Question

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

`tan(\theta)=\cfrac{\stackrel{opposite}{√(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 (√(15))/(2)}{1-\left( (√(15))/(2) \right)^2}\implies tan(2\theta)=\cfrac{√(15)}{1-(15)/(4)}\implies tan(2\theta)=\cfrac{√(15)}{~~(4-15)/(4)~~} \\\\\\ tan(2\theta)=\cfrac{√(15)}{~~ (-11)/(4)~~}\implies tan(2\theta)=-\cfrac{4√(15)}{11}`