Question

an equation shown below for the terminal side of an angle θ in standard position is given with a restriction on x. sketch the least positive such angle θ, and find the values of the six trigonometric functions of θ. 4x + 8y = 0, x ≤ 0. choose the graph of the given equation with the least positive angle θ.

Explanation:

Step1: Rewrite the equation

Rewrite $4x + 8y=0$ as $y=-\frac{1}{2}x$. Since $x\leq0$, the terminal - side of the angle lies in the second quadrant. The slope of the line is $m =-\frac{1}{2}=\tan\theta$.

Step2: Find a point on the line

Let $x=-2$, then $y = 1$. So, a point on the terminal - side of the angle $\theta$ is $(-2,1)$.

Step3: Calculate $r$

Using the formula $r=\sqrt{x^{2}+y^{2}}$, with $x=-2$ and $y = 1$, we have $r=\sqrt{(-2)^{2}+1^{2}}=\sqrt{4 + 1}=\sqrt{5}$.

Step4: Calculate the six trigonometric functions

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

To sketch the angle: The line $y =-\frac{1}{2}x$ with $x\leq0$ starts from the origin and goes through the point $(-2,1)$. The least - positive angle $\theta$ in standard position has its terminal side in the second quadrant.

Since we don't have the actual visual content of the graphs A, B, and C to choose from in a non - digital way, we can analyze the properties of the line $y =-\frac{1}{2}x$ with $x\leq0$. The line has a negative slope and for the least positive angle, the terminal side is in the second quadrant.

Answer:

$\sin\theta=\frac{\sqrt{5}}{5},\cos\theta=-\frac{2\sqrt{5}}{5},\tan\theta=-\frac{1}{2},\csc\theta=\sqrt{5},\sec\theta=-\frac{\sqrt{5}}{2},\cot\theta=-2$