Question

The value of \( \cos\left(\frac{\pi}{4} + x\right) + \cos\left(\frac{\pi}{4} - x\right) \) is:

• A. \( \sqrt{2}\sin^2 x \)
• B. \( \sqrt{2}\sin x \)
• C. \( \sqrt{2}\cos^2 x \)
• D. \( \sqrt{3}\cos x \)
• E. \( \sqrt{2}\cos x \)

Hint:

Alternatively, you can expand using \( \cos(A+B) + \cos(A-B) = 2\cos A\cos B \). Both methods yield the same result, but sum-to-product is often faster.

Answer 1

Answer: (E)

Concept: We utilize the sum-to-product identity: \( \cos A + \cos B = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) \). This identity simplifies the addition of two trigonometric functions with different angles into a single product.

Step 1: Determining the sum and difference of the angles.
Let \( A = \frac{\pi}{4} + x \) and \( B = \frac{\pi}{4} - x \). Sum: \( A + B = \frac{\pi}{4} + x + \frac{\pi}{4} - x = \frac{\pi}{2} \). Therefore, \( \frac{A+B}{2} = \frac{\pi}{4} \). Difference: \( A - B = \left(\frac{\pi}{4} + x\right) - \left(\frac{\pi}{4} - x\right) = 2x \). Therefore, \( \frac{A-B}{2} = x \).

Step 2: Evaluating the final product.
\[ \text{Expression} = 2\cos\left(\frac{\pi}{4}\right)\cos(x) \] \[ = 2\left(\frac{1}{\sqrt{2}}\right)\cos x = \sqrt{2}\cos x \]