Question

Sendo t=sen²x/1+cosx, mostre que senx=2t/1+t².

Answer 1

Creio que haja um erro no enunciado.

A expressão correta para $t$ é

$\boxed{\begin{array}{c}t=\dfrac{\mathrm{sen\,}x}{1+\cos x} \end{array}}~~~~~~(\text{com }\cos x \ne -1)$

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Vamos partir de $\mathrm{sen\,}x.$

Supondo $(1+\cos x)\ne 0,\,$ vale que

$\mathrm{sen\,}x=\mathrm{sen\,}x\cdot \dfrac{1+\cos x}{1+\cos x}\\\\\\ =\dfrac{\mathrm{sen\,}x\,(1+\cos x)}{1+\cos x}$


Multiplicando e dividindo por $2,$

$=\dfrac{2\,\mathrm{sen\,}x\,(1+\cos x)}{2\,(1+\cos x)}\\\\\\ =\dfrac{2\,\mathrm{sen\,}x\,(1+\cos x)}{2+2\cos x}\\\\\\ =\dfrac{2\,\mathrm{sen\,}x\,(1+\cos x)}{1+2\cos x+1}~~~~~~(\text{mas }1=\mathrm{sen^2\,}x+\cos^2 x)\\\\\\ =\dfrac{2\,\mathrm{sen\,}x\,(1+\cos x)}{(\mathrm{sen^2\,}x+\cos^2 x)+2\cos x+1}\\\\\\ =\dfrac{2\,\mathrm{sen\,}x\,(1+\cos x)}{\mathrm{sen^2\,}x+(\cos^2 x+2\cos x+1)}\\\\\\ =\dfrac{2\,\mathrm{sen\,}x\,(1+\cos x)}{\mathrm{sen^2\,}x+(\cos x+1)^2}\\\\\\ =\dfrac{2\,\mathrm{sen\,}x\,(1+\cos x)}{(1+\cos x)^2+\mathrm{sen^2\,}x}$


Colocando $(1+\cos x)^2\ne 0$ em evidência no numerador e no denominador, e simplificando,

$=\dfrac{(1+\cos x)^2\cdot \left(\frac{2\,\mathrm{sen\,}x}{1+\cos x} \right )}{(1+\cos x)^2\cdot \left(1+\frac{\mathrm{sen^2\,}x}{(1+\cos x)^2}\right)}\\\\\\ =\dfrac{2\cdot \frac{\mathrm{sen\,}x}{1+\cos x} }{1+\left(\frac{\mathrm{sen\,}x}{1+\cos x} \right )^{\!2}}\\\\\\ =\dfrac{2t}{1+t^2}\\\\\\\\ \therefore~~\boxed{\begin{array}{c} \mathrm{sen\,}x=\dfrac{2t}{1+t^2} \end{array}}$

como queríamos demonstrar.


Bons estudos! :-)