Question

Sendo x um arco do primeiro quadrante e sen x = 0,3. O valor sen2x vale:

a) $0,06\sqrt{91}$
b) $0,6\sqrt{91}$
c) $0,3\sqrt{91}$
d) $0,03\sqrt{91}$
e) $-0,06\sqrt{91}$

Answer 1

Primeiro vamos descobrir o cos x


$\mathsf{sen^{2}~x+cos^{2}~x=1}\\\\\\ \mathsf{0,3^{2}+cos^{2}~x=1}\\\\\\ \mathsf{cos^{2}~x=1-0,09}\\\\\\ \mathsf{cos^{2}~x=\dfrac{91}{100}}\\\\\\ \mathsf{cos~x=\sqrt{\dfrac{91}{100}}}\\\\\\ \mathsf{cos~x=0,1\cdot \sqrt{91}}$


Agora podemos encontrar o sen 2x


$\mathsf{sen~(2x)=2\cdot sen~x\cdot cos~x}\\\\\\ \mathsf{sen~(2x)=2\cdot 0,3\cdot 0,1\cdot \sqrt{91}}\\\\\\ \boxed{\begin{array}{c} \mathsf{sen~(2x)=0,06\cdot \sqrt{91}}\end{array}}$


Alternativa: A


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