shawn4998144
Answered

Question 1 (5 points)

Suppose [tex]\theta[/tex] is an angle in the standard position whose terminal side is in Quadrant IV and [tex]\cot \theta = -\frac{2}{17}[/tex]. Find the exact values of the five remaining trigonometric functions of [tex]\theta[/tex].

A. [tex]\sin \theta = -\frac{2}{\sqrt{293}}, \cos \theta = \frac{17}{\sqrt{293}}, \csc \theta = -\frac{\sqrt{293}}{2}, \sec \theta = \frac{\sqrt{293}}{17}, \tan \theta = -\frac{17}{2}[/tex]
B. [tex]\sin \theta = -\frac{17}{\sqrt{293}}, \cos \theta = \frac{2}{\sqrt{293}}, \csc \theta = -\frac{\sqrt{293}}{17}, \sec \theta = \frac{\sqrt{293}}{2}, \tan \theta = -\frac{17}{2}[/tex]
C. [tex]\sin \theta = -\frac{17}{\sqrt{293}}, \cos \theta = \frac{2}{\sqrt{293}}, \csc \theta = -\frac{\sqrt{293}}{2}, \sec \theta = \frac{\sqrt{293}}{17}, \tan \theta = -\frac{17}{2}[/tex]
D. [tex]\sin \theta = \frac{\sqrt{293}}{17}, \cos \theta = -\frac{\sqrt{293}}{2}, \csc \theta = \frac{17}{\sqrt{293}}, \sec \theta = -\frac{2}{\sqrt{293}}, \tan \theta = -\frac{2}{17}[/tex]



Answer :

GinnyAnswer
To solve for the remaining trigonometric functions of [tex]\(\theta\)[/tex] given [tex]\(\cot \theta = -\frac{2}{17}\)[/tex] and knowing that [tex]\(\theta\)[/tex] is in Quadrant IV, we will follow these steps:

1. Understanding the Relationship:
[tex]\(\cot \theta = \frac{\cos \theta}{\sin \theta}\)[/tex]. Given that [tex]\(\cot \theta = -\frac{2}{17}\)[/tex], this implies:
[tex]\[ \frac{\cos \theta}{\sin \theta} = -\frac{2}{17} \][/tex]
Since [tex]\(\theta\)[/tex] is in Quadrant IV, [tex]\(\cos \theta\)[/tex] must be positive and [tex]\(\sin \theta\)[/tex] must be negative.

2. Expressing [tex]\(\cos \theta\)[/tex] and [tex]\(\sin \theta\)[/tex] in terms of a constant [tex]\(k\)[/tex]:
We can let [tex]\(\cos \theta = 17k\)[/tex] and [tex]\(\sin \theta = -2k\)[/tex]. Keep in mind that [tex]\(|\cot \theta|\)[/tex] ratio holds.

3. Applying the Pythagorean Identity:
The Pythagorean identity states:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
Substituting [tex]\(\cos \theta = 17k\)[/tex] and [tex]\(\sin \theta = -2k\)[/tex]:
[tex]\[ (17k)^2 + (-2k)^2 = 1 \][/tex]
Simplifying:
[tex]\[ 289k^2 + 4k^2 = 1 \][/tex]
[tex]\[ 293k^2 = 1 \][/tex]
[tex]\[ k^2 = \frac{1}{293} \][/tex]
[tex]\[ k = \sqrt{\frac{1}{293}} = \frac{1}{\sqrt{293}} \][/tex]

4. Determining the Values of [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
[tex]\[ \cos \theta = 17k = 17 \cdot \frac{1}{\sqrt{293}} = \frac{17}{\sqrt{293}} \][/tex]
[tex]\[ \sin \theta = -2k = -2 \cdot \frac{1}{\sqrt{293}} = -\frac{2}{\sqrt{293}} \][/tex]

5. Calculating the Remaining Trigonometric Functions:
- Cosecant:
[tex]\[ \csc \theta = \frac{1}{\sin \theta} = \frac{1}{-\frac{2}{\sqrt{293}}} = -\frac{\sqrt{293}}{2} \][/tex]
- Secant:
[tex]\[ \sec \theta = \frac{1}{\cos \theta} = \frac{1}{\frac{17}{\sqrt{293}}} = \frac{\sqrt{293}}{17} \][/tex]
- Tangent:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\frac{2}{\sqrt{293}}}{\frac{17}{\sqrt{293}}} = -\frac{2}{17} \][/tex]

Thus, the exact values of the five remaining trigonometric functions for the given [tex]\(\theta\)[/tex] are:

1. [tex]\(\sin \theta = -\frac{2}{\sqrt{293}} \approx -0.1168412475673972\)[/tex]
2. [tex]\(\cos \theta = \frac{17}{\sqrt{293}} \approx 0.9931506043228762\)[/tex]
3. [tex]\(\csc \theta = -\frac{\sqrt{293}}{2} \approx -8.558621384311845\)[/tex]
4. [tex]\(\sec \theta = \frac{\sqrt{293}}{17} \approx 1.0068966334484524\)[/tex]
5. [tex]\(\tan \theta = -\frac{2}{17} \approx -0.11764705882352941\)[/tex]

From the choices given in the question, the correct option matches the calculations:

[tex]\[ \text{Option 0:} \quad \sin \theta = -\frac{2}{\sqrt{293}}, \cos \theta = \frac{17}{\sqrt{293}}, \csc \theta = -\frac{\sqrt{293}}{2}, \sec \theta = \frac{\sqrt{293}}{17}, \tan \theta = -\frac{2}{17} \][/tex]

Thus, the correct answer is indeed:

[tex]\[ \boxed{0} \][/tex]