Question

5 0 = - in = in quadrant II Given cos = Find sin 3

Answer 1

Answer:

sinΘ = $\frac{2}{3}$

Step-by-step explanation:

using the identity

sin²x + cos²x = 1  ( subtract cos²x from both sides )

sin²x = 1 - cos²x ( take square root of both sides )

sinx = ± $\sqrt{1-cos^2x}$

given

cosΘ = - $\frac{\sqrt{5} }{3}$ , then

sinΘ = ± $\sqrt{1-(-\frac{\sqrt{5} }{3})^2 }$

        = ± $\sqrt{1-\frac{5}{9} }$

        = ± $\sqrt{\frac{4}{9} }$

        = ± $\frac{2}{3}$

since Θ is in quadrant II where sinΘ > 0 , then

sinΘ = $\frac{2}{3}$