Question 6.
Using the pythagorean theorem to find the value of r:
[tex]r=\sqrt{x^2+y^2}=\sqrt{3^2+1^2}=\sqrt{9+1}=\sqrt{10}[/tex]
Next, using:
x = 3
y = 1
The six trigonometric functions are given by:
[tex]\begin{gathered} \sin\theta=\frac{y}{r}=\frac{1}{\sqrt{10}} \\ \tan\theta=\frac{y}{x}=\frac{1}{3} \\ sec\theta=\frac{r}{x}=\frac{\sqrt{10}}{3} \\ \cos\theta=\frac{x}{r}=\frac{3}{\sqrt{10}} \\ csc\theta=\frac{r}{y}=\frac{\sqrt{10}}{1} \\ cot\theta=\frac{x}{y}=\frac{3}{1}=3 \end{gathered}[/tex]
