let's first off notice something, the hypotenuse is never negative, since it's just a radius unit, so if the cosine is -(4/7), the hypotenuse of 7, is not the negative one, is the 4 above, so is really (-4)/7 in the fraction.
[tex]\bf cos(\theta )=\cfrac{\stackrel{adjacent}{-4}}{\stackrel{hypotenuse}{7}}~\hspace{5em}\textit{let's find they \underline{opposite side}}
\\\\\\
\textit{using the pythagorean theorem}
\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b
\qquad
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}[/tex]
[tex]\bf \pm\sqrt{7^2-(-4)^2}=b\implies \pm\sqrt{49-16}=b\implies \boxed{\pm\sqrt{33}=b}
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~\dotfill\\\\
sin(\theta )=\cfrac{\stackrel{opposite}{\pm \sqrt{33}}}{\stackrel{hypotenuse}{7}}~\hspace{7em}tan(\theta )=\cfrac{\stackrel{opposite}{\pm \sqrt{33}}}{\stackrel{adjacent}{-4}}\implies tan(\theta )=\cfrac{\stackrel{opposite}{\mp \sqrt{33}}}{\stackrel{adjacent}{4}}[/tex]
