We have a terminal point (-3,-5).
We have to find cos(θ), csc (θ and tan)(θ.)
We can locate the terminal point and the angle as:
The cosine of this angle will be negative, as ( is located in the third quadrant)
The hypotenuse of this right triangle will be called R and we can calculate it using the Pythagorean theorem:
[tex]\begin{gathered} R^2=x^2+y^2 \\ R=\sqrt[]{x^2+y^2} \end{gathered}[/tex]
We can estimate the cosine as:
[tex]\begin{gathered} \cos (\theta)=\frac{x}{R} \\ \cos (\theta)=\frac{-3}{\sqrt[]{(-3)^2+(-5)^2}} \\ \cos (\theta)=\frac{-3}{\sqrt[]{9+25}} \\ \cos (\theta)=\frac{-3}{\sqrt[]{34}} \\ \cos (\theta)=\frac{-3\sqrt[]{34}}{34} \end{gathered}[/tex]
We can now relate this to the csc(θ) as:
[tex]\begin{gathered} \csc (\theta)=\frac{1}{\sin (\theta)} \\ \csc (\theta)=\frac{1}{\frac{y}{R}} \\ \csc (\theta)=\frac{R}{y} \\ \csc (\theta)=\frac{\sqrt[]{34}}{-5} \\ \csc (\theta)=-\frac{\sqrt[]{34}}{5} \end{gathered}[/tex]
Finally, we can calculate the tangent as:
[tex]\begin{gathered} \tan (\theta)=\frac{y}{x} \\ \tan (\theta)=\frac{-5}{-3} \\ \tan (\theta)=\frac{5}{3} \end{gathered}[/tex]
Answer:
cos(θ) = -3√34/34
csc(θ) = -√34/5
tan(θ) = 5/3
