Question

Find the exact value of tan (arcsin (two fifths)). For full credit, explain your reasoning.

Answer 1

$\bf sin^{-1}(some\ value)=\theta \impliedby \textit{this simply means} \\\\\\ sin(\theta )=some\ value\qquad \textit{now, also bear in mind that} \\\\\\ sin(\theta)=\cfrac{opposite}{hypotenuse}\qquad \qquad % tangent tan(\theta)=\cfrac{opposite}{adjacent}\\\\ -------------------------------$

$\bf sin^{-1}\left( \frac{2}{5} \right)=\theta \impliedby \textit{this simply means that} \\\\\\ sin(\theta )=\cfrac{2}{5}\cfrac{\leftarrow opposite}{\leftarrow hypotenuse}\qquad \textit{now let's find the adjacent side} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}$

$\bf \pm \sqrt{5^2-2^2}=a\implies \pm\sqrt{21}=a \\\\\\ \textit{we don't know if it's +/-, so we'll assume is the + one}\quad \sqrt{21}=a\\\\ -------------------------------\\\\ tan(\theta)=\cfrac{opposite}{adjacent}\qquad \qquad tan(\theta)=\cfrac{2}{\sqrt{21}} \\\\\\ \textit{and now, let's rationalize the denominator} \\\\\\ \cfrac{2}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies \cfrac{2\sqrt{21}}{(\sqrt{21})^2}\implies \cfrac{2\sqrt{21}}{21}$