3)
we know that θ is the IV quadrant, well, there the sine(y) is negative and the cosine(x) is positive, we have a tangent(sine/cosine) of -1/2, so one of those is the negative, in the IV Quadrant it has to be the sine, so the 1 is negative, so the fraction is (-1)/2.
[tex] \bf tan(\theta )=\cfrac{\stackrel{opposite}{-1}}{\stackrel{adjacent}{2}}\impliedby \textit{let's find the \underline{hypotenuse}}
\\\\\\
\textit{using the pythagorean theorem}
\\\\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}
\qquad
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases} [/tex]
[tex] \bf c=\sqrt{2^2+(-1)^2}\implies c=\sqrt{5}
\\\\\\
\stackrel{\textit{double-angle identities}}{cos(2\theta )=2cos^2(\theta )-1}\implies 2[cos(\theta )]^2-1\implies 2\left[ \cfrac{\stackrel{adjacent}{2}}{\stackrel{hypotenuse}{\sqrt{5}}} \right]^2-1
\\\\\\
2\cdot \cfrac{2^2}{(\sqrt{5})^2}-1\implies 2\cdot \cfrac{4}{5}-1\implies \cfrac{8}{5}-1\implies \cfrac{3}{5} [/tex]
4)
recall that the hypotenuse is just a length unit, and is never negative, therefore on that sine ratio of opposite/hypotenuse, the negative is the 4, so the fraction is really (-4)/5,
also let's recall that the cosine(x) is also negative in the III Quadrant.
[tex] \bf sin(\alpha )=\cfrac{\stackrel{opposite}{-4}}{\stackrel{hypotenuse}{5}}\impliedby \textit{let's find the \underline{adjacent}}
\\\\\\
\textit{using the pythagorean theorem}
\\\\
c^2=a^2+b^2\implies \sqrt{c^2-b^2}=a
\qquad
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases} [/tex]
[tex] \bf \sqrt{5^2-(-4)^2}=a\implies \sqrt{25-16}=a\implies \pm 3=a\implies \stackrel{III~quadrant}{-3=a}
\\\\\\
\stackrel{\textit{double-angle identities}}{sin(2\alpha)= 2sin(\alpha)cos(\alpha)}\implies 2\left( \cfrac{-4}{5} \right)\left( \cfrac{-3}{5} \right)
\implies
\cfrac{24}{25} [/tex]
