Complex Number Trigonometric Identities Proof
Problem Statement
Given:
- \( m, n \) are integers
- \( x = \cos\alpha + i\sin\alpha \)
- \( y = \cos\beta + i\sin\beta \)
Prove that:
- \( x^m y^n + \frac{1}{x^m y^n} = 2\cos(m\alpha + n\beta) \)
- \( x^m y^n - \frac{1}{x^m y^n} = 2i\sin(m\alpha + n\beta) \)
Proof
Step 1: Express in Exponential Form
Using Euler's formula:
\[ x = \cos\alpha + i\sin\alpha = e^{i\alpha} \]
\[ y = \cos\beta + i\sin\beta = e^{i\beta} \]
Step 2: Compute \( x^m y^n \)
\[ x^m = (e^{i\alpha})^m = e^{im\alpha} \]
\[ y^n = (e^{i\beta})^n = e^{in\beta} \]
\[ x^m y^n = e^{im\alpha} \cdot e^{in\beta} = e^{i(m\alpha + n\beta)} \]
Step 3: Compute \( \frac{1}{x^m y^n} \)
\[ \frac{1}{x^m y^n} = e^{-i(m\alpha + n\beta)} \]
Step 4: Prove the First Identity
\[ x^m y^n + \frac{1}{x^m y^n} = e^{i(m\alpha + n\beta)} + e^{-i(m\alpha + n\beta)} \]
\[ = 2\cos(m\alpha + n\beta) \]
(Using Euler's formula: \( e^{i\theta} + e^{-i\theta} = 2\cos\theta \))
Step 5: Prove the Second Identity
\[ x^m y^n - \frac{1}{x^m y^n} = e^{i(m\alpha + n\beta)} - e^{-i(m\alpha + n\beta)} \]
\[ = 2i\sin(m\alpha + n\beta) \]
(Using Euler's formula: \( e^{i\theta} - e^{-i\theta} = 2i\sin\theta \))
Final Results
\[ \boxed{x^m y^n + \frac{1}{x^m y^n} = 2\cos(m\alpha + n\beta)} \]
\[ \boxed{x^m y^n - \frac{1}{x^m y^n} = 2i\sin(m\alpha + n\beta)} \]