Question

Considere que tg β = 0,7 e 0&lt; β &lt;3$\pi$/2 . ComConsidere que tg β = 0,7 e 0< β <3/2   . Com isso, o valor de cos β – sen β vale

Answer 1

$\pi<\beta<3\pi/2$

$\pi~rad=180\º\\3\pi/2~rad=270\º$

0º < β < 270º ----> β está no terceiro quadrante
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$tg~\beta=0,7~~~\therefore~~~\dfrac{sen~\beta}{cos~\beta}=0,7~~~\therefore~~~\boxed{sen~\beta=0,7cos~\beta}$

Pela relação fundamental da trigonometria, sabemos que:

$sen^{2}\beta+cos^{2}\beta=1$

Como sen β = 0,7cosβ:

$(0,7cos\beta)^{2}+cos^{2}\beta=1\\0,49cos^{2}\beta+cos^{2}\beta=1\\1,49cos^{2}\beta=1\\cos^{2}\beta=1/1,49\\cos^{2}\beta=100/149\\cos~\beta=\pm\sqrt{100/149}\\cos~\beta=\pm10/\sqrt{149}\\cos~\beta=\pm10\sqrt{149}/149$

Como o ângulo β está no terceiro quadrante, seu cosseno é negativo:

$\boxed{\boxed{cos~\beta=-\dfrac{10\sqrt{149}}{149}}}$

$sen~\beta=0,7*cos~\beta\\sen~\beta=0,7*(-10\sqrt{149}/149)\\sen~\beta=-7\sqrt{149}/149$
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$cos~\beta-sen~\beta=-\dfrac{10\sqrt{149}}{149}-(-\dfrac{7\sqrt{149}}{149})\\\\\\cos~\beta-sen~\beta=-\dfrac{10\sqrt{149}}{149}+\dfrac{7\sqrt{149}}{149}$

$cos~\beta-sen~\beta=\dfrac{-10\sqrt{149}+7\sqrt{149}}{149}\\\\\\\boxed{\boxed{cos~\beta-sen~\beta=-\dfrac{3\sqrt{149}}{149}}}$