz analogno... --------------------

NeverMore21	Feb 10 2008, 06:43 PM Post #5
Group: Članovi Joined: 21-September 06 From: 21. блок Member No.: 115 Status: Učenik MGa Ime i prezime: Bojan Zukic Škola/Razred: Matematicka gimnazija IVb	Nenade, imash od mene 2h u forumu ili 300g fornetija u ponedeljak ! --------------------

Puzzler	Feb 10 2008, 07:42 PM Post #6
Group: Članovi Joined: 2-July 06 Member No.: 74 Status: Van MGa Škola/Razred: MG/IIb (proud of it)	Papir "16" 7. $cos2x-sqrt{2}sinx+sin2x=0$ $sqrt{2}sin(2x+{\frac{\pi}{4}})-sqrt{2}sinx=0$ $sin(2x+{\frac{\pi}{4}})-sinx=0$ $2sin({\frac{x}{2}}+{\frac{\pi}{8}})cos({\frac{3x}{2}}+{\frac{\pi}{8}})=0$ $sin({\frac{x}{2}}+{\frac{\pi}{8}})=0 {\vee} cos({\frac{3x}{2}}+{\frac{\pi}{8}})=0$ $...$ ${\frac{1}{cosx}}+{\frac{1}{sinx}}=2sqrt{2}$ $[D: x{\neq}k{\frac{\pi}{2}}, k {\in} Z]$ $sinx+cosx=2sqrt{2}sinxcosx$ $sqrt{2}sin(x+{\frac{\pi}{4}})=sqrt{2}sin2x$ $sin(x+{\frac{\pi}{4}}) - sin2x=0$ $2sin(-{\frac{x}{2}}+{\frac{\pi}{8}})cos({\frac{3x}{2}}+{\frac{\pi}{8}})=0$ $sin({\frac{x}{2}}-{\frac{\pi}{8}})=0 {\vee} cos({\frac{3x}{2}}+{\frac{\pi}{8}})=0$ $...$ $sin3x=msinx$ $3sinx-4sin^3x=msinx$ $4sin^3x+(m-3)sinx=0$ $sinx(4sin^2x+m-3)=0$ $sinx=0 {\vee} 4sin^2x+m-3 =0$ Drugi slučaj (prvi je trivijalan): $4sin^2x {\in} [0,4] {\rightarrow} m {\in} [-1,3]$ za $m {\notin} [-1,3]$ nema rešenja, a u suprotnom $sin^2x={\frac{3-m}{4}}$ Ne trebaju meni fornetti ni forum, dosta je i prijateljstvo Ili da mi klikneš na potpis This post has been edited by Puzzler: Feb 10 2008, 07:43 PM --------------------

Puzzler	Feb 10 2008, 07:49 PM Post #7
Group: Članovi Joined: 2-July 06 Member No.: 74 Status: Van MGa Škola/Razred: MG/IIb (proud of it)	Papir "16" 8. $cos x cos y={\frac{1}{2}}$ $tg x+tg y=2$ $cos x cos y={\frac{1}{2}}$ ${\frac{sinxcosy+sinycosx}{cosxcosy}}=2$ $cos x cos y={\frac{1}{2}}$ ${\frac{sin(x+y)}{cosxcosy}}=2$ $cos x cos y={\frac{1}{2}}$ $sin(x+y)=1$ $cos x cos y={\frac{1}{2}}$ $cos(x+y)=0$ $cos x cos y={\frac{1}{2}}$ $cos x cos y - sin x sin y=0$ $cos x cos y={\frac{1}{2}}$ $sin x sin y={\frac{1}{2}}$ $cosxcosy-sinx siny=0$ $cosxcosy+sinxsiny=1$ $cos(x+y)=0$ $cos(x-y)=1$ $...$ --------------------

Puzzler	Feb 10 2008, 08:05 PM Post #8
Group: Članovi Joined: 2-July 06 Member No.: 74 Status: Van MGa Škola/Razred: MG/IIb (proud of it)	Papir "11" 3. $sin2{\alpha}+cos2{\alpha}=a \ \ \ \ \ \ 2{\alpha} {\in} ({\frac{\pi}{4}},{\frac{\pi}{2}}) {\Rightarrow} 4{\alpha} {\in} ({\frac{\pi}{2}},{\pi})$ $a^2=sin^22{\alpha}+cos^22{\alpha}+2sin2{\alpha}cos2{\alpha}$ $sin4{\alpha}=a^2-1$ $sin^24{\alpha}+cos^24{\alpha}=1$ $cos^24{\alpha}=1-sin^24{\alpha}$ $cos^24{\alpha}=1-a^4+2a^2-1$ $cos4{\alpha}=\|a sqrt{2-a^2}\| \ \ \ \ {\wedge} \ \ \ \ cos4{\alpha}<0 \ \ \ \ {\Rightarrow} cos4{\alpha}=-a sqrt{2-a^2}$ $sin4{\alpha}+cos4{\alpha}=a^2-1-a sqrt{2-a^2}$ --------------------

maxydelanoche	Feb 10 2008, 08:09 PM Post #9
Group: Članovi Joined: 3-May 06 From: Zion Member No.: 61 Status: Van MGa	Zar je toliko tesko bilo otvoriti temu Trigonometrija ZABOGA!!! Ae sada modovi neka lepo izmeste sve postove u zasebnu temu jer je ovo o FIZICI -------------------- Mi znamo sta se desava sa ljudima koji zastanu nasred puta. Bivaju pregazeni. Nista nije nemoguce. Za nemoguce je samo potrebno malo vise vremena. I'm doing the best I ever did, I'm doing the best that I can. www.viva-fizika.org

pyost	Feb 10 2008, 08:10 PM Post #10
Deus Ex Makina Group: Administratori Joined: 25-January 06 From: Beograd Member No.: 2 Status: Bivši učenik MGa Škola/Razred: RAF	A zar ne bi to moglo preko PMa? Naspamovaste topic... -------------------- Baby, it's a violent world. Registrovani korisnik Linuxa broj 460770 [Ubuntu 7.10]

RZA	Feb 10 2008, 08:11 PM Post #11
Njeno Ljubičanstvo Group: Članovi Joined: 7-September 06 Member No.: 92 Status: Bivši učenik MGa	A ne.. trebace i meni! -------------------- cold

pyost	Feb 10 2008, 08:13 PM Post #12
Deus Ex Makina Group: Administratori Joined: 25-January 06 From: Beograd Member No.: 2 Status: Bivši učenik MGa Škola/Razred: RAF	Evo vam na, nov topic -------------------- Baby, it's a violent world. Registrovani korisnik Linuxa broj 460770 [Ubuntu 7.10]

Puzzler	Feb 10 2008, 08:15 PM Post #13
Group: Članovi Joined: 2-July 06 Member No.: 74 Status: Van MGa Škola/Razred: MG/IIb (proud of it)	Papir "11" 4. $1+tg{\alpha}+tg^2{\alpha}+tg^3{\alpha}=4$ $tg^3{\alpha}+tg^2{\alpha}+tg{\alpha}-3=0$ $tg^3{\alpha}-tg^2{\alpha}+2tg^2{\alpha}-2tg{\alpha}+3tg{\alpha}-3=0$ $(tg{\alpha}-1)(tg^2{\alpha}+2tg{\alpha}+3)=0$ $tg{\alpha}=1$ , jer je $tg^2{\alpha}+2tg{\alpha}+3=(tg{\alpha}+1)^2+2>0$ $sin2{\alpha}={\frac{2tg{\alpha}}{1+tg^2{\alpha}}}=1$ $cos2{\alpha}={\frac{1-tg^2{\alpha}}{1+tg^2{\alpha}}}=0$ This post has been edited by Puzzler: Feb 10 2008, 08:21 PM --------------------

Anchi	Feb 10 2008, 08:17 PM Post #14
Group: Članovi Joined: 27-July 06 Member No.: 76 Status: Učenik MGa Škola/Razred: MG	Jeeeeeeeej extra

Puzzler	Feb 10 2008, 08:21 PM Post #15
Group: Članovi Joined: 2-July 06 Member No.: 74 Status: Van MGa Škola/Razred: MG/IIb (proud of it)	QUOTE(pyost @ Feb 10 2008, 08:10 PM) A zar ne bi to moglo preko PMa? Naspamovaste topic... Sorry, ne bi moglo, probao sam jednom drugu i ranije da pošaljem, i ovi tagovi ne rade... Izvini još jednom, pyoste. --------------------

pyost	Feb 10 2008, 08:22 PM Post #16
Deus Ex Makina Group: Administratori Joined: 25-January 06 From: Beograd Member No.: 2 Status: Bivši učenik MGa Škola/Razred: RAF	Meni rade -------------------- Baby, it's a violent world. Registrovani korisnik Linuxa broj 460770 [Ubuntu 7.10]

Anchi	Feb 10 2008, 08:29 PM Post #17
Group: Članovi Joined: 27-July 06 Member No.: 76 Status: Učenik MGa Škola/Razred: MG	Neka, bolje ovako da svi vidimo I meni su neka resenja bila neophodna

T-Rex	Feb 10 2008, 08:43 PM Post #18
Group: Članovi Joined: 5-December 07 Member No.: 790 Status: Učenik MGa Ime i prezime: Чедо Ш. Škola/Razred: MG IIб	Pazite ljudi shpijunira Ljubinka This post has been edited by T-Rex: Feb 10 2008, 08:44 PM

Anchi	Feb 10 2008, 08:53 PM Post #19
Group: Članovi Joined: 27-July 06 Member No.: 76 Status: Učenik MGa Škola/Razred: MG	QUOTE(T-Rex @ Feb 10 2008, 08:43 PM) Pazite ljudi shpijunira Ljubinka Jos samo kad bi znala da upali komp i pokrene Firefox (ili IE...)

T-Rex	Feb 10 2008, 09:05 PM Post #20
Group: Članovi Joined: 5-December 07 Member No.: 790 Status: Učenik MGa Ime i prezime: Чедо Ш. Škola/Razred: MG IIб	Kako sam glup brate meni ova reshenja nishta ne znache... Ja sam sutra ugasio, bice iz keca u kec...

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