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.
-5x - 4y = 0, x ≤ 0
choose the graph of the given equation with the least positive angle θ.
select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. sin θ =
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)
b. the function is undefined.
select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. cos θ =

Explanation:

Step1: Rewrite the line - equation

Rewrite $-5x - 4y=0$ as $y =-\frac{5}{4}x$. Since $x\leq0$, the terminal - side of the angle $\theta$ is in the second quadrant.

Step2: Find a point on the terminal - side

Let $x=-4$, then $y = 5$ (by substituting $x=-4$ into $y =-\frac{5}{4}x$). The point $(-4,5)$ lies on the terminal - side of the angle $\theta$.

Step3: Calculate the distance $r$ from the origin to the point

Use the formula $r=\sqrt{x^{2}+y^{2}}$. Substitute $x=-4$ and $y = 5$ into it, we get $r=\sqrt{(-4)^{2}+5^{2}}=\sqrt{16 + 25}=\sqrt{41}$.

Step4: Calculate the six trigonometric functions

• $\sin\theta=\frac{y}{r}=\frac{5}{\sqrt{41}}=\frac{5\sqrt{41}}{41}$
• $\cos\theta=\frac{x}{r}=\frac{-4}{\sqrt{41}}=-\frac{4\sqrt{41}}{41}$
• $\tan\theta=\frac{y}{x}=\frac{5}{-4}=-\frac{5}{4}$
• $\csc\theta=\frac{r}{y}=\frac{\sqrt{41}}{5}$
• $\sec\theta=\frac{r}{x}=-\frac{\sqrt{41}}{4}$
• $\cot\theta=\frac{x}{y}=-\frac{4}{5}$

For the graph: The line $y =-\frac{5}{4}x$ with $x\leq0$ has a negative slope and lies in the second quadrant. The correct graph is the one where the line passes through the origin and has a negative - sloped part in the second quadrant.

Answer:

For the graph: (Assuming the correct graph among A, B, C is the one with a negative - sloped line passing through the origin and having the part for $x\leq0$ in the second quadrant)
For $\sin\theta$: A. $\sin\theta=\frac{5\sqrt{41}}{41}$
For $\cos\theta$: A. $\cos\theta=-\frac{4\sqrt{41}}{41}$