Answer :
To find \(\cos(\theta)\) and \(\csc(\theta)\) given \(\cot(\theta) = \frac{5}{9}\), where \(\theta\) lies in the interval \([\pi, \frac{3\pi}{2}]\), follow these steps:
1. Identify the quadrant:
Since \(\theta\) is in the interval \([\pi, \frac{3\pi}{2}]\), it's located in the third quadrant. In the third quadrant, both \(\sin(\theta)\) and \(\cos(\theta)\) are negative.
2. Relate \(\cot(\theta)\) to trigonometric functions:
Recall that \(\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}\). Given that \(\cot(\theta) = \frac{5}{9}\), we have:
[tex]\[ \frac{\cos(\theta)}{\sin(\theta)} = \frac{5}{9} \][/tex]
3. Use the Pythagorean identity:
Consider the identity \(\sin^2(\theta) + \cos^2(\theta) = 1\).
Let \(\sin(\theta) = -y\) and \(\cos(\theta) = -x\) (since both are negative in the third quadrant).
From the cotangent relation:
[tex]\[ \frac{\cos(\theta)}{\sin(\theta)} = \frac{5}{9} \implies \frac{-x}{-y} = \frac{5}{9} \implies \frac{x}{y} = \frac{5}{9} \][/tex]
So, \(x = \frac{5}{9} y\).
4. Substitute into the Pythagorean identity:
Substitute \(x = \frac{5}{9} y\) into the identity \(\sin^2(\theta) + \cos^2(\theta) = 1\):
[tex]\[ y^2 + \left(\frac{5}{9} y\right)^2 = 1 \][/tex]
Simplify:
[tex]\[ y^2 + \frac{25}{81} y^2 = 1 \][/tex]
[tex]\[ y^2 \left(1 + \frac{25}{81}\right) = 1 \][/tex]
[tex]\[ y^2 \left(\frac{81 + 25}{81}\right) = 1 \][/tex]
[tex]\[ y^2 \left(\frac{106}{81}\right) = 1 \][/tex]
[tex]\[ y^2 = \frac{81}{106} \][/tex]
[tex]\[ y = \sqrt{\frac{81}{106}} = \frac{9}{\sqrt{106}} \][/tex]
Given \(y\) is in the third quadrant (where sine is negative):
[tex]\[ y = -\frac{9}{\sqrt{106}} \][/tex]
So:
[tex]\[ \sin(\theta) = -\frac{9}{\sqrt{106}} \][/tex]
5. Calculate \(\cos(\theta)\):
Substitute \(y = -\frac{9}{\sqrt{106}}\) back to find \(x\):
[tex]\[ \cos(\theta) = -x = -\left(\frac{5}{9} \cdot \frac{9}{\sqrt{106}}\right) = -\frac{5}{\sqrt{106}} \][/tex]
6. Calculate \(\csc(\theta)\):
The cosecant is the reciprocal of sine:
[tex]\[ \csc(\theta) = \frac{1}{\sin(\theta)} = -\frac{\sqrt{106}}{9} \][/tex]
Therefore, the values are:
[tex]\[ \cos(\theta) \approx -0.48564 \][/tex]
[tex]\[ \csc(\theta) \approx -1.14396 \][/tex]
These values match the previously given numerical results:
[tex]\[ \cos(\theta) \approx -0.4856429311786321 \][/tex]
[tex]\[ \csc(\theta) \approx -1.1439589045541112 \][/tex]
1. Identify the quadrant:
Since \(\theta\) is in the interval \([\pi, \frac{3\pi}{2}]\), it's located in the third quadrant. In the third quadrant, both \(\sin(\theta)\) and \(\cos(\theta)\) are negative.
2. Relate \(\cot(\theta)\) to trigonometric functions:
Recall that \(\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}\). Given that \(\cot(\theta) = \frac{5}{9}\), we have:
[tex]\[ \frac{\cos(\theta)}{\sin(\theta)} = \frac{5}{9} \][/tex]
3. Use the Pythagorean identity:
Consider the identity \(\sin^2(\theta) + \cos^2(\theta) = 1\).
Let \(\sin(\theta) = -y\) and \(\cos(\theta) = -x\) (since both are negative in the third quadrant).
From the cotangent relation:
[tex]\[ \frac{\cos(\theta)}{\sin(\theta)} = \frac{5}{9} \implies \frac{-x}{-y} = \frac{5}{9} \implies \frac{x}{y} = \frac{5}{9} \][/tex]
So, \(x = \frac{5}{9} y\).
4. Substitute into the Pythagorean identity:
Substitute \(x = \frac{5}{9} y\) into the identity \(\sin^2(\theta) + \cos^2(\theta) = 1\):
[tex]\[ y^2 + \left(\frac{5}{9} y\right)^2 = 1 \][/tex]
Simplify:
[tex]\[ y^2 + \frac{25}{81} y^2 = 1 \][/tex]
[tex]\[ y^2 \left(1 + \frac{25}{81}\right) = 1 \][/tex]
[tex]\[ y^2 \left(\frac{81 + 25}{81}\right) = 1 \][/tex]
[tex]\[ y^2 \left(\frac{106}{81}\right) = 1 \][/tex]
[tex]\[ y^2 = \frac{81}{106} \][/tex]
[tex]\[ y = \sqrt{\frac{81}{106}} = \frac{9}{\sqrt{106}} \][/tex]
Given \(y\) is in the third quadrant (where sine is negative):
[tex]\[ y = -\frac{9}{\sqrt{106}} \][/tex]
So:
[tex]\[ \sin(\theta) = -\frac{9}{\sqrt{106}} \][/tex]
5. Calculate \(\cos(\theta)\):
Substitute \(y = -\frac{9}{\sqrt{106}}\) back to find \(x\):
[tex]\[ \cos(\theta) = -x = -\left(\frac{5}{9} \cdot \frac{9}{\sqrt{106}}\right) = -\frac{5}{\sqrt{106}} \][/tex]
6. Calculate \(\csc(\theta)\):
The cosecant is the reciprocal of sine:
[tex]\[ \csc(\theta) = \frac{1}{\sin(\theta)} = -\frac{\sqrt{106}}{9} \][/tex]
Therefore, the values are:
[tex]\[ \cos(\theta) \approx -0.48564 \][/tex]
[tex]\[ \csc(\theta) \approx -1.14396 \][/tex]
These values match the previously given numerical results:
[tex]\[ \cos(\theta) \approx -0.4856429311786321 \][/tex]
[tex]\[ \csc(\theta) \approx -1.1439589045541112 \][/tex]
