Question

sendo tgx= 3/4 calcule o seno 2x considere 90 graus

Answer 1

$tg~x=\dfrac{3}{4}~~~~~~(elevando~os~dois~lados~ao~quadrado)\\\\\\(tg~x)^{2}=\left(\dfrac{3}{4}\right)^{2}\\\\\\tg^{2}x=\dfrac{9}{16}~~~~~~(somando~1~nos~dois~lados)\\\\\\tg^{2}x+1=\dfrac{9}{16}+1~~~~~~\left(sabe-se~que~tg^{2}x+1=sec^{2}x=\dfrac{1}{cos^{2}x}\right)\\\\\\\dfrac{1}{cos^{2}x}=\dfrac{9+16}{16}\\\\\\\dfrac{1}{cos^{2}x}=\dfrac{25}{16}$

Invertendo os dois lados da equação:

$cos^{2}x=\dfrac{16}{25}~~~~~~(raiz~quadrada~nos~dois~lados)\\\\\\\sqrt{cos^{2}x}=\sqrt{\dfrac{16}{25}}\\\\\\cos~x=\pm\dfrac{4}{5}$

Como x está entre 90º e 270º, cos x é negativo.

$\boxed{\boxed{cos~x=-\dfrac{4}{5}}}$

Achando sen x:

$tg~x=\dfrac{3}{4}\\\\\\\dfrac{sen~x}{cos~x}=\dfrac{3}{4}\\\\\\sen~x=\dfrac{3}{4}\cdot cos~x\\\\\\sen~x=\dfrac{3}{4}\cdot\left(-\dfrac{4}{5}\right)\\\\\\\boxed{\boxed{sen~x=-\dfrac{3}{5}}}$
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Achando sen (2x):

$sen~(2x)=sen~(x+x)\\sen~(2x)=sen~x\cdot cos~x+sen~x\cdot cos~x\\sen~(2x)=2\cdot sen~x\cdot cos~x$

Substituindo sen x e cos x:

$sen~(2x)=2\cdot\left(-\dfrac{3}{5}\right)\cdot\left(-\dfrac{4}{5}\right)\\\\\\sen~(2x)=\dfrac{2\cdot3\cdot4}{5\cdot5}\\\\\\\boxed{\boxed{sen~(2x)=\dfrac{24}{25}}}$