Question

Evaluate: \[ \cos x \cos 7x - \cos 5x \cos 13x \]

• A. \( 2\cos^2 6x \cos 12x \)
• B. \( 2\sin^2 6x \cos 6x \)
• C. \( 2\sin 6x \sin 12x \)
• D. \( 2\sin 6x \cos 12x \)

Hint:

Always convert products of trigonometric functions into sums before simplifying complex expressions.

Answer 1

The Correct Option is B

Step 1: Use the product-to-sum identity.
\[ \cos A \cos B = \frac{1}{2}[\cos(A+B) + \cos(A-B)] \]

Step 2: Apply the identity to each term.
\[ \cos x \cos 7x = \frac{1}{2}[\cos 8x + \cos 6x] \] \[ \cos 5x \cos 13x = \frac{1}{2}[\cos 18x + \cos 8x] \]

Step 3: Subtract the expressions.
\[ \cos x \cos 7x - \cos 5x \cos 13x \] \[ = \frac{1}{2}(\cos 8x + \cos 6x - \cos 18x - \cos 8x) \] \[ = \frac{1}{2}(\cos 6x - \cos 18x) \]

Step 4: Use the identity for difference of cosines.
\[ \cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2} \] \[ = -2\sin 12x \sin(-6x) \] \[ = 2\sin 12x \sin 6x \]

Step 5: Simplify the expression.
\[ 2\sin 6x \sin 12x = 2\sin^2 6x \cos 6x \]

Step 6: Conclusion.
\[ \boxed{2\sin^2 6x \cos 6x} \]