Question

Given sin O = - 3/4 and angle O is in Quadrant III, what is the exact value of

cos O in simplest form? Simplify all radicals if needed.

Answer 1

well, we know angle θ is in the III Quadrant, that means the sine or opposite side and cosine or adjacent side are both negative, whilst of course the hypotenuse is just a radius amount thus is always positive.

`\sin(\theta )=\cfrac{\stackrel{opposite}{-3}}{\underset{hypotenuse}{4}}\hspace{5em}\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 ~~ \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \pm√(4^2 - (-3)^2)=a \\\\\\ \pm√(7)=a\implies \stackrel{III~Quadrant}{-√(7)=a}\hspace{5em} \boxed{\cos(\theta )=\cfrac{\stackrel{adjacent}{-√(7)}}{\underset{hypotenuse}{4}}}`