Derivation of the Relationship

Oct 13, 2025 by recoveryshake.com 107 views

Introduction

In this article, we will explore a problem that involves trigonometric identities and their applications. The problem requires us to prove a relationship between two given expressions, m and n, and a trigonometric function, A. We will use various trigonometric identities and formulas to derive the required relationship.

Given Expressions

We are given two expressions:

$\cos A - \sin A = m$ ... (1) $\cos A + \sin A = n$ ... (2)

Our objective is to show that:

$\frac{m^2 - n^2}{m^2 + n^2} = -2 \sin A \cdot \cos A = -\frac{2}{\tan A + \cot A}$

Derivation of the Relationship

To derive the required relationship, we will start by manipulating the given expressions (1) and (2).

Squaring the Expressions

We will square both expressions (1) and (2) to obtain:

$(\cos A - \sin A)^2 = m^2$ ... (3) $(\cos A + \sin A)^2 = n^2$ ... (4)

Expanding the Squared Expressions

We will expand the squared expressions (3) and (4) using the formula $(a-b)^2 = a^2 - 2ab + b^2$ and $(a+b)^2 = a^2 + 2ab + b^2$ .

$(\cos A - \sin A)^2 = \cos^2 A - 2\cos A \sin A + \sin^2 A = m^2$ ... (5) $(\cos A + \sin A)^2 = \cos^2 A + 2\cos A \sin A + \sin^2 A = n^2$ ... (6)

Subtracting the Expanded Expressions

We will subtract the expanded expression (5) from the expanded expression (6) to obtain:

$(\cos^2 A + 2\cos A \sin A + \sin^2 A) - (\cos^2 A - 2\cos A \sin A + \sin^2 A) = n^2 - m^2$

Simplifying the expression, we get:

$4\cos A \sin A = n^2 - m^2$ ... (7)

Dividing the Expression by m^2 + n^2

We will divide the expression (7) by $m^2 + n^2$ to obtain:

$\frac{4\cos A \sin A}{m^2 + n^2} = \frac{n^2 - m^2}{m^2 + n^2}$

Simplifying the expression, we get:

$\frac{m^2 - n^2}{m^2 + n^2} = -2 \sin A \cdot \cos A$ ... (8)

Derivation of the Second Part of the Relationship

We will use the trigonometric identity $\tan A = \frac{\sin A}{\cos A}$ and $\cot A = \frac{\cos A}{\sin A}$ to derive the second part of the relationship.

$\frac{m^2 - n^2}{m^2 + n^2} = -2 \sin A \cdot \cos A = -\frac{2}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}}$

Simplifying the expression, we get:

$\frac{m^2 - n^2}{m^2 + n^2} = -\frac{2}{\tan A + \cot A}$ ... (9)

Conclusion

In this article, we have derived the required relationship between m, n, and A using various trigonometric identities and formulas. We have shown that:

$\frac{m^2 - n^2}{m^2 + n^2} = -2 \sin A \cdot \cos A = -\frac{2}{\tan A + \cot A}$

This relationship can be used to solve various problems in trigonometry and other branches of mathematics.

References

[1] Trigonometry by Michael Corral
[2] Calculus by Michael Spivak
[3] Algebra by Michael Artin

Keywords

Trigonometric identities
Trigonometric functions
Algebraic manipulations
Mathematical derivations
Trigonometry
Algebra
Calculus

Discussion

This problem requires a good understanding of trigonometric identities and their applications. The solution involves various algebraic manipulations and mathematical derivations. The problem can be used to test the understanding of trigonometric functions and their relationships.

Solutions

[1] Solution to problem 1
[2] Solution to problem 2
[3] Solution to problem 3

Introduction

In this article, we will provide a Q&A section to help clarify any doubts or questions that readers may have regarding the problem of proving the relationship between m, n, and A.

Q&A

Q1: What is the significance of the given expressions (1) and (2)?

A1: The given expressions (1) and (2) are used to derive the required relationship between m, n, and A. They are the building blocks of the solution and are used to manipulate the trigonometric functions.

Q2: How do we square the expressions (1) and (2)?

A2: We square the expressions (1) and (2) using the formula $(a-b)^2 = a^2 - 2ab + b^2$ and $(a+b)^2 = a^2 + 2ab + b^2$ . This allows us to expand the expressions and simplify them.

Q3: What is the purpose of subtracting the expanded expressions (5) and (6)?

A3: The purpose of subtracting the expanded expressions (5) and (6) is to eliminate the common terms and obtain the required expression (7). This allows us to derive the relationship between m, n, and A.

Q4: How do we divide the expression (7) by $m^2 + n^2$ ?

A4: We divide the expression (7) by $m^2 + n^2$ to obtain the required relationship between m, n, and A. This allows us to simplify the expression and derive the final result.

Q5: What is the significance of the trigonometric identity $\tan A = \frac{\sin A}{\cos A}$ and $\cot A = \frac{\cos A}{\sin A}$ ?

A5: The trigonometric identity $\tan A = \frac{\sin A}{\cos A}$ and $\cot A = \frac{\cos A}{\sin A}$ is used to derive the second part of the relationship between m, n, and A. It allows us to simplify the expression and obtain the final result.

Q6: How do we simplify the expression (9)?

A6: We simplify the expression (9) by using the trigonometric identity $\tan A = \frac{\sin A}{\cos A}$ and $\cot A = \frac{\cos A}{\sin A}$ . This allows us to obtain the final result and derive the relationship between m, n, and A.

Common Mistakes

Not squaring the expressions (1) and (2) correctly
Not expanding the squared expressions (5) and (6) correctly
Not subtracting the expanded expressions (5) and (6) correctly
Not dividing the expression (7) by $m^2 + n^2$ correctly
Not simplifying the expression (9) correctly

Tips and Tricks

Make sure to square the expressions (1) and (2) correctly
Expand the squared expressions (5) and (6) correctly
Subtract the expanded expressions (5) and (6) correctly
Divide the expression (7) by $m^2 + n^2$ correctly
Simplify the expression (9) correctly

Conclusion

In this article, we have provided a Q&A section to help clarify any doubts or questions that readers may have regarding the problem of proving the relationship between m, n, and A. We have also provided common mistakes and tips and tricks to help readers avoid mistakes and simplify the solution.

References

[1] Trigonometry by Michael Corral
[2] Calculus by Michael Spivak
[3] Algebra by Michael Artin

Keywords

Trigonometric identities
Trigonometric functions
Algebraic manipulations
Mathematical derivations
Trigonometry
Algebra
Calculus

Discussion

Solutions

[1] Solution to problem 1
[2] Solution to problem 2
[3] Solution to problem 3