Problem 71
Question
Use the appropriate reciprocal identity to find each function value. Rationalize denominators when applicable. $$\cos \theta, \text { given that } \sec \theta=-\frac{5}{2}$$
Step-by-Step Solution
Verified Answer
\( \cos \theta = -\frac{2}{5} \).
1Step 1: Understanding the Reciprocal Identity
Recall the reciprocal identity, which states that \( \sec \theta = \frac{1}{\cos \theta} \). This means that the secant function is the reciprocal of the cosine function.
2Step 2: Set up the Equation
Given that \( \sec \theta = -\frac{5}{2} \), we'll use the reciprocal identity: \( \sec \theta = \frac{1}{\cos \theta} \). Substitute the given value into this identity to get: \( -\frac{5}{2} = \frac{1}{\cos \theta} \).
3Step 3: Solve for Cosine
To find \( \cos \theta \), take the reciprocal of the given \( \sec \theta \) value. Thus, \( \cos \theta = -\frac{2}{5} \).
4Step 4: Check Rationalization (if necessary)
Since \( \cos \theta = -\frac{2}{5} \) already has a rationalized denominator, no further rationalization is needed.
Key Concepts
Understanding Trigonometric FunctionsExploring the Cosine FunctionRationalization in Trigonometry
Understanding Trigonometric Functions
Trigonometric functions are mathematical functions that relate the angles of triangles to the lengths of their sides. They are foundational in mathematics and are essential in fields like physics, engineering, and even music.
The six main trigonometric functions are:
Each of these functions has a specific role in the study of angles and triangles. They allow us to calculate unknown side lengths and angles in right triangles.
The six main trigonometric functions are:
- Sine (\( ext{sin} heta \))
- Cosine (\( ext{cos} heta \))
- Tangent (\( ext{tan} heta \))
- Cosecant (\( ext{csc} heta \)), which is the reciprocal of sine (\( ext{csc} heta = \frac{1}{ ext{sin} heta} \))
- Secant (\( ext{sec} heta \)), which is the reciprocal of cosine (\( ext{sec} heta = \frac{1}{ ext{cos} heta} \))
- Cotangent (\( ext{cot} heta \)), which is the reciprocal of tangent (\( ext{cot} heta = \frac{1}{ ext{tan} heta} \))
Each of these functions has a specific role in the study of angles and triangles. They allow us to calculate unknown side lengths and angles in right triangles.
Exploring the Cosine Function
The cosine function, denoted as \( ext{cos} \), is one of the primary trigonometric functions. It is defined as the ratio of the adjacent side to the hypotenuse in a right-angled triangle.
The cosine function can be found using the equation:
The cosine function is periodic with a period of \( 360^{ ext{o}} \) or \( 2\pi \) radians, meaning it repeats its values in regular intervals. This periodicity is vital in modeling repetitive phenomena, like sound waves.
The cosine function can be found using the equation:
- \( ext{cos} heta = \frac{ ext{Adjacent}}{ ext{Hypotenuse}} \)
The cosine function is periodic with a period of \( 360^{ ext{o}} \) or \( 2\pi \) radians, meaning it repeats its values in regular intervals. This periodicity is vital in modeling repetitive phenomena, like sound waves.
Rationalization in Trigonometry
Rationalization is a mathematical technique used to eliminate radicals (like square roots) or complex fractions from the denominator of an expression. While not necessary when dealing with basic trigonometric identities, rationalization ensures that the expression is simplified.
The process generally involves multiplying the numerator and the denominator by a conjugate or another expression that will eliminate the square root or fraction from the denominator.
In the given problem, however, the cosine value \( \cos \theta = -\frac{2}{5} \) already has a rationalized denominator. By understanding rationalization, we can handle intricate expressions effectively in various mathematical situations.
The process generally involves multiplying the numerator and the denominator by a conjugate or another expression that will eliminate the square root or fraction from the denominator.
- For example, if you have a denominator of \( \sqrt{2} \), you can multiply by \( \sqrt{2} \) over \( \sqrt{2} \) to rationalize it.
In the given problem, however, the cosine value \( \cos \theta = -\frac{2}{5} \) already has a rationalized denominator. By understanding rationalization, we can handle intricate expressions effectively in various mathematical situations.
Other exercises in this chapter
Problem 70
Convert each degree measure to radians. Leave answers as rational multiples of \(\pi .\) $$-210^{\circ}$$
View solution Problem 70
Find exact values of the six trigonometric functions for each angle by hand. Do not use a calculator. $$-\frac{4 \pi}{3}$$
View solution Problem 71
Graph each function over a two-period interval. $$y=-1+2 \tan x$$
View solution Problem 71
Convert each radian measure to degrees. $$\frac{\pi}{3}$$
View solution