Respuesta :
now, we know that 90°< θ <180°, that simply means the angle θ is in the II quadrant, where sine is positive and cosine is negative.
[tex]\bf sin(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{hypotenuse}{13}}\impliedby \textit{now let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}[/tex]
[tex]\bf \pm\sqrt{13^2-5^2}=a\implies \pm\sqrt{144}=a\implies \pm 12 =a \implies \stackrel{II~quadrant}{-12=a} \\\\\\ therefore \qquad cos(\theta )=\cfrac{\stackrel{adjacent}{-12}}{\stackrel{hypotenuse}{13}} \\\\ -------------------------------\\\\ sin(2\theta )\implies 2sin(\theta )cos(\theta )\implies 2\left(\frac{5}{13} \right)\left( \frac{-12}{13} \right)\implies -\cfrac{120}{169}[/tex]
[tex]\bf sin(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{hypotenuse}{13}}\impliedby \textit{now let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}[/tex]
[tex]\bf \pm\sqrt{13^2-5^2}=a\implies \pm\sqrt{144}=a\implies \pm 12 =a \implies \stackrel{II~quadrant}{-12=a} \\\\\\ therefore \qquad cos(\theta )=\cfrac{\stackrel{adjacent}{-12}}{\stackrel{hypotenuse}{13}} \\\\ -------------------------------\\\\ sin(2\theta )\implies 2sin(\theta )cos(\theta )\implies 2\left(\frac{5}{13} \right)\left( \frac{-12}{13} \right)\implies -\cfrac{120}{169}[/tex]
