sin^2 a + cos^2 (a + b) + 2 sin a sin b cos(a + b) is independent of
This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
Step 1: Simplify the given expression using trigonometric identities.
The given expression is:
$$E = \sin^2 \alpha + \cos^2(\alpha + \beta) + 2\sin \alpha \sin \beta \cos(\alpha + \beta)$$
We use the product-to-sum identity $2\sin A \sin B = \cos(A-B) - \cos(A+B)$.
Substitute $A = \alpha$ and $B = \beta$:
$$2\sin \alpha \sin \beta = \cos(\alpha - \beta) - \cos(\alpha + \beta)$$
Substitute this into the expression for $E$:
$$E = \sin^2 \alpha + \cos^2(\alpha + \beta) + (\cos(\alpha - \beta) - \cos(\alpha + \beta))\cos(\alpha + \beta)$$
Distribute $\cos(\alpha + \beta)$ in the last term:
$$E = \sin^2 \alpha + \cos^2(\alpha + \beta) + \cos(\alpha - \beta)\cos(\alpha + \beta) - \cos^2(\alpha + \beta)$$
The terms $\cos^2(\alpha + \beta)$ and $-\cos^2(\alpha + \beta)$ cancel out:
$$E = \sin^2 \alpha + \cos(\alpha - \beta)\cos(\alpha + \beta)$$
Step 2: Simplify the product of cosines.
We use the product-to-sum identity $2\cos A \cos B = \cos(A+B) + \cos(A-B)$.
Substitute $A = \alpha - \beta$ and $B = \alpha + \beta$:
$$\cos(\alpha - \beta)\cos(\alpha + \beta) = \frac{1}{2}[\cos((\alpha - \beta) + (\alpha + \beta)) + \cos((\alpha - \beta) - (\alpha + \beta))]$$
$$\cos(\alpha - \beta)\
