Respuesta :

Given an angle θ in standard position passing through (x, y), the following relationship stands:

[tex]\tan\theta=\frac{y}{x}[/tex]

Our angle has a terminal side that passes through (-6, -3). Since both x and y are negative, the angle lies in quadrant III where the sine and the cosine are negative.

Substituting the given values:

[tex]\tan\theta=\frac{-3}{-6}=\frac{1}{2}[/tex]

Now we use the following identity:

[tex]\sec^2\theta=1+\tan^2\theta[/tex]

Substituting:

[tex]\begin{gathered} \sec^2\theta=1+(\frac{1}{2})^2 \\ \text{ Calculating:} \\ \sec^2\theta=1+\frac{1}{4}=\frac{5}{4} \end{gathered}[/tex]

The secant is also negative in quadrant III, so:

[tex]\begin{gathered} \sec\theta=-\sqrt{\frac{5}{4}} \\ \sec\theta=-\frac{\sqrt{5}}{2} \end{gathered}[/tex]

The cosine is the reciprocal of the secant:

[tex]\cos\theta=\frac{1}{\sec\theta}=\frac{1}{(-\frac{\sqrt{5}}{2})}=-\frac{2}{\sqrt{5}}[/tex]

Rationalizing:

[tex]\boxed{\cos\theta={}-\frac{2\sqrt{5}}{5}}[/tex]

The sine can be calculated as:

[tex]\begin{gathered} \sin\theta=\tan\theta\cdot\cos\theta \\ \sin\theta=\frac{1}{2}\cdot(-\frac{2\sqrt{5}}{5}) \\ \sin\theta=-\frac{\sqrt{5}}{5} \end{gathered}[/tex][tex]\boxed{\sin\theta=-\frac{\sqrt{5}}{5}}[/tex]

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