Respuesta :
let's recall that the angle θ is in the IV Quadrant, meaning the x/adjacent is positive whilst the y/opposite is negative, thus
[tex]\bf tan(\theta )=\cfrac{\stackrel{opposite}{-3}}{\stackrel{adjacent}{4}}\qquad \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} \\\\\\ c = \sqrt{4^2+(-3)^2}\implies c = \sqrt{16+9}\implies c = \sqrt{25}\implies c = 5 \\\\[-0.35em] ~\dotfill[/tex]
[tex]\bf tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\implies tan(2\theta ) = \cfrac{2\cdot \frac{-3}{4}}{1-\left( \frac{-3}{4} \right)^2}\implies tan(2\theta)=\cfrac{-\frac{3}{2}}{1-\frac{9}{16}} \\\\\\ tan(2\theta)=\cfrac{-\frac{3}{2}}{\frac{7}{16}}\implies tan(2\theta)=-\cfrac{3}{~~\begin{matrix} 2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\cdot \cfrac{\stackrel{8}{~~\begin{matrix} 16 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}{7}\implies tan(2\theta)=-\cfrac{24}{7}[/tex]
[tex]\bf cos(2\theta )=2cos^2(\theta )-1\implies cos(2\theta )=2\left[ \cfrac{\stackrel{adjacent}{4}}{\stackrel{hypotenuse}{5}} \right]^2 - 1 \\\\\\ cos(2\theta )=2\left( \cfrac{16}{25} \right) - 1\implies cos(2\theta )=\cfrac{32}{25}-1\implies cos(2\theta )=\cfrac{7}{25}[/tex]
