Answer :
To find [tex]\(\csc(\theta)\)[/tex] given that [tex]\(\cos(\theta) = -\frac{13}{15}\)[/tex] and the terminal side of [tex]\(\theta\)[/tex] lies in quadrant III, we will use the Pythagorean identity. Let's go through the steps in detail.
1. Identify the known cosine value and quadrant:
- Given: [tex]\(\cos(\theta) = -\frac{13}{15}\)[/tex]
- The terminal side of [tex]\(\theta\)[/tex] is in quadrant III.
2. Pythagorean identity:
- We know that [tex]\(\cos^2(\theta) + \sin^2(\theta) = 1\)[/tex].
3. Calculate [tex]\(\sin^2(\theta)\)[/tex]:
- [tex]\(\cos(\theta) = -\frac{13}{15}\)[/tex], so [tex]\(\cos^2(\theta) = \left(-\frac{13}{15}\right)^2 = \frac{169}{225}\)[/tex].
- Using the Pythagorean identity:
[tex]\[ \sin^2(\theta) = 1 - \cos^2(\theta) = 1 - \frac{169}{225} = \frac{225}{225} - \frac{169}{225} = \frac{56}{225}. \][/tex]
4. Determine [tex]\(\sin(\theta)\)[/tex]:
- Since the terminal side of [tex]\(\theta\)[/tex] lies in quadrant III, both sine and cosine are negative in this quadrant.
- Thus, [tex]\(\sin(\theta) = -\sqrt{\sin^2(\theta)} = -\sqrt{\frac{56}{225}} = -\frac{\sqrt{56}}{15}\)[/tex].
- Simplifying the square root, we get [tex]\(\sqrt{56} = \sqrt{4 \cdot 14} = 2\sqrt{14}\)[/tex], so [tex]\(\sin(\theta) = -\frac{2\sqrt{14}}{15}\)[/tex].
5. Calculate [tex]\(\csc(\theta)\)[/tex]:
- [tex]\(\csc(\theta) = \frac{1}{\sin(\theta)} = \frac{1}{-\frac{2\sqrt{14}}{15}} = -\frac{15}{2\sqrt{14}}\)[/tex].
- Rationalizing the denominator, we multiply the numerator and the denominator by [tex]\(\sqrt{14}\)[/tex]:
[tex]\[ \csc(\theta) = -\frac{15 \cdot \sqrt{14}}{2 \cdot 14} = -\frac{15 \sqrt{14}}{28}. \][/tex]
So, [tex]\(\csc(\theta) = -\frac{15 \sqrt{14}}{28}\)[/tex].
1. Identify the known cosine value and quadrant:
- Given: [tex]\(\cos(\theta) = -\frac{13}{15}\)[/tex]
- The terminal side of [tex]\(\theta\)[/tex] is in quadrant III.
2. Pythagorean identity:
- We know that [tex]\(\cos^2(\theta) + \sin^2(\theta) = 1\)[/tex].
3. Calculate [tex]\(\sin^2(\theta)\)[/tex]:
- [tex]\(\cos(\theta) = -\frac{13}{15}\)[/tex], so [tex]\(\cos^2(\theta) = \left(-\frac{13}{15}\right)^2 = \frac{169}{225}\)[/tex].
- Using the Pythagorean identity:
[tex]\[ \sin^2(\theta) = 1 - \cos^2(\theta) = 1 - \frac{169}{225} = \frac{225}{225} - \frac{169}{225} = \frac{56}{225}. \][/tex]
4. Determine [tex]\(\sin(\theta)\)[/tex]:
- Since the terminal side of [tex]\(\theta\)[/tex] lies in quadrant III, both sine and cosine are negative in this quadrant.
- Thus, [tex]\(\sin(\theta) = -\sqrt{\sin^2(\theta)} = -\sqrt{\frac{56}{225}} = -\frac{\sqrt{56}}{15}\)[/tex].
- Simplifying the square root, we get [tex]\(\sqrt{56} = \sqrt{4 \cdot 14} = 2\sqrt{14}\)[/tex], so [tex]\(\sin(\theta) = -\frac{2\sqrt{14}}{15}\)[/tex].
5. Calculate [tex]\(\csc(\theta)\)[/tex]:
- [tex]\(\csc(\theta) = \frac{1}{\sin(\theta)} = \frac{1}{-\frac{2\sqrt{14}}{15}} = -\frac{15}{2\sqrt{14}}\)[/tex].
- Rationalizing the denominator, we multiply the numerator and the denominator by [tex]\(\sqrt{14}\)[/tex]:
[tex]\[ \csc(\theta) = -\frac{15 \cdot \sqrt{14}}{2 \cdot 14} = -\frac{15 \sqrt{14}}{28}. \][/tex]
So, [tex]\(\csc(\theta) = -\frac{15 \sqrt{14}}{28}\)[/tex].
