Question

$\textnormal{Dac\u a } x\in (\frac{3\pi }{2} , 2\pi ) \textnormal{ \c si } cos x=\frac{1}{3}, \textnormal{ atunci } sin 2x \textnormal{ este egal cu:}$

Answer 1

Răspuns:

sin 2x=2sinxcosx

sin²x+cos²x=1

sin²x=1-cos²x

sin²x=1-(1/3)²

sin²x=1-1/9

sin²x=8/9

sinx=√8/9=±2√2/3

deoarece x∈cadranul 4   sinusul e  negativ. se ia  valoarea  negativa

sinx= -2√2/3

=>sin2x=2*(-2√2/3)*1/3= -4√3/9

Explicație pas cu pas:

Answer 2

$\it x\in\Big(\dfrac{3\pi}{2},\ 2\pi\Big) \Rightarrow sinx=-\sqrt{1-cos^2x}=-\sqrt{1-\dfrac{1}{9}}=-\sqrt{\dfrac{8}{9}}=-\dfrac{2\sqrt2}{3}\\ \\ \\ sin2x=2sinxcosx=2\cdot(-\dfrac{2\sqrt2}{3})\cdot\dfrac{1}{3}=-\dfrac{4\sqrt2}{9}$