ТРИГОНОМЕТРИЧНИ ФУНКЦИИ НА ОБОБЩЕН ЪГЪЛ (5)

[ганка симеонова]

пратих ти лично съобщение..

[Реклама]

[mathinvalidnik]

[tex]\frac{(tg\alpha+cotg\alpha)^2-(tg\alpha-cotg\alpha)^2 }{\frac{1}{sin^2\alpha*cos^2\alpha}-tg^2\alpha-cotg^2\alpha}[/tex]

[mathinvalidnik]

ouuuuuuu yeaaahhhhhhhhhhhhhhh

[mathinvalidnik]

[ганка симеонова]

е, това е! браво!!!

[martosss]

[mathinvalidnik]

[tex]\frac{(tg\alpha+cotg\alpha-tg\alpha+cotg\alpha)(tg\alpha+cptg\alpha+tg\alpha-cotg\alpha}{\frac{1}{sin^2\alpha*cos^2\alpha }-(tg^2\alpha+cotg^2\alpha }=\frac{2cotg\alpha*2tg\alpha}{(\frac{1}{sin\alpha*cos\alpha)^2 }-((tg\alpha+cotg\alpha)^2-2tg\alpha*cotg\alpha) }= \frac{2cotg\alpha*2tg\alpha}{(\frac{1}{sin\alpha*cos\alpha)^2 }-(tg\alpha+cotg\alpha)^2+2tg\alpha*cotg\alpha }=\frac{2cotg\alpha*2tg\alpha}{(\frac{1}{sin\alpha*cos\alpha }-tg\alpha-cotg\alpha)(\frac{1}{sin\alpha*cos\alpha }+tg\alpha+cotg\alpha)+2tg\alpha*cotg\alpha }=\frac{2cotg\alpha*2tg\alpha}{(\frac{1}{sin\alpha*cos\alpha }-\frac{sin\alpha}{cos\alpha }-\frac{cos\alpha}{sin\alpha })(\frac{1}{sin\alpha*cos\alpha }+\frac{sin\alpha}{cos\alpha }+\frac{cos\alpha}{sin\alpha }+2tg\alpha*cotg\alpha }=\frac{2cotg\alpha*2tg\alpha}{(\frac{1-sin^2\alpha-cos^2\alpha}{sin\alpha*cos\alpha)(\frac{1+sin^2\alpha+cos^2\alpha}{sin\alpha*cos\alpha })+2tg\alpha*cotg\alpha } }=\frac{2cotg\alpha*2tg\alpha}{\frac{(1-(sin^2\alpha+cos^2\alpha)}{sin\alpha*cos\alpha })(\frac{1+(sin^2\alpha+cos^2\alpha)}{sin\alpha*cos\alpha })+2tg\alpha*cotg\alpha }=\frac{2cotg\alpha*2tg\alpha}{(\frac{1-1}{sin\alpha*cos\alpha)(\frac{1+1}{sin\alpha*cos\alpha })+2tg\alpha*cotg\alpha } }=\frac{2cotg\alpha*2tg\alpha}{0*\frac{2}{sin\alpha*cos\alpha }+2tg\alpha*cotg\alpha } =2[/tex]

Няма да успея да ви покажа примера

[mathinvalidnik]

Някой ако иска да ме цитира да опита да се сети за какво става дма и коа може да ми поправи грешките.Искам да знам дали е вярно решението

[mathinvalidnik]

[martosss]

[tex]\frac{(tg\alpha+cotg\alpha-tg\alpha+cotg\alpha)(tg\alpha+cptg\alpha+tg\alpha cotg\alpha}{\frac{1}{sin^2\alpha*cos^2\alpha }-(tg^2\alpha+cotg^2\alpha } = \frac{2cotg\alpha*2tg\alpha}{(\frac{1}{sin\alpha*cos\alpha)^2 }-((tg\alpha+cotg\alpha)^2-2tg\alpha*cotg\alpha) } = \frac{2cotg\alpha*2tg\alpha}{(\frac{1}{sin\alpha*cos\alpha)^2 }-(tg\alpha+cotg\alpha)^2+2tg\alpha*cotg\alpha } =[/tex]

[tex]= \frac{2cotg\alpha*2tg\alpha}{(\frac{1}{sin\alpha*cos\alpha }-tg\alpha-cotg\alpha)(\frac{1}{sin\alpha*cos\alpha }+tg\alpha+cotg\alpha)+2tg\alpha*cotg\alpha }=\frac{2cotg\alpha*2tg\alpha}{(\frac{1}{sin\alpha*cos\alpha }-\frac{sin\alpha}{cos\alpha }-\frac{cos\alpha}{sin\alpha })(\frac{1}{sin\alpha*cos\alpha }+\frac{sin\alpha}{cos\alpha }+\frac{cos\alpha}{sin\alpha }+2tg\alpha*cotg\alpha }=[/tex]

=[tex]\frac{2cotg\alpha*2tg\alpha}{(\frac{1-sin^2\alpha-cos^2\alpha}{sin\alpha*cos\alpha)(\frac{1+sin^2\alpha+cos^2\alpha}{sin\alpha*cos\alpha })+2tg\alpha*cotg\alpha } }=\frac{2cotg\alpha*2tg\alpha}{\frac{(1-(sin^2\alpha+cos^2\alpha)}{sin\alpha*cos\alpha })(\frac{1+(sin^2\alpha+cos^2\alpha)}}sin\alpha*cos\alpha })+2tg\alpha*cotg\alpha } =[/tex]

=[tex] \frac{2cotg\alpha*2tg\alpha}{(\frac{1-1}{sin\alpha*cos\alpha)(\frac{1+1}{sin\alpha*cos\alpha })+2tg\alpha*cotg\alpha } }=\frac{2cotg\alpha*2tg\alpha}{0*\frac{2}{sin\alpha*cos\alpha }+2tg\alpha*cotg\alpha } =2[/tex]

Сега можеш да си го поправиш

[ганка симеонова]

вярно е

[mathinvalidnik]

[tex]\frac{(tg\alpha*cotg\alpha-tg\alpha+cotg\alpha)(tg\alpha+cotg\alpha+tg\alpha-cotg\alpha)}{\frac{1}{sin^2\alpha*cos^2\alpha }-(tg^2\alpha+cotg^2\alpha) }=\frac{2cotg\alpha*2tg\alpha}{(\frac{1}{sin\alpha*cos\alpha)^2 }-((tg\alpha+cotg\alpha)^2-2tg\alpha*cotg\alpha ) }=\frac{2cotg\alpha*2tg\alpha}{(\frac{1}{sin\alpha*cos\alpha })^2-(tg\alpha+cotg\alpha)^2+2tg\alpha*cotg\alpha }=\frac{2cotg\alpha*2tg\alpha}{(\frac{1}{sin\alpha*cos\alpha }-tg\alpha-cotg\alpha)(\frac{1}{sin\alpha*cos\alpha }+tg\alpha+cotg\alpha)+2tg\alpha*cotg\alpha }=\frac{2cotg\alpha*2tg\alpha}{(\frac{1}{sin\alpha*cos\alpha }-\frac{sin\alpha}{cos\alpha }-\frac{cos\alpha}{sin\alpha })(\frac{1}{ sin\alpha*cos\alpha}+\frac{sin\alpha}{cos\alpha } +\frac{cos\alpha}{sin\alpha }+2tg\alpha*cotg\alpha }=\frac{2cotg\alpha*2tg\alpha}{(\frac{1-sin^2\alpha-cos^2\alpha}{sin\alpha*cos\alpha })(\frac{1+sin^2\alpha+cos^2\alpha}{ sin\alpha*cos\alpha} )+2tg\alpha.cotg\alpha }=\frac{2tg\alpha*2cotg\alpha}{(\frac{1-(sin^2\alpha+cos^2\alpha)}{sin\alpha*cos\alpha } )(\frac{1+(sin^2\alpha+cos^2\alpha)}{sin\alpha*cos\alpha } )+2tg\alpha*cotg\alpha }=\frac{2cotg\alpha*2tg\alpha}{(\frac{1-1}{sin\alpha*cos\alpha })(\frac{1+1}{sin\alpha*cos\alpha })+2tg\alpha*cotg\alpha }[/tex]

По-нататък ми излиза прецакано

[mathinvalidnik]

[tex]\frac{2cotg\alpha*2tg\alpha}{0*\frac{2}{sin\alpha*cos\alpha }+2tg\alpha*cotg\alpha }=2[/tex]

Най-накрая.Това е

[mathinvalidnik]

Днес научих доста откъм вмъкване на задачите

[martosss]

Оффф, вземи да си го разделиш на части, примерно числител и знаменател, или пък просто тия дроби им давай 2-3 на ред и после на нов ред, защото така виж как се изписват лошо