Question

calcule o sen x cos x , sendo que π/2 < X<  π , onde sen x = - 2 cos x                                                                           

Answer 1

Pela relação fundamental, $\text{sen}^2~x+\text{cos}^2~x=1$, donde, $\text{sen}~x=\sqrt{1-\text{cos}^2~x}$.

Substituindo em $\text{sen}~x=-2\text{cos}~x$:

$\sqrt{1-\text{cos}^2~x}=-2\text{cos}~x$.

Elevando os dois lados ao quadrado:

$1-\text{cos}^2~x=4\text{cos}^2~x$.

$5\text{cos}^2~x=1$

$\text{cos}^2~x=\dfrac{1}{5}$

$\text{cos}~x=-\dfrac{\sqrt{5}}{5}$

$\text{sen}~x=\sqrt{1-\dfrac{5}{25}}=\sqrt{\dfrac{20}{25}}=\dfrac{2\sqrt{5}}{5}$.

Assim:

$\text{sen}~x\cdot\text{cos}~x=\dfrac{2\sqrt{5}}{5}\cdot\left(\dfrac{-\sqrt{5}}{5}\right)=\dfrac{-10}{25}=\dfrac{-2}{5}$