To prove the identity
[tex]\[
(\cos 2\alpha - \cos 2\beta)^2 + (\sin 2\alpha + \sin 2\beta)^2,
\][/tex]
let's explore each term step-by-step and eventually simplify them to show the result.
Let's consider the expressions separately:
1. [tex]\((\cos 2\alpha - \cos 2\beta)^2\)[/tex]
2. [tex]\((\sin 2\alpha + \sin 2\beta)^2\)[/tex]
First, expand both terms individually.
### Expanding [tex]\((\cos 2\alpha - \cos 2\beta)^2\)[/tex]
The term [tex]\((\cos 2\alpha - \cos 2\beta)^2\)[/tex] can be expanded using the formula [tex]\((a - b)^2 = a^2 - 2ab + b^2\)[/tex]:
[tex]\[
(\cos 2\alpha - \cos 2\beta)^2 = \cos^2 2\alpha - 2 \cos 2\alpha \cos 2\beta + \cos^2 2\beta
\][/tex]
### Expanding [tex]\((\sin 2\alpha + \sin 2\beta)^2\)[/tex]
Similarly, the term [tex]\((\sin 2\alpha + \sin 2\beta)^2\)[/tex] can be expanded using the formula [tex]\((a + b)^2 = a^2 + 2ab + b^2\)[/tex]:
[tex]\[
(\sin 2\alpha + \sin 2\beta)^2 = \sin^2 2\alpha + 2 \sin 2\alpha \sin 2\beta + \sin^2 2\beta
\][/tex]
### Adding the expanded terms
Now, add the two expanded terms together:
[tex]\[
\cos^2 2\alpha - 2 \cos 2\alpha \cos 2\beta + \cos^2 2\beta + \sin^2 2\alpha + 2 \sin 2\alpha \sin 2\beta + \sin^2 2\beta
\][/tex]
### Simplifying the expression
Combine the like terms:
[tex]\[
(\cos^2 2\alpha + \sin^2 2\alpha) + (\cos^2 2\beta + \sin^2 2\beta) - 2 \cos 2\alpha \cos 2\beta + 2 \sin 2\alpha \sin 2\beta
\][/tex]
Recall the Pythagorean identity for trigonometric functions, [tex]\(\cos^2 x + \sin^2 x = 1\)[/tex]:
[tex]\[
(\cos^2 2\alpha + \sin^2 2\alpha) = 1 \quad \text{and} \quad (\cos^2 2\beta + \sin^2 2\beta) = 1
\][/tex]
Thus, we can rewrite the expression as:
[tex]\[
1 + 1 - 2 \cos 2\alpha \cos 2\beta + 2 \sin 2\alpha \sin 2\beta
\][/tex]
Combine the constants:
[tex]\[
2 - 2 \cos 2\alpha \cos 2\beta + 2 \sin 2\alpha \sin 2\beta
\][/tex]
To simplify [tex]\(- 2 \cos 2\alpha \cos 2\beta + 2 \sin 2\alpha \sin 2\beta\)[/tex], utilize the sum-to-product identities:
[tex]\[
\cos A \cos B - \sin A \sin B = \cos (A + B)
\][/tex]
for [tex]\(A = 2\alpha\)[/tex] and [tex]\(B = 2\beta\)[/tex]:
[tex]\[
-2 (\cos 2\alpha \cos 2\beta) + 2 (\sin 2\alpha \sin 2\beta) = -2(\cos 2\alpha \cos 2\beta - \sin 2\alpha \sin 2\beta) = -2 \cos (2\alpha + 2\beta)
\][/tex]
Finally, the expression simplifies to:
[tex]\[
2 - 2 \cos (2\alpha + 2\beta)
\][/tex]
So, we have shown that:
[tex]\[
(\cos 2\alpha - \cos 2\beta)^2 + (\sin 2\alpha + \sin 2\beta)^2 = 2 - 2 \cos (2\alpha + 2\beta)
\][/tex]
Thus we have successfully proven the given identity.
