Question

Please Can anyone help me with Calculate the six trigonometric function values of an angle θ in standard position whose terminal side goes through the point (−4,3). Make sure your fractions are fully reduced.

Answer 1

Check the picture below.

so we know from the point the adjacent and opposites sides, let's find the hypotenuse.

`\begin{array}{llll} \textit{using the pythagorean theorem} \\\\ c^2=a^2+o^2\implies c=√(a^2 + o^2) \end{array} \qquad \begin{cases} c=hypotenuse\\ a=\stackrel{adjacent}{-4}\\ o=\stackrel{opposite}{3} \end{cases} \\\\\\ c=√( -4^2 + 3^2)\implies c=√( 16 + 9 ) \implies c=√( 25 )\implies c=5 \\\\[-0.35em] ~\dotfill`

`\sin(\theta )=\cfrac{\stackrel{opposite}{3}}{\underset{hypotenuse}{5}}~\hfill \cos(\theta )=\cfrac{\stackrel{adjacent}{-4}}{\underset{hypotenuse}{5}}~\hfill \tan(\theta )=\cfrac{\stackrel{opposite}{3}}{\underset{adjacent}{-4}} \\\\\\ \cot(\theta )=\cfrac{\stackrel{adjacent}{-4}}{\underset{opposite}{3}}~\hfill \sec(\theta )=\cfrac{\stackrel{hypotenuse}{5}}{\underset{adjacent}{-4}}~\hfill \csc(\theta )=\cfrac{\stackrel{hypotenuse}{5}}{\underset{opposite}{3}}`