To solve these trigonometric identities, we use the following sum-to-product formulas:

1. For $\sin A + \sin B$: $\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$

2. For $\cos A - \cos B$: $\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$

3. For $\cos A + \cos B$: $\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$

4. For $\sin A - \sin B$: $\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$

Now, let's apply these formulas to each part:

(a) $\sin(7x) + \sin(x)$:

$\sin(7x) + \sin(x) = 2 \sin\left(\frac{7x + x}{2}\right) \cos\left(\frac{7x - x}{2}\right)$ $= 2 \sin(4x) \cos(3x)$

(b) $\cos(7x) - \cos(x)$:

$\cos(7x) - \cos(x) = -2 \sin\left(\frac{7x + x}{2}\right) \sin\left(\frac{7x - x}{2}\right)$ $= -2 \sin(4x) \sin(3x)$

(c) $\cos(7x) + \cos(x)$:

$\cos(7x) + \cos(x) = 2 \cos\left(\frac{7x + x}{2}\right) \cos\left(\frac{7x - x}{2}\right)$ $= 2 \cos(4x) \cos(3x)$

(d) $\sin(7x) - \sin(x)$:

$\sin(7x) - \sin(x) = 2 \cos\left(\frac{7x + x}{2}\right) \sin\left(\frac{7x - x}{2}\right)$ $= 2 \cos(4x) \sin(3x)$

Final Answers:

• (a) $\sin(7x) + \sin(x) = 2 \sin(4x) \cos(3x)$
• (b) $\cos(7x) - \cos(x) = -2 \sin(4x) \sin(3x)$
• (c) $\cos(7x) + \cos(x) = 2 \cos(4x) \cos(3x)$
• (d) $\sin(7x) - \sin(x) = 2 \cos(4x) \sin(3x)$

Would you like detailed steps for any part? Here are five related questions to deepen your understanding:

1. What are the derivations of the sum-to-product formulas?
2. Can the product-to-sum formulas be applied in reverse to these equations?
3. How do these trigonometric transformations simplify integrals or derivatives?
4. What is the geometric interpretation of sum-to-product identities?
5. How do these identities apply to solving trigonometric equations?

Tip: Memorize these formulas and practice applying them to various problems to develop speed and accuracy!