Question

Can you show the work as well?

Answer 1

let's keep in mind that the sine is positive on the II Quadrant whilst cosine is negative on the II Quadrant.

`\sin( \theta )=\cfrac{\stackrel{opposite}{√(15)}}{\underset{hypotenuse}{4}} \hspace{5em}\textit{let's find the \underline{adjacent side}} \\\\\\ \begin{array}{llll} \textit{using the pythagorean theorem} \\\\ a^2+o^2=c^2\implies a=√(c^2 - o^2) \end{array} \qquad \begin{cases} c=\stackrel{hypotenuse}{4}\\ a=adjacent\\ o=\stackrel{opposite}{√(15)} \end{cases}`

`a=\pm\sqrt{ 4^2 - (√(15))^2}\implies a=\pm√( 16 - 15 ) \implies a=\pm 1\implies \stackrel{\textit{II Quadrant} }{a=-1} \\\\\\ ~\hfill~\cos( \theta )=\cfrac{\stackrel{adjacent}{-1}}{\underset{hypotenuse}{4}}~\hfill~`