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What is the value of sin θ, given that cosθ=-2/5 and sin θ>0

Guest Apr 14, 2017

Best Answer 

 #1
avatar+7266 
+2

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 =  \(+\sqrt{1-(-2/5)^2}=\sqrt{1-\frac4{25}}=\sqrt{\frac{21}{25}}= \frac{\sqrt{21}}{5}\)

hectictar  Apr 14, 2017
 #1
avatar+7266 
+2
Best Answer

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 =  \(+\sqrt{1-(-2/5)^2}=\sqrt{1-\frac4{25}}=\sqrt{\frac{21}{25}}= \frac{\sqrt{21}}{5}\)

hectictar  Apr 14, 2017

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