Question

*50 pontos*
É dado sen x = 3/5, com 0<x<π/2, calcule:

a)cos x

b)sen (x-π/6)

c)cos (π/3+x)​

Answer 1

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$\tt a)~\sf sen(x)=\dfrac{3}{5}\implies sen^2(x)=\dfrac{9}{25}\\\sf cos^2(x)=\dfrac{25}{25}-\dfrac{9}{25}=\dfrac{16}{25}\\\sf cos(x)=\sqrt{\dfrac{16}{25}}\\\huge\boxed{\boxed{\boxed{\boxed{\sf cos(x)=\dfrac{4}{5}\checkmark}}}}$

$\tt b)~\sf sen\left(x-\dfrac{\pi}{6}\right)=sen(x)\cdot cos\left(\dfrac{\pi}{6}\right)+sen\left(\dfrac{\pi}{6}\right)\cdot cos(x)\\\sf sen\left(x-\dfrac{\pi}{6}\right)=\dfrac{3}{5}\cdot\dfrac{\sqr{3}}{2}+\dfrac{1}{2}\cdot\dfrac{4}{5}\\\sf sen\left(x-\dfrac{\pi}{6}\right)=\dfrac{9}{10}+\dfrac{4}{10}\\\huge\boxed{\boxed{\boxed{\boxed{\boxed{\sf sen\left(x-\dfrac{\pi}{6}\right)=\dfrac{13}{10}\checkmark}}}}}$

$\tt c)~\sf cos\left(\dfrac{\pi}{3}+x\right)=cos\left(\dfrac{\pi}{3}\right)\cdot cos(x)- sen\left(\dfrac{\pi}{3}\right)\cdot sen(x)\\\sf cos\left(\dfrac{\pi}{3}+x\right)=\dfrac{1}{2}\cdot\dfrac{4}{5}-\dfrac{\sqrt{3}}{2}\cdot\dfrac{3}{5}\\\huge\boxed{\boxed{\boxed{\boxed{\sf cos\left(\dfrac{\pi}{3}+x\right)=\dfrac{4-3\sqrt{3}}{10}\checkmark}}}}$

Answer 2

Resposta:

Explicação passo-a-passo:

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