Question

Cos²a-sin²b=cos(a-b) cos(a+b)а-альфаb-бета​

Answer 1

Ответ:

............................

Answer 2

To prove the identity $\cos^2(\alpha) - \sin^2(\beta) = \cos(\alpha - \beta) \cos(\alpha + \beta)$, you can use trigonometric identities and basic trigonometric principles.

Starting with the left side of the equation:

$\cos^2(\alpha) - \sin^2(\beta)$

You can use the Pythagorean identity $\sin^2(x) + \cos^2(x) = 1$ to rewrite $\sin^2(\beta)$ as $1 - \cos^2(\beta)$:

$\cos^2(\alpha) - (1 - \cos^2(\beta))$

Now, distribute the negative sign:

$\cos^2(\alpha) - 1 + \cos^2(\beta)$

Combine like terms:

$\cos^2(\alpha) + \cos^2(\beta) - 1$

Now, let's work on the right side of the equation:

$\cos(\alpha - \beta) \cos(\alpha + \beta)$

Use the angle addition and subtraction formulas for cosine:

$\cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)$ $\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)$

Now, multiply these two expressions together:

$\cos(\alpha - \beta) \cos(\alpha + \beta) = (\cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta))(\cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta))$

Use the difference of squares formula $(a - b)(a + b) = a^2 - b^2$:

$\cos(\alpha - \beta) \cos(\alpha + \beta) = \cos^2(\alpha)\cos^2(\beta) - \sin^2(\alpha)\sin^2(\beta)$

Now, combine like terms:

$\cos^2(\alpha)\cos^2(\beta) - \sin^2(\alpha)\sin^2(\beta)$

So, we have now shown that:

$\cos^2(\alpha) - \sin^2(\beta) = \cos(\alpha - \beta) \cos(\alpha + \beta)$

This proves the identity.