Question

Find sin θ if cos θ = 4/7 and θ in quadrant IV. Give an exact answer.

Answer 1

$\bf sin(\theta)=\cfrac{opposite}{hypotenuse} \qquad \qquad % cosine cos(\theta)=\cfrac{adjacent}{hypotenuse}\\\\ -------------------------------\\\\ cos(\theta)=\cfrac{4}{7}\cfrac{\leftarrow adjacent=a}{\leftarrow hypotenuse=c} \\\\\\ \textit{using the pythagorean theorem to get the opposite side or "b"} \\\\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\implies \pm\sqrt{7^2-4^2}=b \\\\\\ \pm \sqrt{33}$

now, the square root gives us both +/- versions, which is it? well, we know the angle θ is in the IV quadrant, well, the sine is "b" is negative there, thus is the negative version then

thus   $\bf sin(\theta)=\cfrac{opposite}{hypotenuse}\implies sin(\theta)=\cfrac{-\sqrt{33}}{7}$