+9466 If cosθ = -2/5 and sin θ > 0, this is the triangle ∠θ makes:
Use the Pythagorean theorem to find sin θ .
(-2/5)2 + sin2 θ = 12
sin2 θ = 1 - (-2/5)2
Take the + square root of both sides.
sin θ = \(+\sqrt{1-(-2/5)^2}=\sqrt{1-\frac4{25}}=\sqrt{\frac{21}{25}}= \frac{\sqrt{21}}{5}\)
+9466 If cosθ = -2/5 and sin θ > 0, this is the triangle ∠θ makes:
Use the Pythagorean theorem to find sin θ .
(-2/5)2 + sin2 θ = 12
sin2 θ = 1 - (-2/5)2
Take the + square root of both sides.
sin θ = \(+\sqrt{1-(-2/5)^2}=\sqrt{1-\frac4{25}}=\sqrt{\frac{21}{25}}= \frac{\sqrt{21}}{5}\)
