Question

If sinθ = -1/2 and θ is in Quadrant III, then tanθ

Answer 1

let's recall that on the III Quadrant sine/y is negative and cosine/y is negative, now, the hypotenuse/radius is never negative, since it's just a radius unit.

`\bf sin(\theta )=\cfrac{\stackrel{opposite}{-1}}{\stackrel{hypotenuse}{2}}\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=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm√(2^2-(-1)^2)=a\implies \pm√(4-1)=a\implies \pm√(3)=a\implies \stackrel{\textit{III Quadrant}}{-√(3)=a} \\\\[-0.35em] ~\dotfill`

`\bf tan(\theta )=\cfrac{\stackrel{opposite}{-1}}{\stackrel{adjacent}{-√(3)}}\implies \stackrel{\textit{rationalizing the denominator}}{tan(\theta )=\cfrac{-1}{-√(3)}\cdot \cfrac{√(3)}{√(3)} }\implies tan(\theta )=\cfrac{√(3)}{3}`