Question

Sabendo que sen=3/5 e x pertence ao 1° quadrante , calcular sen 2x

Answer 1

$sen^{2}x+cos^{2}x=1\\\\\left(\dfrac{3}{5}\right)^{2}+cos^{2}x=1\\\\\\cos^{2}x=1-\dfrac{9}{25}\\\\\\cos^{2}x=\dfrac{25-9}{25}\\\\\\cos^{2}x=\dfrac{16}{25}\\\\\\cos~x=\pm\sqrt{\dfrac{16}{25}}\\\\\\cos~x=\pm\dfrac{4}{5}$

Como x pertence ao primeiro quadrante, seu cosseno é positivo

$\boxed{\boxed{cos~x=\dfrac{4}{5}}}$
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$sen~2x=sen~(x+x)\\sen~2x=sen~x\cdot cos~x+sen~x\cdot cos~x\\sen~2x=2\cdot sen~x\cdot cos~x$

Substituindo os valores de seno e cosseno:

$sen~x=2\cdot\dfrac{3}{5}\cdot\dfrac{4}{5}\\\\\\\boxed{\boxed{sen~2x=\dfrac{24}{25}}}$