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.

$\bf sin(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{hypotenuse}{13}}\impliedby \textit{now let's find the \underline{adjacent side}}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}$

$\bf \pm\sqrt{13^2-5^2}=a\implies \pm\sqrt{144}=a\implies \pm 12 =a \implies \stackrel{II~quadrant}{-12=a}
\\\\\\
therefore \qquad cos(\theta )=\cfrac{\stackrel{adjacent}{-12}}{\stackrel{hypotenuse}{13}}
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sin(2\theta )\implies 2sin(\theta )cos(\theta )\implies 2\left(\frac{5}{13} \right)\left( \frac{-12}{13} \right)\implies -\cfrac{120}{169}$

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4 years ago