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Chapter 5: Problem 55

Write the first expression in terms of the second if the terminal point determined by \(t\) is in the given quadrant. $$\tan t, \sin t ; \quad \text { Quadrant IV }$$

Short Answer

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

Step by step solution

01

Understand the Properties of Quadrant IV

In Quadrant IV, the x-coordinate (cosine) is positive and the y-coordinate (sine) is negative. Therefore, \( an t = \frac{\sin t}{\cos t} = -\frac{\sin t}{|\cos t|}\).
02

Express \(\tan t\) in terms of \(\sin t\)

Since \(\tan t = \frac{\sin t}{\cos t}\), we need to use the identity \(\cos^2 t = 1 - \sin^2 t\) to express \(\cos t\) in terms of \(\sin t\). Thus, \(\cos t = \sqrt{1 - \sin^2 t}\).
03

Insert Quadrant IV Properties

In Quadrant IV, \(\cos t\) is positive, so \(\cos t = \sqrt{1 - \sin^2 t}\) holds true without any sign change. Since \(\tan t = \frac{\sin t}{\cos t}\), substitute for \(\cos t\): \(\tan t = \frac{\sin t}{\sqrt{1 - \sin^2 t}}\).
04

Account for the Negative Sine Component

As sine is negative in Quadrant IV, ensure \(\tan t\) reflects this. Therefore, \(\tan t = \frac{-|\sin t|}{\sqrt{1 - \sin^2 t}}\), further simplified to \(\tan t = -\frac{|\sin t|}{\sqrt{1 - \sin^2 t}}\). However since \(\sin t < 0\), it simplifies directly to \(\tan t = -\frac{\sin t}{\sqrt{1 - \sin^2 t}}\).
05

Final Expression

The expression of \(\tan t\) in terms of \(\sin t\) in Quadrant IV is: \(\tan t = -\frac{\sin t}{\sqrt{1 - \sin^2 t}}\).

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Key Concepts

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

Quadrant Analysis in Trigonometry
Understanding how angles and their corresponding values behave in different quadrants is crucial in trigonometry. The coordinate plane is divided into four quadrants, each having distinct characteristics affecting trigonometric functions. In Quadrant IV, the x-coordinate (cosine) remains positive while the y-coordinate (sine) becomes negative.

This has implications on the overall sign of trigonometric functions in this quadrant. When dealing with trigonometric functions, such as sine, cosine, and tangent, it's important to know:
  • Sine (\(\sin t\)) is negative.
  • Cosine (\(\cos t\)) is positive.
  • Tangent (\(\tan t\)) combines both sine and cosine, resulting in a negative value (\(\tan t = \frac{\sin t}{\cos t}\)).
A firm grasp of these quadrant rules ensures accurate calculations and simplifications in trigonometric expressions, regardless of the given quadrant.
Tangent-Sine Relationship
The relationship between the tangent and sine functions is an interesting aspect of trigonometry. Tangent, denoted as \(\tan t\), is calculated as the ratio of sine to cosine: \(\tan t = \frac{\sin t}{\cos t}\). This definition highlights that tangent relies on both sine and cosine for its value.

In scenarios where we need to express tangent in terms of sine, especially when cosine is not readily given, we use the Pythagorean identity associated with trigonometric functions. The identity \(\cos^2 t = 1 - \sin^2 t\) allows us to substitute expression for cosine directly. This provides an equation:
  • \(\cos t = \sqrt{1 - \sin^2 t}\)
  • Therefore, \(\tan t = \frac{\sin t}{\sqrt{1 - \sin^2 t}}\).
Taking quadrant properties into account ensures that the correct signs are applied for values in these formulas, leading to the final expression for \(\tan t\) in the intended terms.
Trigonometric Functions in Quadrants
Each quadrant on the coordinate plane affects the trigonometric functions differently. Remember, each quadrant has its properties mainly regarding the signs of the trigonometric functions, which leads us to evaluate their expressions uniquely.

For instance:
  • In Quadrant I, all trigonometric functions are positive.
  • In Quadrant II, sine is positive, but cosine and tangent are negative.
  • In Quadrant III, tangent is positive, while sine and cosine are negative.
  • In Quadrant IV, as analyzed, cosine is positive, and sine is negative, resulting in a negative tangent.
Understanding these quadrant-based variations helps in predicting and computing the signs and values of the trigonometric ratios. Proper assessment of these relationships ensures correct manipulation of trigonometric equations and solutions in various quadratic contexts.

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Most popular questions from this chapter

Graph the three functions on a common screen. How are the graphs related? $$y=\frac{1}{1+x^{2}}, \quad y=-\frac{1}{1+x^{2}}, \quad y=\frac{\cos 2 \pi x}{1+x^{2}}$$

When a car with its horn blowing drives by an observer, the pitch of the horn seems higher as it approaches and lower as it recedes (see the figure below). This phenomenon is called the Doppler effect. If the sound source is moving at speed \(v\) relative to the observer and if the speed of sound is \(v_{0}\), then the perceived frequency \(f\) is related to the actual frequency \(f_{0}\) as follows. $$f=f_{0}\left(\frac{v_{0}}{v_{0} \pm v}\right)$$ We choose the minus sign if the source is moving toward the observer and the plus sign if it is moving away. Suppose that a car drives at \(110 \mathrm{ft} / \mathrm{s}\) past a woman standing on the shoulder of a highway, blowing its horn, which has a frequency of \(500 \mathrm{Hz}\). Assume that the speed of sound is \(1130 \mathrm{ft} / \mathrm{s} .\) (This is the speed in dry air at \(70^{\circ} \mathrm{F}\).) (a) What are the frequencies of the sounds that the woman hears as the car approaches her and as it moves away from her? (b) Let \(A\) be the amplitude of the sound. Find functions of the form $$y=A \sin \omega t$$ that model the perceived sound as the car approaches the woman and as it recedes. (Figure cant copy)

For each sine curve find the amplitude, period, phase, and horizontal shift. $$y=5 \sin \left(2 t-\frac{\pi}{2}\right)$$

An initial amplitude \(k\), damping constant \(c,\) and frequency \(f\) or period \(p\) are given. (Recall that frequency and period are related by the equation \(f=1 / p .\) ) (a) Find a function that models the damped harmonic motion. Use a function of the form \(y=k e^{-c t} \cos \omega t\) in Exercises 21-24 and of the form \(y=k e^{-c t} \sin \omega t\) in Exercises 25-28 (b) Graph the function. $$k=100, \quad c=0.05, \quad p=4$$

A pair of sine curves with the same period is given. (a) Find the phase of each curve. (b) Find the phase difference between the curves. (c) Determine whether the curves are in phase or out of phase. (d) Sketch both curves on the same axes. $$y_{1}=10 \sin \left(3 t-\frac{\pi}{2}\right) ; \quad y_{2}=10 \sin \left(3 t-\frac{5 \pi}{2}\right)$$
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