Question

If the function \(y = \sin(x)(1 - \cos x)\) is defined in the interval \([-\pi, \pi]\), then y is strictly increasing in the interval

• A. \(\left(-\pi, -\dfrac{\pi}{3} \right) \cup \left(\dfrac{\pi}{3}, \pi \right)\)
• B. \(\left( \dfrac{\pi}{6}, \dfrac{\pi}{2} \right)\)
• C. \(\left(-\dfrac{\pi}{3}, 0 \right) \cup \left(0, \dfrac{\pi}{3} \right)\)
• D. \(\left(-\dfrac{\pi}{6}, 0 \right) \cup \left(0, \dfrac{\pi}{6} \right)\)

Hint:

To test increasing/decreasing behavior, compute the derivative and test sign in sub-intervals.

Answer 1

The Correct Option is C

Let \(f(x) = \sin x (1 - \cos x)\) Differentiate: \[ f'(x) = \cos x (1 - \cos x) + \sin^2 x = \cos x - \cos^2 x + \sin^2 x = \cos x + (1 - 2\cos^2 x) = \cos x + 1 - 2 \cos^2 x \] So, \(f'(x)>0\) in regions where this expression is positive. Use sign test in interval: it is positive in \(\left(-\dfrac{\pi}{3}, 0 \right)\cup \left(0, \dfrac{\pi}{3} \right)\)