Problem 42
Question
Write the first trigonometric function in terms of the second for \(\theta\) in the given quadrant. $$ \sec \theta, \quad \sin \theta ; \quad \theta \text { in Quadrant } \mathrm{I} $$
Step-by-Step Solution
Verified Answer
\( \sec \theta = \frac{1}{\sqrt{1 - \sin^2 \theta}} \).
1Step 1: Identify the relationship between sec and sin
The secant function, \( \sec \theta \), is the reciprocal of the cosine function, so \( \sec \theta = \frac{1}{\cos \theta} \). Additionally, in trigonometry, the Pythagorean identity is \( \sin^2 \theta + \cos^2 \theta = 1 \) in all quadrants.
2Step 2: Express cos θ in terms of sin θ
Using the Pythagorean identity, solve for \( \cos \theta \). We have \( \cos^2 \theta = 1 - \sin^2 \theta \), and knowing \( \theta \) is in Quadrant I, \( \cos \theta \) is positive. Thus, \( \cos \theta = \sqrt{1 - \sin^2 \theta} \).
3Step 3: Substitute cos θ in sec θ expression
Since \( \sec \theta = \frac{1}{\cos \theta} \), substitute \( \cos \theta \) from the previous step into the expression for secant. Therefore, \( \sec \theta = \frac{1}{\sqrt{1 - \sin^2 \theta}} \).
4Step 4: Conclusion
The expression for \( \sec \theta \) in terms of \( \sin \theta \) is \( \sec \theta = \frac{1}{\sqrt{1 - \sin^2 \theta}} \), valid in Quadrant I where \( \cos \theta > 0 \).
Key Concepts
Secant FunctionSine FunctionPythagorean Identity
Secant Function
The secant function, denoted as \( \sec \theta \), plays a crucial role in trigonometry. It is defined as the reciprocal of the cosine function. This means that any reference to the secant of an angle \( \theta \) is simply \( \sec \theta = \frac{1}{\cos \theta} \).
This relationship implies that whenever you know the cosine of an angle, you can easily find its secant.
In right-angle trigonometry, secant is particularly useful because:
This relationship implies that whenever you know the cosine of an angle, you can easily find its secant.
In right-angle trigonometry, secant is particularly useful because:
- It helps in determining lengths of sides in relation to the hypotenuse.
- It facilitates transformations between different trigonometric equations.
Sine Function
The sine function, denoted \( \sin \theta \), is one of the fundamental trigonometric functions, representing the y-coordinate on the unit circle.
When considering an angle \( \theta \) in standard position, the sine of \( \theta \) describes the height above the x-axis:
This makes sine particularly useful in various fields involving waves, like sound or light.
Some properties of the sine function include:
When considering an angle \( \theta \) in standard position, the sine of \( \theta \) describes the height above the x-axis:
This makes sine particularly useful in various fields involving waves, like sound or light.
Some properties of the sine function include:
- Sine is always between -1 and 1, because it represents a ratio of the lengths of the sides of a right triangle to its hypotenuse.
- In the first quadrant, sine values are positive, meaning the y-coordinate is positive.
Pythagorean Identity
The Pythagorean identity is one of the most essential relationships in trigonometry. It’s expressed as \( \sin^2 \theta + \cos^2 \theta = 1 \).
This identity is powerful because it relates sine and cosine in a simple equation that holds for any angle \( \theta \).
Here's why it matters:
For instance, knowing \( \sin \theta \) allows us to solve for \( \cos \theta \) using the Pythagorean identity, as \( \cos^2 \theta = 1 - \sin^2 \theta \).
This relationship is the backbone of many trigonometric transformations and applications.
This identity is powerful because it relates sine and cosine in a simple equation that holds for any angle \( \theta \).
Here's why it matters:
- You can shift between sine and cosine values easily.
- It's fundamental to solving trigonometric equations where one value is known.
For instance, knowing \( \sin \theta \) allows us to solve for \( \cos \theta \) using the Pythagorean identity, as \( \cos^2 \theta = 1 - \sin^2 \theta \).
This relationship is the backbone of many trigonometric transformations and applications.
Other exercises in this chapter
Problem 41
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