Answer :
To determine the value of [tex]\(\cos(\theta)\)[/tex] given that [tex]\(\tan(\theta) = \frac{3}{5}\)[/tex] and that [tex]\(\theta\)[/tex] is an angle in the first quadrant, we will follow these steps:
1. Understand the relationship of [tex]\(\tan(\theta)\)[/tex], [tex]\(\sin(\theta)\)[/tex], and [tex]\(\cos(\theta)\)[/tex]:
[tex]\(\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\)[/tex]. Given [tex]\(\tan(\theta) = \frac{3}{5}\)[/tex], we can write:
[tex]\[ \frac{\sin(\theta)}{\cos(\theta)} = \frac{3}{5} \][/tex]
2. Express [tex]\(\sin(\theta)\)[/tex] in terms of [tex]\(\cos(\theta)\)[/tex]:
[tex]\[ \sin(\theta) = 3k \quad \text{and} \quad \cos(\theta) = 5k \][/tex]
Here, [tex]\(k\)[/tex] is a proportionality constant.
3. Use the Pythagorean identity [tex]\(\sin^2(\theta) + \cos^2(\theta) = 1\)[/tex]:
Substitute [tex]\(\sin(\theta)\)[/tex] and [tex]\(\cos(\theta)\)[/tex] into the Pythagorean identity:
[tex]\[ (3k)^2 + (5k)^2 = 1 \][/tex]
Simplify the equation:
[tex]\[ 9k^2 + 25k^2 = 1 \][/tex]
Combine the terms:
[tex]\[ 34k^2 = 1 \][/tex]
Solve for [tex]\(k^2\)[/tex]:
[tex]\[ k^2 = \frac{1}{34} \][/tex]
Therefore,
[tex]\[ k = \sqrt{\frac{1}{34}} = \frac{1}{\sqrt{34}} \][/tex]
4. Find [tex]\(\cos(\theta)\)[/tex]:
Recall that [tex]\(\cos(\theta) = 5k\)[/tex]:
[tex]\[ \cos(\theta) = 5 \cdot \frac{1}{\sqrt{34}} = \frac{5}{\sqrt{34}} \][/tex]
Rationalize the denominator by multiplying the numerator and the denominator by [tex]\(\sqrt{34}\)[/tex]:
[tex]\[ \cos(\theta) = \frac{5}{\sqrt{34}} \cdot \frac{\sqrt{34}}{\sqrt{34}} = \frac{5\sqrt{34}}{34} \][/tex]
Thus, the value of [tex]\(\cos(\theta)\)[/tex] when [tex]\(\tan(\theta) = \frac{3}{5}\)[/tex] in the first quadrant is:
[tex]\[ \cos(\theta) = \frac{5\sqrt{34}}{34} \][/tex]
1. Understand the relationship of [tex]\(\tan(\theta)\)[/tex], [tex]\(\sin(\theta)\)[/tex], and [tex]\(\cos(\theta)\)[/tex]:
[tex]\(\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\)[/tex]. Given [tex]\(\tan(\theta) = \frac{3}{5}\)[/tex], we can write:
[tex]\[ \frac{\sin(\theta)}{\cos(\theta)} = \frac{3}{5} \][/tex]
2. Express [tex]\(\sin(\theta)\)[/tex] in terms of [tex]\(\cos(\theta)\)[/tex]:
[tex]\[ \sin(\theta) = 3k \quad \text{and} \quad \cos(\theta) = 5k \][/tex]
Here, [tex]\(k\)[/tex] is a proportionality constant.
3. Use the Pythagorean identity [tex]\(\sin^2(\theta) + \cos^2(\theta) = 1\)[/tex]:
Substitute [tex]\(\sin(\theta)\)[/tex] and [tex]\(\cos(\theta)\)[/tex] into the Pythagorean identity:
[tex]\[ (3k)^2 + (5k)^2 = 1 \][/tex]
Simplify the equation:
[tex]\[ 9k^2 + 25k^2 = 1 \][/tex]
Combine the terms:
[tex]\[ 34k^2 = 1 \][/tex]
Solve for [tex]\(k^2\)[/tex]:
[tex]\[ k^2 = \frac{1}{34} \][/tex]
Therefore,
[tex]\[ k = \sqrt{\frac{1}{34}} = \frac{1}{\sqrt{34}} \][/tex]
4. Find [tex]\(\cos(\theta)\)[/tex]:
Recall that [tex]\(\cos(\theta) = 5k\)[/tex]:
[tex]\[ \cos(\theta) = 5 \cdot \frac{1}{\sqrt{34}} = \frac{5}{\sqrt{34}} \][/tex]
Rationalize the denominator by multiplying the numerator and the denominator by [tex]\(\sqrt{34}\)[/tex]:
[tex]\[ \cos(\theta) = \frac{5}{\sqrt{34}} \cdot \frac{\sqrt{34}}{\sqrt{34}} = \frac{5\sqrt{34}}{34} \][/tex]
Thus, the value of [tex]\(\cos(\theta)\)[/tex] when [tex]\(\tan(\theta) = \frac{3}{5}\)[/tex] in the first quadrant is:
[tex]\[ \cos(\theta) = \frac{5\sqrt{34}}{34} \][/tex]