f $$ heta $$ satisfying the conditions. (H) 15. $$ an heta=\frac{7}{3}, \cos heta

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f $$ 	heta $$ satisfying the conditions. (H) 15. $$ 	an 	heta=\frac{7}{3}, \cos 	heta<0 $$

UpStudy AI · Accepted Answer

Let
$$
\tan \theta = \frac{7}{3}.
$$
Since $$ \tan \theta = \frac{\sin \theta}{\cos \theta} $$ and we are given $$ \cos \theta < 0 $$, we must have both $$\sin \theta$$ and $$\cos \theta$$ negative (because $$\tan \theta$$ is positive only when $$\sin \theta$$ and $$\cos \theta$$ have the same sign). This places $$\theta$$ in the third quadrant.

1. Assume a right triangle where the "opposite" side is $$7$$ and the "adjacent" side is $$3$$ (ignoring signs for now). Compute the hypotenuse:
   $$
   r = \sqrt{3^2 + 7^2} = \sqrt{9 + 49} = \sqrt{58}.
   $$

2. Now assign the proper signs based on the quadrant. In the third quadrant:
   $$
   \sin \theta = -\frac{\text{opposite}}{r} = -\frac{7}{\sqrt{58}}, \quad \cos \theta = -\frac{\text{adjacent}}{r} = -\frac{3}{\sqrt{58}}.
   $$

Thus, the values that satisfy the given condition are:
$$
\sin \theta = -\frac{7}{\sqrt{58}}, \qquad \cos \theta = -\frac{3}{\sqrt{58}}.
$$
$$ \sin \theta = -\frac{7}{\sqrt{58}}, \quad \cos \theta = -\frac{3}{\sqrt{58}}. $$

f satisfying the conditions. (H)
15.

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f satisfying the conditions. (H) 15.