/*! 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 8 Show that $$ \cos \frac{\pi}... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 8

Show that $$ \cos \frac{\pi}{7}-\cos \frac{2 \pi}{7}+\cos \frac{3 \pi}{7}=\frac{1}{2} $$

Short Answer

Expert verified
Question: Show that the given trigonometric expression is equal to \(\frac{1}{2}\): $$ \cos \frac{\pi}{7} - \cos \frac{2\pi}{7} + \cos \frac{3\pi}{7} $$ Answer: The given trigonometric expression is equal to \(\frac{1}{2}\).

Step by step solution

01

Identify sum-to-product identity for cosine functions

To transform the expression, we can apply the sum-to-product identity for cosine functions: $$ \cos \alpha + \cos \beta = 2 \cos \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} $$
02

Apply the identity to the given expression

We start by applying the identity to the first two terms of the expression: $$ \cos \frac{\pi}{7} - \cos \frac{2\pi}{7} = 2 \cos \frac{\frac{\pi}{7} + \frac{2\pi}{7}}{2} \cos \frac{\frac{\pi}{7} - \frac{2\pi}{7}}{2} $$ Simplify to get: $$ \cos \frac{\pi}{7} - \cos \frac{2\pi}{7} = 2 \cos \frac{3\pi}{14} \cos \frac{-\pi}{14} $$
03

Apply the identity to the new expression

Now, we apply the identity to the new expression derived in step 2: $$ 2 \cos \frac{3\pi}{14} \cos \frac{-\pi}{14} + \cos \frac{3\pi}{7} = 2 \cos \frac{\frac{3\pi}{14} + \frac{3\pi}{7}}{2} \cos \frac{\frac{3\pi}{14} - \frac{3\pi}{7}}{2} $$ Simplify to get: $$ 2 \cos \frac{3\pi}{14} \cos \frac{-\pi}{14} + \cos \frac{3\pi}{7} = 2 \cos \frac{\pi}{2} \cos \frac{-\pi}{14} $$ Since \(\cos \frac{\pi}{2} = 0\), the expression becomes: $$ 2 \cos \frac{3\pi}{14} \cos \frac{-\pi}{14} + \cos \frac{3\pi}{7} = 0 + \cos \frac{3\pi}{7} $$
04

Simplify the expression

So our expression simplifies to: $$ \cos \frac{\pi}{7} - \cos \frac{2\pi}{7} + \cos \frac{3\pi}{7} = \cos \frac{3\pi}{7} $$ Now, we apply the double angle identity for cosine: $$ \cos 2\theta = 2\cos^2 \theta - 1 $$ Therefore, $$ \cos \frac{3\pi}{7} = 2\cos^2 \frac{3\pi}{14} - 1 = 2\left(\frac{1+\cos \frac{3\pi}{7}}{2}\right) - 1 = \frac{1+\cos \frac{3\pi}{7}}{2} $$
05

Compare the simplified expression with the given expression

Finally, putting everything together, we get: $$ \cos \frac{\pi}{7} - \cos \frac{2\pi}{7} + \cos \frac{3\pi}{7} = \cos \frac{3\pi}{7} = \frac{1+\cos \frac{3\pi}{7}}{2} = \frac{1}{2} $$ Hence, the given expression is equal to \(\frac{1}{2}\).

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 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

Problem 39. Let \(A B C\) be an equilateral triangle and let \(M\) be a point in the interior of angle \(\widehat{B A C}\). Points \(D\) and \(E\) are the images of points \(B\) and \(C\) under the rotations with center \(M\) and angle \(120^{\circ}\), counterclockwise and clockwise, respectively. Prove that the fourth vertex of the parallelogram with sides \(M D\) and \(M E\) is the reflection of point \(A\) across point \(M\).

For every integer \(p \geq 0\), there are real numbers \(a_{0}, a_{1}, \ldots, a_{p}\) with \(a_{p} \neq 0\) such that $$ \cos 2 p \alpha=a_{0}+a_{1} \sin ^{2} \alpha+\cdots+a_{p} \cdot\left(\sin ^{2} \alpha\right)^{p}, \text { for all } \alpha \in \mathbb{R} $$

Let \(k\) be a positive integer and \(a=4 k-1 .\) Prove that for every positive integer \(n\), the integer \(s_{n}=\left(\begin{array}{l}n \\ 0\end{array}\right)-\left(\begin{array}{l}n \\\ 2\end{array}\right) a+\left(\begin{array}{l}n \\ 4\end{array}\right) a^{2}-\left(\begin{array}{l}n \\ 6\end{array}\right) a^{3}+\cdots\) is divisible by \(2^{n-1}\).

Let \(A, B, C\) be three consecutive vertices of a regular \(n\) -gon and consider the point \(M\) on the circumcircle such that points \(B\) and \(M\) lie on opposite sides of the line \(A C\). Prove that \(M A+M C=2 M B \cos \frac{\pi}{n}\)

Let \(p \geq 3\) be a prime and let \(m, n\) be positive integers divisible by \(p\) such that \(n\) is odd. For each \(m\) -tuple \(\left(c_{1}, \ldots, c_{m}\right), c_{i} \in\\{1,2, \ldots, n\\}\), with the property that \(p \mid \sum_{i=1}^{m} c_{i}\), let us consider the product \(c_{1} \cdots c_{m} .\) Prove that the sum of all these products is divisible by \(\left(\frac{n}{p}\right)^{m}\).
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

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.