Да се опрости тригонометричен израз

[Меди]

При $\alpha\in(0^\circ;90^\circ)$ да се опрости изразът $A=\dfrac{\sin^6\alpha+\cos^6\alpha-1}{\sin^2\alpha\cdot\cos^2\alpha}.$

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Отг. $-3$

[Davids]

При $\alpha\in(0^\circ;90^\circ)$ да се опрости изразът $A=\dfrac{\sin^6\alpha+\cos^6\alpha-1}{\sin^2\alpha\cdot\cos^2\alpha}.$

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Отг. $-3$

$A = \frac{sin^6\alpha + cos^6\alpha - 1}{sin^2\alpha.cos^2\alpha}$

Хващаме числителя:
$sin^6\alpha + cos^6\alpha - 1 =$
$= sin^6\alpha + cos^6\alpha - (sin^2\alpha + cos^2\alpha) =$
$= sin^6\alpha - sin^2\alpha + cos^6\alpha - cos^2\alpha =$
$= sin^2\alpha(sin^4\alpha - 1) + cos^2\alpha(cos^4\alpha - 1) =$
$= -sin^2\alpha(1 - sin^2\alpha)(1 + sin^2\alpha) - cos^2\alpha(1 - cos^2\alpha)(1 + cos^2\alpha)$
$= -sin^2\alpha cos^2\alpha(1 + sin^2\alpha) - -sin^2\alpha cos^2\alpha(1 + cos^2\alpha) = $
$= -sin^2\alpha cos^2\alpha(1 + sin^2\alpha + 1 + cos^2\alpha) =$
$= -3sin^2\alpha cos^2\alpha$

Бухаш върху знаменателя и получаваш -3

[Евва]

(2 начин)
A=[tex]\frac{sin^{6}\alpha+cos^{6}\alpha-1}{sin^{2}\alpha.cos^{2}\alpha}[/tex]

A=[tex]\frac{(sin^{2}\alpha+cos^{2}\alpha)^{3}-3sin^{4}(\alpha)cos^{2}\alpha-3sin^{2}(\alpha)cos^{4}\alpha-1}{sin^{2}\alpha.cos^{2}\alpha}[/tex]

A=[tex]\frac{1-3sin^{2}\alpha.cos^{2}\alpha(sin^{2}\alpha+cos^{2}\alpha)-1}{sin^{2}\alpha.cos^{2}\alpha}[/tex]

A=-3

[S.B.]

При $\alpha\in(0^\circ;90^\circ)$ да се опрости изразът $A=\dfrac{\sin^6\alpha+\cos^6\alpha-1}{\sin^2\alpha\cdot\cos^2\alpha}.$

Скрит текст: покажи

Отг. $-3$

Трети начин

$A = \frac{sin^{6}\alpha + cos^{6}\alpha - 1}{sin^{2}\alpha.cos^{2}\alpha} = \frac{(sin^{2}\alpha)^{3} + (cos^{2}\alpha)^{3} - 1}{sin^{2}\alpha .cos^{2}\alpha} =$

$\frac{(sin^{2}\alpha + cos^{2}\alpha)(sin^{4}\alpha - sin^{2}\alpha.cos^{2}\alpha + cos^{4}\alpha) - 1}{sin^{2}\alpha.cos^{2}\alpha} =$

[tex]\frac{( sin^{2}\alpha + cos^{2}\alpha)^{2} - 2sin^{2}\alpha.cos^{2}\alpha - sin^{2}\alpha.cos^{2}\alpha - 1}{sin^{2}\alpha.cos^{2}\alpha} = \frac{-3sin^{2}\alpha.cos^{2}\alpha}{sin^{2}\alpha.cos^{2}\alpha} = - 3[/tex]

[Меди]

Благодаря на всички за отговорите!

При $\alpha\in(0^\circ;90^\circ)$ да се опрости изразът $A=\dfrac{\sin^6\alpha+\cos^6\alpha-1}{\sin^2\alpha\cdot\cos^2\alpha}.$

Скрит текст: покажи

Отг. $-3$

Трети начин

$A = \frac{sin^{6}\alpha + cos^{6}\alpha - 1}{sin^{2}\alpha.cos^{2}\alpha} = \frac{(sin^{2}\alpha)^{3} + (cos^{2}\alpha)^{3} - 1}{sin^{2}\alpha .cos^{2}\alpha} =$

$\frac{\color{red}{(sin^{2}\alpha + cos^{2}\alpha)(sin^{4}\alpha - sin^{2}\alpha.cos^{2}\alpha + cos^{4}\alpha)}\color{black}{ - 1}}{sin^{2}\alpha.cos^{2}\alpha}$

[tex]\frac{\color{green}{( sin^{2}\alpha + cos^{2}\alpha)^{2} - 2sin^{2}\alpha.cos^{2}\alpha - sin^{2}\alpha.cos^{2}\alpha}\color{black}{ - 1}}{sin^{2}\alpha.cos^{2}\alpha} = \frac{-3sin^{2}\alpha.cos^{2}\alpha}{sin^{2}\alpha.cos^{2}\alpha} = - 3[/tex]

Може ли да попитам как правим прехода между втория и третия ред? Как червеното е равно на зеленото?

[S.B.]

$\frac{\color{red}{(sin^{2}\alpha + cos^{2}\alpha)(sin^{4}\alpha - sin^{2}\alpha.cos^{2}\alpha + cos^{4}\alpha)}\color{black}{ - 1}}{sin^{2}\alpha.cos^{2}\alpha}$

[tex]\frac{\color{green}{( sin^{2}\alpha + cos^{2}\alpha)^{2} - 2sin^{2}\alpha.cos^{2}\alpha - sin^{2}\alpha.cos^{2}\alpha}\color{black}{ - 1}}{sin^{2}\alpha.cos^{2}\alpha} = \frac{-3sin^{2}\alpha.cos^{2}\alpha}{sin^{2}\alpha.cos^{2}\alpha} = - 3[/tex]

Може ли да попитам как правим прехода между втория и третия ред? Как червеното е равно на зеленото?[/quote]

[tex](sin^{2}\alpha + cos^{2}\alpha)(sin^{4}\alpha - sin^{2}\alpha.cos^{2}\alpha + cos^{4}\alpha) =[/tex]

$ 1 . (sin^{4}\alpha - sin^{2}\alpha.cos^{2}\alpha + cos^{4}\alpha) $

Сигурна съм,че знаеш формулата :$a^{2} + b^{2} = (a + b)^{2} - 2ab$
Тук $a = sin^{2}\alpha $, $ b = cos^{2}\alpha$

$(sin^{2}\alpha)^{2} + (cos^{2}\alpha)^{2} - sin^{2}\alpha.cos^{2}\alpha = $

$(sin^{2}\alpha + cos^{2}\alpha)^{2} - 2sin^{2}\alpha.cos^{2}\alpha - sin^{2}\alpha.cos^{2}\alpha = $

$1 - 3sin^{2}\alpha.cos^{2}\alpha$