Question

Find the six trigonometric function values of the angle θ in standard position, if the terminal side of θ is defined by x + 2y = 0, x ≥ 0.

Answer 1

Answer:

$\sin \theta = \frac{y}r} = \frac{-1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{\sqrt{5}}{5}\\\\\cos \theta = \frac{x}{r} = \frac{2}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{2\sqrt{5}}{5} \\\\\tan \theta = \frac{y}{x} = \frac{-1}{2} = -\frac{1}{2} \\\\\cot \theta = \frac{x}{y} = \frac{2}{-1} = -2\\\\\sec \theta = \frac{r}{x} = \frac{\sqrt{5}}{2} \\\\\csc \theta = \frac{r}{y} = \frac{\sqrt{5}}{-1} = -\sqrt{5}$

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 $x + 2y= 0, x\geq 0$

If we choose $x = 2$ (we can do so because of the condition $x \geq 0$, which means that any positive value is suitable for $x$), then we have

$2 +2y = 0\implies 2 = -2y \implies y = -1$

Therefore, the terminal side of the angle $\theta$  is passing through the origin and the point  $(2,-1)$ and now we can draw it.

The angle  $\theta$  is presented below.

The distance of the point  $(2,-1)$ from the origin equals

$r = \sqrt{2^2 + (-1)^2} = \sqrt{5}$

Now, we can determine the values of the six trigonometric function, by using their definitions.

$\sin \theta = \frac{y}r} = \frac{-1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{\sqrt{5}}{5}\\\\\cos \theta = \frac{x}{r} = \frac{2}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{2\sqrt{5}}{5} \\\\\tan \theta = \frac{y}{x} = \frac{-1}{2} = -\frac{1}{2} \\\\\cot \theta = \frac{x}{y} = \frac{2}{-1} = -2\\\\\sec \theta = \frac{r}{x} = \frac{\sqrt{5}}{2} \\\\\csc \theta = \frac{r}{y} = \frac{\sqrt{5}}{-1} = -\sqrt{5}$