Certainly! Let's solve the problem step by step.
Given:
[tex]\[
\cos x = \frac{5}{13}
\][/tex]
### Step 1: Find [tex]\(\sin x\)[/tex]
We will use the Pythagorean identity:
[tex]\[
\sin^2 x + \cos^2 x = 1
\][/tex]
Substitute [tex]\(\cos x = \frac{5}{13}\)[/tex] into the identity:
[tex]\[
\sin^2 x + \left(\frac{5}{13}\right)^2 = 1
\][/tex]
First, calculate [tex]\(\left(\frac{5}{13}\right)^2\)[/tex]:
[tex]\[
\left(\frac{5}{13}\right)^2 = \frac{25}{169}
\][/tex]
Now, substitute this value back into the Pythagorean identity:
[tex]\[
\sin^2 x + \frac{25}{169} = 1
\][/tex]
Next, solve for [tex]\(\sin^2 x\)[/tex]:
[tex]\[
\sin^2 x = 1 - \frac{25}{169}
\][/tex]
Convert 1 to a fraction with the same denominator:
[tex]\[
1 = \frac{169}{169}
\][/tex]
Then, subtract the fractions:
[tex]\[
\sin^2 x = \frac{169}{169} - \frac{25}{169} = \frac{144}{169}
\][/tex]
Take the square root of both sides to find [tex]\(\sin x\)[/tex]:
[tex]\[
\sin x = \sqrt{\frac{144}{169}}
\][/tex]
Simplify the square root:
[tex]\[
\sin x = \frac{\sqrt{144}}{\sqrt{169}} = \frac{12}{13}
\][/tex]
Since the goal is to compute the principal value, we take the positive root:
[tex]\[
\sin x = \frac{12}{13}
\][/tex]
### Step 2: Find [tex]\(\tan x\)[/tex]
We know that:
[tex]\[
\tan x = \frac{\sin x}{\cos x}
\][/tex]
Substitute the values of [tex]\(\sin x\)[/tex] and [tex]\(\cos x\)[/tex]:
[tex]\[
\tan x = \frac{\frac{12}{13}}{\frac{5}{13}}
\][/tex]
Simplify the fraction:
[tex]\[
\tan x = \frac{12}{13} \times \frac{13}{5} = \frac{12 \times 13}{13 \times 5} = \frac{12}{5}
\][/tex]
Thus:
[tex]\[
\tan x = 2.4
\][/tex]
### Final Answers:
[tex]\[
\sin x = \frac{12}{13} \approx 0.9230769230769231
\][/tex]
[tex]\[
\tan x = 2.4
\][/tex]
In summary:
- [tex]\(\sin x \approx 0.923\)[/tex]
- [tex]\(\tan x = 2.4\)[/tex]
