Problem 62
Question
Write the first expression in terms of the second if the terminal point determined by \(t\) is in the given quadrant. \(\sin t,\) sec \(t ; \quad\) Quadrant IV
Step-by-Step Solution
Verified Answer
\(\text{sin } t = -\sqrt{1 - \frac{1}{\text{sec}^2 t}}\) in Quadrant IV.
1Step 1: Understand the Quadrant IV Trigonometric Relations
In Quadrant IV, the sine function is negative because it corresponds to the y-coordinate of the point on the unit circle. The secant, which is the reciprocal of the cosine, is positive because cosine corresponds to the x-coordinate, which is positive in Quadrant IV.
2Step 2: Use the Pythagorean Identity
Recall the identity involving sine and cosine: \(egin{equation} ext{sin}^2 t + ext{cos}^2 t = 1. ag{1} ext{cos } t = + rac{1}{ ext{sec } t} ext{ (since sec } t ext { is positive in Quadrant IV)} ext{So, } ext{cos}^2 t = rac{1}{ ext{sec}^2 t}. ext{Substitute this into (1) to get:} ext{sin}^2 t = 1 - rac{1}{ ext{sec}^2 t} ext{Therefore, } ext{sin} t = - rac{ ext{sec}t}{ ext{sec}^2 t} = - rac{1}{ ext{sec } t} ext{ (taking the negative root as } ext{sin } t ext{ is negative in Quadrant IV)}.\)
3Step 3: Final Answer
Therefore, the expression of \( ext{sin } t \) in terms of \( ext{sec } t \) when \(t\) is in Quadrant IV is \( ext{sin } t = -\sqrt{1 - \frac{1}{\text{sec}^2 t}}\). This is derived by substituting the relation \( ext{cos} t = \frac{1}{\text{sec} t}\) into the Pythagorean Identity and solving for \( ext{sin } t\).
Key Concepts
Quadrant IV TrigonometrySine and Secant RelationshipPythagorean Identity
Quadrant IV Trigonometry
Trigonometry in Quadrant IV is all about understanding the behavior of different trigonometric functions when angles terminate in this specific quadrant. In the unit circle, Quadrant IV spans angles between 270° and 360°, or in radians, between \(\frac{3\pi}{2}\) and \(2\pi\). Here, the x-component or cosine is positive, while the y-component or sine is negative. This occurs because as you move clockwise from the positive y-axis downwards, you enter Quadrant IV.
Knowing this, it's clear why sine has a negative value here – it matches the direction of the y-axis, which drops below zero. Meanwhile, secant, the reciprocal of cosine, remains positive as it feeds from the non-negative x-values. Remember, in any quadrant, the signs of trigonometric functions are crucial – they help you determine the exact values and relationships needed for solving problems correctly.
Knowing this, it's clear why sine has a negative value here – it matches the direction of the y-axis, which drops below zero. Meanwhile, secant, the reciprocal of cosine, remains positive as it feeds from the non-negative x-values. Remember, in any quadrant, the signs of trigonometric functions are crucial – they help you determine the exact values and relationships needed for solving problems correctly.
Sine and Secant Relationship
The sine and secant functions are intricately connected in Quadrant IV, but understanding this relationship requires a bit of exploration. Let's begin with secant. Defined as the reciprocal of cosine, secant is positive in Quadrant IV. This is because cosine, representing the x-coordinate on the unit circle, is also positive here.
Now, dealing with sine in Quadrant IV, we know sine is defined by the y-coordinate, which is negative. The mathematical relationship between sine and secant is particularly useful, especially when working within specific constraints such as quadrant limitations. Expressing sine in terms of secant involves using trigonometric identities like the Pythagorean identity, and understanding the positive or negative nature of functions due to their quadrant positions can simplify complex expressions.
Now, dealing with sine in Quadrant IV, we know sine is defined by the y-coordinate, which is negative. The mathematical relationship between sine and secant is particularly useful, especially when working within specific constraints such as quadrant limitations. Expressing sine in terms of secant involves using trigonometric identities like the Pythagorean identity, and understanding the positive or negative nature of functions due to their quadrant positions can simplify complex expressions.
Pythagorean Identity
The Pythagorean identity is a cornerstone in trigonometry, connecting sine, cosine, and secant in profound ways. The identity is expressed as \(\sin^2 t + \cos^2 t = 1\). This formula allows us to find unknown values of sine or cosine, given one but not the other.
In Quadrant IV, since \(\cos t = \frac{1}{\text{sec } t}\), we substitute this into the Pythagorean identity to solve for \(\sin^2 t\). This yields \(\sin^2 t = 1 - \frac{1}{\text{sec}^2 t}\). Given that sine is negative in Quadrant IV, we take the negative root to find \(\sin t = -\sqrt{1 - \frac{1}{\text{sec}^2 t}}\). This method highlights the utility of the Pythagorean identity, especially when functions like secant provide straightforward values for substitution, making calculations easier and more intuitive.
In Quadrant IV, since \(\cos t = \frac{1}{\text{sec } t}\), we substitute this into the Pythagorean identity to solve for \(\sin^2 t\). This yields \(\sin^2 t = 1 - \frac{1}{\text{sec}^2 t}\). Given that sine is negative in Quadrant IV, we take the negative root to find \(\sin t = -\sqrt{1 - \frac{1}{\text{sec}^2 t}}\). This method highlights the utility of the Pythagorean identity, especially when functions like secant provide straightforward values for substitution, making calculations easier and more intuitive.
Other exercises in this chapter
Problem 61
Write the first expression in terms of the second if the terminal point determined by \(t\) is in the given quadrant. \(\tan t, \sec t ; \quad\) Quadrant III
View solution Problem 61
Graph the three functions on a common screen. How are the graphs related? $$y=x^{2}, \quad y=-x^{2}, \quad y=x^{2} \sin x$$
View solution Problem 62
Graph the three functions on a common screen. How are the graphs related? $$y=x, \quad y=-x, \quad y=x \cos x$$
View solution Problem 63
Write the first expression in terms of the second if the terminal point determined by \(t\) is in the given quadrant. \(\tan ^{2} t, \sin t ; \quad\) any quadra
View solution