To find [tex]\(\cos(t)\)[/tex] given [tex]\(\tan(t) = \frac{5}{12}\)[/tex] and [tex]\(0 < t < \frac{\pi}{2}\)[/tex], we can use trigonometric identities.
### Step-by-Step Solution:
1. Understand the Pythagorean Identity:
We know that [tex]\(\tan(t) = \frac{\sin(t)}{\cos(t)}\)[/tex]. Given [tex]\(\tan(t) = \frac{5}{12}\)[/tex], we can set up the relationship:
[tex]\[
\sin(t) = 5k \quad \text{and} \quad \cos(t) = 12k
\][/tex]
for some constant [tex]\(k\)[/tex].
2. Use the Pythagorean Identity:
The Pythagorean identity for sine and cosine is:
[tex]\[
\sin^2(t) + \cos^2(t) = 1
\][/tex]
Substitute [tex]\(\sin(t) = 5k\)[/tex] and [tex]\(\cos(t) = 12k\)[/tex]:
[tex]\[
(5k)^2 + (12k)^2 = 1
\][/tex]
3. Simplify the Equation:
[tex]\[
25k^2 + 144k^2 = 1
\][/tex]
Combine the like terms:
[tex]\[
169k^2 = 1
\][/tex]
4. Solve for [tex]\(k\)[/tex]:
[tex]\[
k^2 = \frac{1}{169}
\][/tex]
[tex]\[
k = \sqrt{\frac{1}{169}}
\][/tex]
[tex]\[
k = \frac{1}{\sqrt{169}}
\][/tex]
[tex]\[
k = \frac{1}{13}
\][/tex]
5. Find [tex]\(\sin(t)\)[/tex] and [tex]\(\cos(t)\)[/tex]:
[tex]\[
\sin(t) = 5k = 5 \left(\frac{1}{13}\right) = \frac{5}{13}
\][/tex]
[tex]\[
\cos(t) = 12k = 12 \left(\frac{1}{13}\right) = \frac{12}{13}
\][/tex]
Thus, [tex]\(\cos(t) = \frac{12}{13}\)[/tex], and the corresponding values of [tex]\(\sin(t)\)[/tex] and [tex]\(\cos(t)\)[/tex] are:
[tex]\[
\sin(t) \approx 0.38461538461538464
\][/tex]
and
[tex]\[
\cos(t) \approx 0.9230769230769231
\][/tex]
