Question

If sin theta = 2/5 and theta is in quadrant I, determine the following.

a. cos theta
b. tan theta
c. sec theta
d. csc theta
e. cot theta

Answer 1

`\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 √(c^2-b^2)=a \qquad \begin{cases} c=\stackrel{hypotenuse}{5}\\ a=adjacent\\ b=\stackrel{opposite}{2}\\ \end{cases} \\\\\\ \pm√(5^2-2^2)=a\implies \pm√(21)=a\implies \stackrel{I~Quadrant}{+√(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}{√(21)}}{\stackrel{hypotenuse}{5}} \\\\\\ tan(\theta)=\cfrac{\stackrel{opposite}{2}}{\stackrel{adjacent}{√(21)}} \qquad \qquad cot(\theta)=\cfrac{\stackrel{adjacent}{√(21)}}{\stackrel{opposite}{2}} \\\\\\ csc(\theta)=\cfrac{\stackrel{hypotenuse}{5}}{\stackrel{opposite}{2}} \qquad \qquad sec(\theta)=\cfrac{\stackrel{hypotenuse}{5}}{\stackrel{adjacent}{√(21)}}`

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

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