Question

The function $ (1 + \cos x) \sin x $ is maximum at

• A. \( x = \frac{\pi}{3} \)
• B. \( x = \pi \)
• C. Both at \( x = \frac{\pi}{3} \) and \( x = \pi \)
• D. \( x = 2\pi \)

Hint:

When differentiating trigonometric functions involving sums and products, use the product rule and apply trigonometric identities to simplify the expression.

Answer 1

The Correct Option is A

To find the maximum of the function \( f(x) = (1 + \cos x) \sin x \), we first compute its derivative and find the critical points.

(1) First, apply the product rule to differentiate the function:
\[ f'(x) = \frac{d}{dx} \left( (1 + \cos x) \sin x \right) \] \[ f'(x) = (1 + \cos x) \cos x - \sin^2 x \] Simplifying: \[ f'(x) = \cos x + \cos^2 x - \sin^2 x \]
(2) Now, set the derivative equal to zero to find the critical points:
\[ \cos x + \cos^2 x - \sin^2 x = 0 \] Using the trigonometric identity \( \cos^2 x - \sin^2 x = \cos(2x) \), we can simplify the equation to: \[ \cos x + \cos(2x) = 0 \]
(3) Solve this equation. By trial or graphing, we find that the function attains a maximum at \( x = \frac{\pi}{3} \).
Thus, the maximum value of \( (1 + \cos x) \sin x \) occurs at \( x = \frac{\pi}{3} \).

Conclusion: The function \( (1 + \cos x) \sin x \) is maximum at \( x = \frac{\pi}{3} \).