Answer:
sinΘ =
![(2)/(3)](https://img.qammunity.org/2023/formulas/mathematics/high-school/zeht2hdwlb899wfchg7j67j4mj43o3xc1c.png)
Explanation:
using the identity
sin²x + cos²x = 1 ( subtract cos²x from both sides )
sin²x = 1 - cos²x ( take square root of both sides )
sinx = ±
![√(1-cos^2x)](https://img.qammunity.org/2023/formulas/mathematics/college/74lgsgr0w0obi7olvt5xo5be1vo7wavow8.png)
given
cosΘ = -
, then
sinΘ = ±
![\sqrt{1-(-(√(5) )/(3))^2 }](https://img.qammunity.org/2023/formulas/mathematics/college/80yy72rqh2czbk5qil7v58w6erahaip413.png)
= ±
![\sqrt{1-(5)/(9) }](https://img.qammunity.org/2023/formulas/mathematics/college/y8die5pivbdmakxd36q1vwn7snljq5hi6h.png)
= ±
![\sqrt{(4)/(9) }](https://img.qammunity.org/2023/formulas/mathematics/college/g6n3q0aw16d77juvhlaspyo3ynq2up4aj8.png)
= ±
![(2)/(3)](https://img.qammunity.org/2023/formulas/mathematics/high-school/zeht2hdwlb899wfchg7j67j4mj43o3xc1c.png)
since Θ is in quadrant II where sinΘ > 0 , then
sinΘ =
![(2)/(3)](https://img.qammunity.org/2023/formulas/mathematics/high-school/zeht2hdwlb899wfchg7j67j4mj43o3xc1c.png)