Given the value of one trigonometric function of an acute angle 0, find the values of the remaining five trigonometric functions of 0. Give the exact answer as a simplified fraction, if necessary. Rationalize denominators if necessary. tan 0= 2 0/0 2 0/6 X Subm Part 1 of 5 Part 2 of 5 cos 0 = S S

Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE: 1. Give the measures of the complement and the supplement of an angle measuring 35°.
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**Title: Finding Trigonometric Functions**

**Objective:**
Given the value of one trigonometric function of an acute angle θ, find the values of the remaining five trigonometric functions of θ. Give the exact answer as a simplified fraction, if necessary. Rationalize denominators if necessary.

**Problem Statement:**

Given: 
\[ \tan \theta = \frac{2}{5} \]

**Tasks:**

**Part 1 of 5:**
Determine the value of \(\sin \theta\).

**Part 2 of 5:**
Determine the value of \(\cos \theta\).

The problem continues with three more parts to find the remaining trigonometric functions: \(\cot \theta\), \(\sec \theta\), and \(\csc \theta\).

**Solution Guide:**

1. **Identify the given function**:
   - Here, \(\tan \theta = \frac{2}{5}\). 
   - This represents \( \tan \theta = \frac{\text{opposite}}{\text{adjacent}} \). So in a right triangle, the opposite side to the angle θ is 2 units, and the adjacent side is 5 units.

2. **Calculate the hypotenuse** using the Pythagorean theorem:
   \[ \text{hypotenuse} = \sqrt{(\text{opposite})^2 + (\text{adjacent})^2} \]
   \[ \text{hypotenuse} = \sqrt{2^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29} \]

3. **Find the trigonometric functions**:

   - **\(\sin \theta\)**: 
     \[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{2}{\sqrt{29}} \]
     Rationalize the denominator:
     \[ \sin \theta = \frac{2}{\sqrt{29}} \cdot \frac{\sqrt{29}}{\sqrt{29}} = \frac{2\sqrt{29}}{29} \]

   - **\(\cos \theta\)**:
     \[ \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{5}{\sqrt{
Transcribed Image Text:**Title: Finding Trigonometric Functions** **Objective:** Given the value of one trigonometric function of an acute angle θ, find the values of the remaining five trigonometric functions of θ. Give the exact answer as a simplified fraction, if necessary. Rationalize denominators if necessary. **Problem Statement:** Given: \[ \tan \theta = \frac{2}{5} \] **Tasks:** **Part 1 of 5:** Determine the value of \(\sin \theta\). **Part 2 of 5:** Determine the value of \(\cos \theta\). The problem continues with three more parts to find the remaining trigonometric functions: \(\cot \theta\), \(\sec \theta\), and \(\csc \theta\). **Solution Guide:** 1. **Identify the given function**: - Here, \(\tan \theta = \frac{2}{5}\). - This represents \( \tan \theta = \frac{\text{opposite}}{\text{adjacent}} \). So in a right triangle, the opposite side to the angle θ is 2 units, and the adjacent side is 5 units. 2. **Calculate the hypotenuse** using the Pythagorean theorem: \[ \text{hypotenuse} = \sqrt{(\text{opposite})^2 + (\text{adjacent})^2} \] \[ \text{hypotenuse} = \sqrt{2^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29} \] 3. **Find the trigonometric functions**: - **\(\sin \theta\)**: \[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{2}{\sqrt{29}} \] Rationalize the denominator: \[ \sin \theta = \frac{2}{\sqrt{29}} \cdot \frac{\sqrt{29}}{\sqrt{29}} = \frac{2\sqrt{29}}{29} \] - **\(\cos \theta\)**: \[ \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{5}{\sqrt{
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