Answer:
[tex]\sin \theta = \frac{y}r} = \frac{-1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{\sqrt{5}}{5}\\\\\cos \theta = \frac{x}{r} = \frac{2}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{2\sqrt{5}}{5} \\\\\tan \theta = \frac{y}{x} = \frac{-1}{2} = -\frac{1}{2} \\\\\cot \theta = \frac{x}{y} = \frac{2}{-1} = -2\\\\\sec \theta = \frac{r}{x} = \frac{\sqrt{5}}{2} \\\\\csc \theta = \frac{r}{y} = \frac{\sqrt{5}}{-1} = -\sqrt{5}[/tex]
Step-by-step explanation:
First, we need to draw the terminal position of the given angle. To do so, we need to find a point that lies on the straight line [tex] x + 2y= 0, x\geq 0 [/tex]
If we choose [tex] x = 2 [/tex] (we can do so because of the condition [tex] x \geq 0 [/tex], which means that any positive value is suitable for [tex] x [/tex]), then we have
[tex] 2 +2y = 0\implies 2 = -2y \implies y = -1 [/tex]
Therefore, the terminal side of the angle [tex] \theta [/tex] is passing through the origin and the point [tex] (2,-1) [/tex] and now we can draw it.
The angle [tex] \theta [/tex] is presented below.
The distance of the point [tex] (2,-1) [/tex] from the origin equals
[tex]r = \sqrt{2^2 + (-1)^2} = \sqrt{5}[/tex]
Now, we can determine the values of the six trigonometric function, by using their definitions.
[tex]\sin \theta = \frac{y}r} = \frac{-1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{\sqrt{5}}{5}\\\\\cos \theta = \frac{x}{r} = \frac{2}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{2\sqrt{5}}{5} \\\\\tan \theta = \frac{y}{x} = \frac{-1}{2} = -\frac{1}{2} \\\\\cot \theta = \frac{x}{y} = \frac{2}{-1} = -2\\\\\sec \theta = \frac{r}{x} = \frac{\sqrt{5}}{2} \\\\\csc \theta = \frac{r}{y} = \frac{\sqrt{5}}{-1} = -\sqrt{5}[/tex]
