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# Help?

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

Guest Apr 14, 2017

#1
+7056
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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
Sort:

#1
+7056
+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