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if cos(t) = 1/3 and sin(t) < 0 find sin(t) and tan(t) using trigonometric identities.

Guest Apr 12, 2017

Best Answer 

 #1
avatar+7339 
+2

If cos(t) = 1/3 and sin(t) is negative,

this is a picture of the triangle ∠t makes:

 

 

Now we can use the Pythagorean theorem to find sin(t).

 

sin2(t) + (1/3)2 = 12

sin2(t) = 1 - 1/9

sin(t) = \(\pm\sqrt{\frac89}=\pm\frac{\sqrt8}{3}\)

The prolem says that sin < 0 , so we know that it is the negative option.

sin(t) = \(-\frac{2\sqrt2}{3}\)

 

tan = sin / cos

tan(t) = sin(t) / cos(t)

tan(t) = \(-\frac{2\sqrt2}{3}/\frac13\)

tan(t) = \(-\frac{2\sqrt2}{3}\cdot\frac31\)

tan(t) = \(-2\sqrt2\)

hectictar  Apr 12, 2017
 #1
avatar+7339 
+2
Best Answer

If cos(t) = 1/3 and sin(t) is negative,

this is a picture of the triangle ∠t makes:

 

 

Now we can use the Pythagorean theorem to find sin(t).

 

sin2(t) + (1/3)2 = 12

sin2(t) = 1 - 1/9

sin(t) = \(\pm\sqrt{\frac89}=\pm\frac{\sqrt8}{3}\)

The prolem says that sin < 0 , so we know that it is the negative option.

sin(t) = \(-\frac{2\sqrt2}{3}\)

 

tan = sin / cos

tan(t) = sin(t) / cos(t)

tan(t) = \(-\frac{2\sqrt2}{3}/\frac13\)

tan(t) = \(-\frac{2\sqrt2}{3}\cdot\frac31\)

tan(t) = \(-2\sqrt2\)

hectictar  Apr 12, 2017

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