Demontre a equação

por Dinheirow Qua 17 Abr 2013, 00:48

Demonstre que

$(\frac{1+sen^{4}x + cos^{4}x}{1 + sen^{4}y+cos^{4}y})^{2}= \frac{1+sen^{8}x + cos^{8}x}{1 + sen^{8}y+cos^{8}y}$

por **fantecele** Ter 29 Ago 2017, 15:04

$\\(1+sen^4x+cos^4x)^2=1+sen^8x+cos^8x+2.(sen^4x+cos^4x+sen^4x.cos^4x)\,\,\,(I)$

Vamos manipular um pouco $\\sen^4x+cos^4x+sen^4x.cos^4x$ :

$\\sen^4x+cos^4x+sen^4x.cos^4x\\1-2sen^2x.cos^2x+sen^4x.cos^4x\\\frac{2-4sen^2x.cos^2x+2sen^4x.cos^4x}{2}\\\frac{2-4sen^2x.cos^2x+4sen^4x.cos^4x-2sen^4x.cos^4x}{2}\\\frac{1+1-4sen^2x.cos^2x+4sen^4x.cos^4x-2sen^4x.cos^4x}{2}\\\frac{1+(1-2sen^2x.cos^2x)^2-2sen^4x.cos^4x}{2}\\\\\frac{1+(sen^4x+cos^4x)^2-2sen^4x.cos^4x}{2}$
$\\\frac{1+sen^8x+2sen^4x.cos^4x+cos^8x-2sen^4x.cos^4x}{2}\\\\\frac{1+sen^8x+cos^8x}{2}$

Portanto:

$\\sen^4x+cos^4x+sen^4x.cos^4x=\frac{1+sen^8x+cos^8x}{2}$

Substituindo a relação encontrada acima em (I):

$\\(1+sen^4x+cos^4x)^2=1+sen^8x+cos^8x+2.(sen^4x+cos^4x+sen^4x.cos^4x)\\(1+sen^4x+cos^4x)^2=1+sen^8x+cos^8x+2.\left ( \frac{1+sen^8x+cos^8x}{2} \right )\\(1+sen^4x+cos^4x)^2=2.(1+sen^8x+cos^8x)\,\,\,(II)$

Analogamente para y:

$\\(1+sen^4y+cos^4y)^2=2.(1+sen^8y+cos^8y)\,\,\,(III)$

Dividindo (II) por (III) iremos encontrar que:

$\\\left ( \frac{1+sen^4x+cos^4x}{1+sen^4y+cos^4y} \right )^2=\left ( \frac{1+sen^8x+cos^8x}{1+sen^8y+cos^8y} \right )$

Demontre a equação

Demontre a equação

Re: Demontre a equação

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