Answer:
[tex]sin\theta_1 = - \frac{2\sqrt{70}}{19}[/tex]
Step-by-step explanation:
We are given that [tex]\theta_1[/tex] is in fourth quadrant.
[tex]cos\theta_1[/tex] is always positive in 4th quadrant and
[tex]sin\theta_1[/tex] is always negative in 4th quadrant.
Also, we know the following identity about [tex]sin\theta[/tex] and [tex]cos\theta[/tex]:
[tex]sin^2\theta + cos^2\theta = 1[/tex]
Using \theta_1 in place of \theta:
[tex]sin^2\theta_1 + cos^2\theta_1 = 1[/tex]
We are given that [tex]cos\theta_1 = \frac{9}{19}[/tex]
[tex]\Rightarrow sin^2\theta_1 + \dfrac{9^2}{19^2} = 1\\\Rightarrow sin^2\theta_1 = 1 - \dfrac{81}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{361-81}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{280}{361}\\\Rightarrow sin\theta_1 = \sqrt{\dfrac{280}{361}}\\\Rightarrow sin\theta_1 = +\dfrac{2\sqrt{70}}{19}, -\dfrac{2\sqrt{70}}{19}[/tex]
[tex]\theta_1[/tex] is in 4th quadrant so [tex]sin\theta_1[/tex] is negative.
So, value of [tex]sin\theta_1 = - \frac{2\sqrt{70}}{19}[/tex]
