Problem 8
Question
In \(3-22 :\) a. Rewrite each function value in terms of its cofunction. b. Find, to four decimal places, the value of the function value found in a. $$ \sec 75^{\circ} $$
Step-by-Step Solution
Verified Answer
\( \sec 75^{\circ} = \csc 15^{\circ} \approx 3.8637 \)
1Step 1: Understanding Cofunction Identities
Cofunction identities relate trigonometric functions whose angles are complementary. For secant, the cofunction identity is given as \( \sec(\theta) = \csc(90^{\circ} - \theta) \). This means that for any angle \( \theta \), the secant of that angle equals the cosecant of its complement.
2Step 2: Applying the Cofunction Identity
We are given \( \sec 75^{\circ} \). Using the cofunction identity: \[ \sec 75^{\circ} = \csc(90^{\circ} - 75^{\circ}) \] Simplifying inside the parentheses gives \( \csc(15^{\circ}) \). Therefore, \( \sec 75^{\circ} \) can be rewritten as \( \csc 15^{\circ} \).
3Step 3: Calculating the Cosecant Value
To find the value of \( \csc 15^{\circ} \) to four decimal places, recognize that \( \csc(\theta) = \frac{1}{\sin(\theta)} \). Therefore, calculate \( \sin(15^{\circ}) \) and use it to find \( \csc(15^{\circ}) \): \[ \csc(15^{\circ}) = \frac{1}{ \sin(15^{\circ}) } \].
4Step 4: Finding the Sine of 15 Degrees
The exact value of \( \sin(15^{\circ}) \) can be calculated using the angle subtraction identity \( \sin(15^{\circ}) = \sin(45^{\circ} - 30^{\circ}) \). Applying the identity, \[ \sin(15^{\circ}) = \sin(45^{\circ}) \cos(30^{\circ}) - \cos(45^{\circ}) \sin(30^{\circ}) \]. \( \sin(45^{\circ}) = \cos(45^{\circ}) = \frac{\sqrt{2}}{2} \), \( \sin(30^{\circ}) = \frac{1}{2} \), and \( \cos(30^{\circ}) = \frac{\sqrt{3}}{2} \). Substitute these values to get: \[ \sin(15^{\circ}) = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} - \sqrt{2}}{4} \].
5Step 5: Calculating Cosecant Using Sine Value
Given \( \sin(15^{\circ}) = \frac{\sqrt{6} - \sqrt{2}}{4} \), find \( \csc(15^{\circ}) \) by taking the reciprocal: \[ \csc(15^{\circ}) = \frac{1}{\sin(15^{\circ})} = \frac{4}{\sqrt{6} - \sqrt{2}} \]. To approximate \( \csc(15^{\circ}) \), use a calculator to find the decimal value. \( \csc(15^{\circ}) \approx 3.8637 \) to four decimal places.
Key Concepts
Secant and Cosecant RelationshipTrigonometric Complementary AnglesSine and Cosecant Calculation
Secant and Cosecant Relationship
Understanding the relationship between the secant and cosecant functions requires familiarity with cofunction identities. Cofunction identities link trigonometric functions involving complementary angles, meaning two angles whose sum is 90 degrees. For example, the identity for the secant function is expressed as:
- \( \sec(\theta) = \csc(90^{\circ} - \theta) \)
Trigonometric Complementary Angles
Complementary angles are two angles whose measurements add up to 90 degrees. They are a fundamental concept in trigonometry because they allow the interchanging of certain trigonometric functions through cofunction identities. Let's look at a few basic pairs:
- \( \sin(\theta) = \cos(90^{\circ} - \theta) \)
- \( \cos(\theta) = \sin(90^{\circ} - \theta) \)
Sine and Cosecant Calculation
To find the value of \( \csc 15^{\circ} \), one essential step involves calculating \( \sin 15^{\circ} \). This is because the cosecant function is the reciprocal of the sine function:
- \( \csc(\theta) = \frac{1}{\sin(\theta)} \)
- \( \csc(15^{\circ}) = \frac{4}{\sqrt{6} - \sqrt{2}} \)
Other exercises in this chapter
Problem 7
In \(3-12\) , find the exact function value of each of the following if the measure of the angle is given in radians. $$ \cos \frac{2 \pi}{3} $$
View solution Problem 7
In \(3-12,\) find the radian measure of each angle whose degree measure is given. \(160^{\circ}\)
View solution Problem 8
In \(3-14,\) find each value of \(\theta : \mathbf{a} .\) in degrees \(\mathbf{b} .\) in radians $$ \theta=\arcsin \left(-\frac{1}{2}\right) $$
View solution Problem 8
For each function value, write the value or tell why it is undefined. Do not use a calculator. \(\cot \frac{\pi}{2}\)
View solution