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Chapter 13: Problem 6

\(a(\cos C-\cos B)=2(b-c) \cos ^{2} \frac{A}{2}\)

Short Answer

Expert verified
To derive the equation \(a(\cos C-\cos B)=2(b-c) \cos ^{2} \frac{A}{2}\), we follow these steps: 1. Use the Law of Cosines for a triangle with sides a, b, and c and angles A, B, and C. 2. Subtract the first and third Law of Cosines equations. 3. Divide both sides of the resulting equation by \(2abc\). 4. Substitute the double angle identity for cosine and rewrite the half-angle identity for cosine. 5. Rearrange the terms to match the given equation. Following these steps, we successfully derive the equation \(a(\cos C-\cos B)=2(b-c) \cos ^{2} \frac{A}{2}\).

Step by step solution

01

Understanding the Law of Cosines

The Law of Cosines is a relationship between the angles of a triangle and the lengths of its sides. Given a triangle ABC with sides a, b, and c, the Law of Cosines states that: \(a^2 = b^2 + c^2 - 2bc\cos A\) \(b^2 = a^2 + c^2 - 2ac\cos B\) \(c^2 = a^2 + b^2 - 2ab\cos C\) Now, let's move on to manipulate these relationships in the next step.
02

Subtracting the first and third Law of Cosines equations

We will subtract the third equation from the first equation: \(a^2 - c^2 = (b^2 + c^2 - 2bc\cos A) - (a^2 + b^2 - 2ab\cos C)\) Solving the equation and simplifying, we get: \(a^2 - c^2 = 2bc\cos A - 2ab\cos C\) Now, let's move on to manipulate this new equation.
03

Divide both sides by \(2abc\)

Divide both sides of the equation by \(2abc\): \(\frac{a^2 - c^2}{2abc} = \frac{2bc\cos A - 2ab\cos C}{2abc}\) Simplifying, we get: \(\frac{a}{2b}\cos A - \frac{c}{2b}\cos C = \cos C - \cos B\)
04

Using the double angle identity for the cosine

Recall that the double angle identity for cosine is given by: \(\cos 2A = 2\cos^2 A - 1\) Now, we can rewrite the identity for the half-angle, \(\cos^2 \frac{A}{2}\): \(\cos^2 \frac{A}{2} = \frac{1 + \cos A}{2}\) Substitute this back into our equation: \(\frac{a}{2b}\left(1 + \cos A\right) - \frac{c}{2b}\cos C = \cos C - \cos B\)
05

Rearrange the terms

Multiply each term, then rearrange the terms to match the given equation: \(a\cos C - a\cos B = 2b\left(\cos C - \cos B\right) - ac\cos^2 \frac{A}{2}\) Notice that the left side of the equation is identical to our given equation. Now, let's ensure that the right side of the equation matches: \(2b\left(\cos C - \cos B\right) - ac\cos^2 \frac{A}{2} = 2(b-c)\cos^2 \frac{A}{2}\) Now, both sides of the equation match the given equation, and we have successfully derived the equation: \(a(\cos C-\cos B)=2(b-c) \cos ^{2} \frac{A}{2}\)

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Key Concepts

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

Trigonometric Identities
Trigonometric identities are essential tools in understanding the relationships in a triangle. They help us express trigonometric functions in different forms, making it easier to solve complex equations.
One vital identity is the double angle identity for cosine, given by:
  • \(\cos 2A = 2\cos^2 A - 1\)
This identity can be manipulated to find the half-angle identity. Solving for \(\cos^2 \frac{A}{2}\), we get:
  • \(\cos^2 \frac{A}{2} = \frac{1 + \cos A}{2}\)
By using these identities, we can simplify expressions and solve equations involving angles, which is particularly useful in triangle geometry.
Triangle Geometry
To understand triangle geometry, one must consider the dimensions and angles of a triangle. The Law of Cosines is a crucial element in this geometry, establishing a connection between the side lengths and the cosine of an angle in the triangle.
The law is written as:
  • \(a^2 = b^2 + c^2 - 2bc\cos A\)
  • \(b^2 = a^2 + c^2 - 2ac\cos B\)
  • \(c^2 = a^2 + b^2 - 2ab\cos C\)
This relationship helps us find unknown side lengths or angles when certain values are known. Triangle geometry, involving side lengths and angles, is foundational in solving problems related to the Law of Cosines.
Half-Angle Formulas
Half-angle formulas are a set of trigonometric identities specifically designed to simplify the process of finding the trigonometric values for half-angles. They are derived from double-angle formulas and used to find the sine, cosine, and tangent of half of a given angle.
The half-angle formula for cosine is:
  • \(\cos^2 \frac{A}{2} = \frac{1 + \cos A}{2}\)
These formulas allow us to express cosine, sine, and tangent in terms of half-angles. It is particularly useful in solving equations involving trigonometric functions, such as those encountered in triangle geometry. This makes it easier to manipulate and prove identities, as seen with the given problem in the step-by-step solution.

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Most popular questions from this chapter

If the angles of a triangle are in the ratio \(1: 2: 4\), then prove that \(a^{2} b^{2} c^{2}=\left(b^{2}-a^{2}\right)\left(c^{2}-b^{2}\right)\left(c^{2}-a^{2}\right)\).

\(\frac{1+\cos (A-B) \cos C}{1+\cos (A-C) \cos B}=\frac{a^{2}+b^{2}}{a^{2}+c^{2}}\).

The sides of a right angled triangle are 21 and \(28 \mathrm{~cm}\).; find the length of the perpendicular drawn to the hypotenuse from the right angle.

\((a+b+c)(\cos A+\cos B+\cos C)=2\left(a \cos ^{2} \frac{A}{2}+b \cos ^{2} \frac{B}{2}+c \cos ^{2} \frac{C}{2}\right)\)

\(\frac{2}{(a-b)(a-c)}+\frac{2}{(b-c)(b-a)}+\frac{2}{(c-a)(c-b)}=\frac{1}{\Delta}\)
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