Respuesta :
In the third quadrant, the value of the cosine function is negative and so is the sine function.
Remember the identity
[tex]\sin ^2\theta+\cos ^2\theta=1[/tex]Then,
[tex]\begin{gathered} \sin ^2\theta=1-\cos ^2\theta=1-(-\frac{4}{7})^2=1-\frac{16}{49}=\frac{33}{49} \\ \Rightarrow\sin ^2\theta=\frac{33}{49} \\ \Rightarrow\sin \theta=\pm\sqrt[]{\frac{33}{49}}=\pm\frac{\sqrt[]{33}}{7} \end{gathered}[/tex]But remember that we are in the third quadrant; therefore, the value of sine has to be negative
[tex]\Rightarrow\sin \theta=-\frac{\sqrt[]{33}}{7}[/tex]The answer is -sqrt(33)/7
