Question

If cosθ = $\frac{2}{\sqrt{19} }$ and angleθ is in Quadrant I, what is the exact value of tan2θ in simplest radical form???

Answer 1

we know that θ is in the I Quadrant, namely that cosine as well as sine are both positive, well, we know that

$cos(\theta )=\cfrac{\stackrel{adjacent}{2}}{\underset{hypotenuse}{\sqrt{19}}}\qquad\qquad \textit{let's get the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{(\sqrt{19})^2-2^2}=b\implies \pm\sqrt{19-4}=b\implies \pm\sqrt{15}=b\implies \stackrel{I~Quadrant}{+\sqrt{15}=b} \\\\[-0.35em] ~\dotfill$

$tan(\theta)=\cfrac{\stackrel{opposite}{\sqrt{15}}}{\underset{adjacent}{2}}~\hspace{10em} tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)} \\\\[-0.35em] ~\dotfill\\\\ tan(2\theta)=\cfrac{2\cdot \frac{\sqrt{15}}{2}}{1-\left( \frac{\sqrt{15}}{2} \right)^2}\implies tan(2\theta)=\cfrac{\sqrt{15}}{1-\frac{15}{4}}\implies tan(2\theta)=\cfrac{\sqrt{15}}{~~\frac{4-15}{4}~~} \\\\\\ tan(2\theta)=\cfrac{\sqrt{15}}{~~ \frac{-11}{4}~~}\implies tan(2\theta)=-\cfrac{4\sqrt{15}}{11}$