/*! 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 81 Show that $$ \cos (\pi-\thet... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 81

Show that $$ \cos (\pi-\theta)=-\cos \theta $$ for every angle \(\theta\).

Short Answer

Expert verified
Using the cosine difference formula, \(\cos(\pi - \theta) = \cos \pi \cos \theta + \sin \pi \sin \theta\). Since \(\cos \pi = -1\) and \(\sin \pi = 0\), we get \(\cos(\pi - \theta) = (-1) \cos \theta + (0) \sin \theta = -\cos \theta\).

Step by step solution

01

Recall the cosine difference formula

The cosine difference formula is: \[ \cos (A - B) = \cos A \cos B + \sin A \sin B \]. In this exercise, we have \(A = \pi\) and \(B = \theta \).
02

Use the cosine difference formula with our angles

From Step 1, substitute \(A = \pi\) and \(B = \theta \) into the formula: \[ \cos (\pi - \theta) = \cos \pi \cos \theta + \sin \pi \sin \theta \]
03

Calculate the sine and cosine of 蟺

Recall that \(\cos \pi = -1\) and \(\sin \pi = 0\). Substitute these values into the formula from Step 2: \[ \cos (\pi - \theta) = (-1) \cos \theta + (0) \sin \theta \]
04

Simplify and finish

Since the \(\sin \theta\) term is multiplied by zero, it will not contribute to the result. Therefore, we are left with: \[ \cos (\pi - \theta) = -\cos \theta \] As we aimed to show, the cosine of the difference between 蟺 and 胃 equals the negative of the cosine of 胃.

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 91影视!

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

Is sine an even function, an odd function, or neither?

Find exact expressions for the indicated quantities, given that $$ \cos \frac{\pi}{12}=\frac{\sqrt{2+\sqrt{3}}}{2} \text { and } \sin \frac{\pi}{8}=\frac{\sqrt{2-\sqrt{2}}}{2} $$ [These values for \(\cos \frac{\pi}{12}\) and \(\sin \frac{\pi}{8}\) will be derived in Examples 4 and 5 in Section 6.3.] $$ \cos \left(-\frac{5 \pi}{12}\right) $$

Show that $$ \tan \left(\theta+\frac{\pi}{2}\right)=-\frac{1}{\tan \theta} $$ for every angle \(\theta\) that is not an integer multiple of \(\frac{\pi}{2} .\) Interpret this result in terms of the characterization of the slopes of perpendicular lines.

Find exact expressions for the indicated quantities. The following information will be useful: $$ \begin{array}{l} \cos 22.5^{\circ}=\frac{\sqrt{2+\sqrt{2}}}{2} \text { and } \sin 22.5^{\circ}=\frac{\sqrt{2-\sqrt{2}}}{2} \\ \cos 18^{\circ}=\sqrt{\frac{\sqrt{5}+5}{8}} \text { and } \sin 18^{\circ}=\frac{\sqrt{5}-1}{4} \end{array} $$ [The value for \(\sin 22.5^{\circ}\) used here was derived in Example 4 in Section \(5.5 ;\) the other values were derived in Exercise 64 and Problems 102 and 103 in Section \(5.5 .]\) $$\cos 82.5^{\circ}$$

In 1768 the Swiss mathematician Johann Lambert proved that if \(\theta\) is a rational number in the interval \(\left(0, \frac{\pi}{2}\right),\) then \(\tan \theta\) is irrational. Explain why this result implies that \(\pi\) is irrational.
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation
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.