$\bf sin(\theta )=\cfrac{\stackrel{opposite}{2}}{\stackrel{hypotenuse}{5}}\impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-b^2}=a \qquad \begin{cases} c=\stackrel{hypotenuse}{5}\\ a=adjacent\\ b=\stackrel{opposite}{2}\\ \end{cases} \\\\\\ \pm\sqrt{5^2-2^2}=a\implies \pm\sqrt{21}=a\implies \stackrel{I~Quadrant}{+\sqrt{21}=a}$

recall that cosine is positive on the I Quadrant, so though we get a ± valid roots, only the positive one applies.

$\bf cos(\theta)=\cfrac{\stackrel{adjacent}{\sqrt{21}}}{\stackrel{hypotenuse}{5}} \\\\\\ tan(\theta)=\cfrac{\stackrel{opposite}{2}}{\stackrel{adjacent}{\sqrt{21}}} \qquad \qquad cot(\theta)=\cfrac{\stackrel{adjacent}{\sqrt{21}}}{\stackrel{opposite}{2}} \\\\\\ csc(\theta)=\cfrac{\stackrel{hypotenuse}{5}}{\stackrel{opposite}{2}} \qquad \qquad sec(\theta)=\cfrac{\stackrel{hypotenuse}{5}}{\stackrel{adjacent}{\sqrt{21}}}$

now, for tangent and secant, let's rationalize the denominator.

$\bf tan(\theta)\implies \cfrac{\stackrel{opposite}{2}}{\stackrel{adjacent}{\sqrt{21}}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies \cfrac{2\sqrt{21}}{21} \\\\\\ sec(\theta)\implies \cfrac{\stackrel{hypotenuse}{5}}{\stackrel{adjacent}{\sqrt{21}}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies \cfrac{5\sqrt{21}}{21}$

5 0

3 years ago