In the second quadrant, the sine function is positive while the cosine function is negative.
[tex]\Rightarrow\cos \theta<0,\sin \theta>0[/tex]Furthermore, we can use the following trigonometric identity.
[tex]\cos ^2\theta+\sin ^2\theta=1[/tex]Therefore,
[tex]\begin{gathered} \Rightarrow\sin ^2\theta=1-\cos ^2\theta \\ \Rightarrow\sin \theta=\pm\sqrt[]{1-\cos ^2\theta} \\ \Rightarrow\sin \theta=\sqrt[]{1-\cos^2\theta} \end{gathered}[/tex]Because sin(theta) has to be positive, as stated before; thus,
[tex]\begin{gathered} \Rightarrow\sin \theta=\sqrt[]{1-(-\frac{3}{7})^2}=\sqrt[]{1-\frac{9}{49}}=\sqrt[]{\frac{40}{49}}=\frac{\sqrt[]{40}}{7}=\frac{2\sqrt[]{10}}{7} \\ \Rightarrow\sin \theta=\frac{2\sqrt[]{10}}{7} \end{gathered}[/tex]Thus, the answer is sinθ=2sqrt(10)/7
