Question

Sabendo que senx= 1/2 , calcule: y= cotgx . (1-cosx). (1 + 1/cosx)

Answer 1

$OLA$ $LULY$

$Senx= \frac{1}{2}$

$DETERMINANDO$  $O$   $Cosx$

$Sen^{2}x+Cos^{2} =1$
$( \frac{1}{2} )^{2} + Cos^{2}x=1$
$\frac{1}{4} + cos^{2}x =1$
$cos^{2} x= 1-\frac{1}{4}$
$cos^{2} x= \frac{4-1}{4}$
$cos^{2} x= \frac{3}{4}$
$cosx= \sqrt{ \frac{3}{4} }$
$cosx= \frac{ \sqrt{3} }{2}$

$DETERMINANDO$  $O$   $Cotgx$

$cotgx= \frac{cosx}{senx}$
$cotgx= \frac{ \frac{ \sqrt{3} }{2} }{ \frac{1}{2} } = \sqrt{3}$

$CALCULO$  $DO$  $PEDIDO$

$y=cotgx*(1-cosx)*(1+ \frac{1}{cosx} )$
$y=\sqrt{3}*(1- \frac{ \sqrt{3} }{2} )*(1+ \frac{1}{ \frac{ \sqrt{3} }{2} } )$
$y= \sqrt{3} *( \frac{2- \sqrt{3} }{2} )*(1+ \frac{2 \sqrt{3} }{3} )$
$y=( \frac{2 \sqrt{3} -3}{2} )*( \frac{3+2 \sqrt{3} }{3} )$
$y= \frac{ (2 \sqrt{3}) ^{2} -3^{2} }{6}$
$y= \frac{4*3-9}{6}$
$y= \frac{12-9}{6}$
$y= \frac{3}{6}$
$y= \frac{1}{2}$