Answer:
[tex]sin\theta_1 = -\frac{\sqrt{731}}{30}[/tex] is the correct answer.
Step-by-step explanation:
It is given that [tex]\theta_{1}[/tex] is in third quadrant.
[tex]cos\theta_{1}[/tex] is always negative in 3rd quadrant and also
[tex]sin\theta_{1}[/tex] is always negative in 3rd quadrant.
Also, we know the following identity about and :
[tex]sin^2\theta + cos^2\theta = 1[/tex]
Put [tex]\theta[/tex] as [tex]\theta_{1}[/tex]:
[tex]sin^2\theta_1 + cos^2\theta_1 = 1[/tex]
We are given that [tex]cos\theta_1 = -\frac{13}{30}[/tex]
[tex]sin^2\theta_1 + (\frac{-13}{30})^2 = 1\\[/tex]
[tex]\Rightarrow sin^2\theta_1 = 1 - \dfrac{169}{900}\\\Rightarrow sin^2\theta_1 = \dfrac{900-169}{900}\\\Rightarrow sin^2\theta_1 = \dfrac{731}{900}\\\Rightarrow sin\theta_1 = +\sqrt{\dfrac{731}{900}}, -\sqrt{\dfrac{731}{900}}\\\Rightarrow sin\theta_1 = +\dfrac{\sqrt{731}}{30}, -\dfrac{\sqrt{731}}{30}[/tex]
[tex]\theta_{1}[/tex] is in 3rd quadrant so [tex]sin\theta_{1}[/tex] is negative.
So [tex]sin\theta_1 = -\frac{\sqrt{731}}{30}[/tex]
