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

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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

#1
+7339
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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
+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