15. $$ an heta=\frac{7}{3}, \cos heta

Question

15. $$ 	an 	heta=\frac{7}{3}, \cos 	heta<0 $$

UpStudy AI · Accepted Answer

We are given
$$
\tan \theta=\frac{7}{3},\quad \cos \theta<0.
$$
Since $$\tan\theta=\frac{\sin\theta}{\cos\theta}$$, we can write
$$
\sin\theta=\frac{7}{3}\cos\theta.
$$

**Step 1. Use the Pythagorean identity**

The identity is
$$
\sin^2\theta+\cos^2\theta=1.
$$
Substitute the expression for $$\sin\theta$$:
$$
\left(\frac{7}{3}\cos\theta\right)^2+\cos^2\theta=1.
$$
Simplify the squares:
$$
\frac{49}{9}\cos^2\theta+\cos^2\theta=1.
$$
Combine like terms:
$$
\left(\frac{49}{9}+1\right)\cos^2\theta=1 \quad \Longrightarrow \quad \frac{49+9}{9}\cos^2\theta=\frac{58}{9}\cos^2\theta=1.
$$

**Step 2. Solve for $$\cos\theta$$**

Isolate $$\cos^2\theta$$:
$$
\cos^2\theta=\frac{9}{58}.
$$
Taking the square root gives
$$
\cos\theta=\pm\frac{3}{\sqrt{58}}.
$$
Since we are given that $$\cos\theta<0$$, we choose the negative value:
$$
\cos\theta=-\frac{3}{\sqrt{58}}.
$$

**Step 3. Find $$\sin\theta$$**

Recall the relation:
$$
\sin\theta=\frac{7}{3}\cos\theta.
$$
Substitute the value found for $$\cos\theta$$:
$$
\sin\theta=\frac{7}{3}\left(-\frac{3}{\sqrt{58}}\right)=-\frac{7}{\sqrt{58}}.
$$

**Final Answer**

$$
\cos\theta=-\frac{3}{\sqrt{58}},\quad \sin\theta=-\frac{7}{\sqrt{58}}.
$$

These values are consistent with an angle in the third quadrant (where both sine and cosine are negative) and satisfy $$\tan\theta=\frac{7}{3}$$.
$$ \cos\theta = -\frac{3}{\sqrt{58}},\quad \sin\theta = -\frac{7}{\sqrt{58}} $$

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