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Chapter 5: Problem 62

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

Short Answer

Expert verified
\(\text{sin } t = -\sqrt{1 - \frac{1}{\text{sec}^2 t}}\) in Quadrant IV.

Step by step solution

01

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.
02

Use the Pythagorean Identity

Recall the identity involving sine and cosine: \(egin{equation} ext{sin}^2 t + ext{cos}^2 t = 1. ag{1} ext{cos } t = + rac{1}{ ext{sec } t} ext{ (since sec } t ext { is positive in Quadrant IV)} ext{So, } ext{cos}^2 t = rac{1}{ ext{sec}^2 t}. ext{Substitute this into (1) to get:} ext{sin}^2 t = 1 - rac{1}{ ext{sec}^2 t} ext{Therefore, } ext{sin} t = - rac{ ext{sec}t}{ ext{sec}^2 t} = - rac{1}{ ext{sec } t} ext{ (taking the negative root as } ext{sin } t ext{ is negative in Quadrant IV)}.\)
03

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 \( ext{sin } t\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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掳 and 360掳, 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 鈥� 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 鈥� 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.
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.

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Most popular questions from this chapter

A Bobbing Cork A cork floating in a lake is bobbing in simple harmonic motion. Its displacement above the bottom of the lake is modeled by $$ y=0.2 \cos 20 \pi t+8 $$ where \(y\) is measured in meters and \(t\) is measured in minutes. (a) Find the frequency of the motion of the cork. (b) Sketch a graph of \(y .\) (c) Find the maximum displacement of the cork above the lake bottom.

(a) To define the inverse sine function, we restrict the domain of sine to the interval _____________________ . On this interval the sine function is one-to- one, and its inverse function \(\sin ^{-1}\) is defined by \(\sin ^{-1} x=y \Leftrightarrow \sin\) _________ \(=\) __________ . For example, \(\sin ^{-1} \frac{1}{2}=\) __________ because sin ___________ = __________. (b) To define the inverse cosine function we restrict the domain of cosine to the interval _________________. On this interval the cosine function is one- to-one and its inverse function \(\cos ^{-1}\) is defined by \(\cos ^{-1} x=y \Leftrightarrow\) cos ____________ = ___________. For example, \(\cos ^{-1} \frac{1}{2}=\) _____________ because cos __________ = ___________.

Blood Pressure Each time your heart beats, your blood pressure first increases and then decreases as the heart rests between beats. The maximum and minimum blood pressures are called the systolic and diastolic pressures, respectively. Your blood pressure reading is written as systolic/diastolic. A reading of \(120 / 80\) is considered normal. A certain person's blood pressure is modeled by the function $$ p(t)=115+25 \sin (160 \pi t) $$ where \(p(t)\) is the pressure in \(\mathrm{mmHg}\) (millimeters of mercury), at time \(t\) measured in minutes. (a) Find the period of \(p\) (b) Find the number of heartbeats per minute. (c) Graph the function \(p\) (d) Find the blood pressure reading. How does this compare to normal blood pressure?

Find the period and graph the function. $$y=\tan \left(x-\frac{\pi}{4}\right)$$

Find a function that models the simple harmonic motion having the given properties. Assume that the displacement is at its maximum at time \(t=0\) amplitude \(60 \mathrm{ft}\), period \(0.5 \mathrm{min}\)
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