Question

Evaluate: \(\displaystyle \int_0^{\frac{\pi}{6}} \cos x \cdot \cos 2x \, dx = \)

• A. \(\frac{6}{5}\)
• B. \(1\)
• C. \(\frac{12}{5}\)
• D. \(-\frac{12}{5}\)

Hint:

Use product-to-sum formulas to simplify products of trigonometric functions before integrating. \[ \cos A \cos B = \frac{1}{2} [\cos(A-B) + \cos(A+B)] \]

Answer 1

The Correct Option is A

Step 1: Use product-to-sum formula for \(\cos x \cos 2x\): \[ \cos x \cos 2x = \frac{1}{2} [\cos (x - 2x) + \cos (x + 2x)] = \frac{1}{2} [\cos(-x) + \cos 3x] = \frac{1}{2} [\cos x + \cos 3x] \] Step 2: Rewrite the integral: \[ \int_0^{\frac{\pi}{6}} \cos x \cdot \cos 2x \, dx = \frac{1}{2} \int_0^{\frac{\pi}{6}} (\cos x + \cos 3x) \, dx = \frac{1}{2} \left( \int_0^{\frac{\pi}{6}} \cos x \, dx + \int_0^{\frac{\pi}{6}} \cos 3x \, dx \right) \] Step 3: Evaluate integrals: \[ \int_0^{\frac{\pi}{6}} \cos x \, dx = \sin x \Big|_0^{\frac{\pi}{6}} = \sin \frac{\pi}{6} - \sin 0 = \frac{1}{2} - 0 = \frac{1}{2} \] \[ \int_0^{\frac{\pi}{6}} \cos 3x \, dx = \frac{\sin 3x}{3} \Big|_0^{\frac{\pi}{6}} = \frac{\sin \frac{\pi}{2}}{3} - 0 = \frac{1}{3} \] Step 4: Sum and multiply by \(\frac{1}{2}\): \[ \frac{1}{2} \left( \frac{1}{2} + \frac{1}{3} \right) = \frac{1}{2} \times \frac{5}{6} = \frac{5}{12} \] Step 5: So the value is \(\frac{5}{12}\), which does not match any option. --- Please verify the options or the integral again. ---