/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 Find the exact value of the trig... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the exact value of the trigonometric function at the given real number. (a) \(\cos \frac{3 \pi}{4}\) (b) \(\cos \frac{5 \pi}{4}\) (c) \(\cos \frac{7 \pi}{4}\)

Short Answer

Expert verified
(a) \(-\frac{\sqrt{2}}{2}\), (b) \(-\frac{\sqrt{2}}{2}\), (c) \(\frac{\sqrt{2}}{2}\)."

Step by step solution

01

Understanding the Unit Circle Position

First, we need to determine the position on the unit circle for each given angle. For (a) \( \frac{3\pi}{4} \), this angle is located in the second quadrant. For (b) \( \frac{5\pi}{4} \), the angle lies in the third quadrant, and for (c) \( \frac{7\pi}{4} \), the angle is in the fourth quadrant.
02

Reference Angle Calculation

A reference angle helps in finding the cosine value based on the angle's symmetry in the unit circle. For (a) \( \frac{3\pi}{4} \), the reference angle is \( \pi - \frac{3\pi}{4} = \frac{\pi}{4} \). For (b) \( \frac{5\pi}{4} \), the reference angle is \( \frac{5\pi}{4} - \pi = \frac{\pi}{4} \). For (c) \( \frac{7\pi}{4} \), the reference angle is \( 2\pi - \frac{7\pi}{4} = \frac{\pi}{4} \).
03

Cosine Value from Reference Angle

The reference angle \( \frac{\pi}{4} \) has a cosine value of \( \frac{\sqrt{2}}{2} \). However, we must consider the sign based on the quadrant. For (a) in the second quadrant, cosine is negative, so \( \cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2} \). For (b) in the third quadrant, cosine is also negative, so \( \cos \frac{5\pi}{4} = -\frac{\sqrt{2}}{2} \). For (c) in the fourth quadrant, cosine is positive, so \( \cos \frac{7\pi}{4} = \frac{\sqrt{2}}{2} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

unit circle
The unit circle is a fundamental concept in trigonometry, revolving around a circle with a radius of 1 centered at the origin of a coordinate plane. This circle allows us to connect angles with their corresponding trigonometric values in a simple and visual manner.

On the unit circle, every point \((x, y)\) corresponds to an angle measured in radians from the positive x-axis. When we talk about angles like \( \frac{3\pi}{4} \), \( \frac{5\pi}{4} \), or \( \frac{7\pi}{4} \), we are essentially finding specific points on this circle.
  • The angle \( \frac{3\pi}{4} \) is located in the second quadrant.
  • The angle \( \frac{5\pi}{4} \) lies in the third quadrant.
  • The angle \( \frac{7\pi}{4} \) exists in the fourth quadrant.
The quadrants where these angles reside help us determine the signs of the trigonometric functions. In this context, the unit circle facilitates the understanding of the behavior of trigonometric functions such as sine, cosine, and tangent, as these values are directly related to the coordinates of the points on the circle.
reference angle
Reference angles are vital for simplifying the analysis of trigonometric functions. A reference angle is the smallest angle between the terminal side of the angle and the x-axis. For any angle \(\theta\) in a standard position, the reference angle is always an acute angle, meaning it is less than or equal to 90 degrees or \((\frac{\pi}{2}) \).

The process of determining a reference angle allows us to take advantage of the symmetry properties of the unit circle. For example:
  • For \( \frac{3\pi}{4} \), the reference angle is calculated as \( \pi - \frac{3\pi}{4} = \frac{\pi}{4} \).
  • For \( \frac{5\pi}{4} \), it is \( \frac{5\pi}{4} - \pi = \frac{\pi}{4} \).
  • For \( \frac{7\pi}{4} \), it becomes \( 2\pi - \frac{7\pi}{4} = \frac{\pi}{4} \).
These reference angles are useful because trigonometric function values for angles with the same reference angle are equivalent in magnitude, differing only in sign depending on their quadrant. This makes it much easier to find exact values of trigonometric functions without needing to memorize large numbers of individual values.
cosine function
The cosine function is one of the primary trigonometric functions and is denoted as \( \cos(\theta) \). On the unit circle, the cosine of an angle is the x-coordinate of the corresponding point. By understanding the cosine function, we can determine how it behaves across different quadrants:

  • In the first and fourth quadrants, the cosine values are positive because the x-coordinates are positive.
  • In the second and third quadrants, cosine values are negative due to negative x-coordinates.
For instance, the reference angle \( \frac{\pi}{4} \) typically has a cosine value of \( \frac{\sqrt{2}}{2} \). However, the quadrant dictates the sign of the cosine:
  • For \( \cos \left( \frac{3\pi}{4} \right) \), it is negative because it is in the second quadrant, resulting in \( -\frac{\sqrt{2}}{2}\).
  • For \( \cos \left( \frac{5\pi}{4} \right) \), it remains negative in the third quadrant, giving the same value of \( -\frac{\sqrt{2}}{2}\).
  • For \( \cos \left( \frac{7\pi}{4} \right) \), it becomes positive in the fourth quadrant, yielding \( \frac{\sqrt{2}}{2}\).
Thus, understanding the cosine function in combination with concepts like the unit circle and reference angles can significantly streamline the computation of trigonometric values.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

When a car hits a certain bump on the road, a shock absorber on the car is compressed a distance of 6 in., then released (see the figure). The shock absorber vibrates in damped harmonic motion with a frequency of 2 cycles per second. The damping constant for this particular shock absorber is 2.8 (a) Find an equation that describes the displacement of the shock absorber from its rest position as a function of time. Take \(t=0\) to be the instant that the shock absorber is released. (b) How long does it take for the amplitude of the vibration to decrease to 0.5 in? (IMAGE CANNOT COPY)

Find the values of the trigonometric functions of \(t\) information. sec \(t=3,\) terminal point of \(t\) is in quadrant IV

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 \(\mathrm{mmHg}\), 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?

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

Find the value of each of the six trigonometric functions (if it is defined) at the given real number \(t\). Use your answers to complete the table. $$\begin{array}{|c|c|c|c|c|c|c|} \hline t & \sin t & \cos t & \tan t & \csc t & \sec t & \cot t \\ \hline 0 & 0 & 1 & & \text { undefined } & & \\ \hline \frac{\pi}{2} & & & & & & \\ \hline \pi & & & 0 & & & \text { undefined } \\ \hline \frac{3 \pi}{2} & & & & & & \\ \hline \end{array}$$ $$t=\pi$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.