MATH SOLVE

4 months ago

Q:
# Angle θ is in standard position. If (8, -15) is on the terminal ray of angle θ, find the values of the trigonometric functions.

Accepted Solution

A:

check the picture below.

[tex]\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=\stackrel{adjacent}{8}\\ b=\stackrel{opposite}{-15}\\ \end{cases} \\\\\\ c=\sqrt{8^2+(-15)^2}\implies c=\sqrt{289}\implies c=17[/tex]

[tex]\bf sin(\theta)=\cfrac{\stackrel{opposite}{-15}}{\stackrel{hypotenuse}{17}} \qquad\qquad cos(\theta)=\cfrac{\stackrel{adjacent}{8}}{\stackrel{hypotenuse}{17}} \\\\\\ tan(\theta)=\cfrac{\stackrel{opposite}{-15}}{\stackrel{adjacent}{8}} \qquad \qquad cot(\theta)=\cfrac{\stackrel{adjacent}{8}}{\stackrel{opposite}{-15}} \\\\\\ csc(\theta)=\cfrac{\stackrel{hypotenuse}{17}}{\stackrel{opposite}{-15}} \qquad \qquad sec(\theta)=\cfrac{\stackrel{hypotenuse}{17}}{\stackrel{adjacent}{8}}[/tex]

[tex]\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=\stackrel{adjacent}{8}\\ b=\stackrel{opposite}{-15}\\ \end{cases} \\\\\\ c=\sqrt{8^2+(-15)^2}\implies c=\sqrt{289}\implies c=17[/tex]

[tex]\bf sin(\theta)=\cfrac{\stackrel{opposite}{-15}}{\stackrel{hypotenuse}{17}} \qquad\qquad cos(\theta)=\cfrac{\stackrel{adjacent}{8}}{\stackrel{hypotenuse}{17}} \\\\\\ tan(\theta)=\cfrac{\stackrel{opposite}{-15}}{\stackrel{adjacent}{8}} \qquad \qquad cot(\theta)=\cfrac{\stackrel{adjacent}{8}}{\stackrel{opposite}{-15}} \\\\\\ csc(\theta)=\cfrac{\stackrel{hypotenuse}{17}}{\stackrel{opposite}{-15}} \qquad \qquad sec(\theta)=\cfrac{\stackrel{hypotenuse}{17}}{\stackrel{adjacent}{8}}[/tex]