Which of the following statements is/are correct for $0 < \theta < \frac{\pi}{2}$?

  • A
    $(\cos \theta)^{1/2} \leq \cos \frac{\theta}{2}$
  • B
    $(\cos \theta)^{3/4} \geq \cos \frac{3\theta}{4}$
  • C
    $\cos \frac{5\theta}{6} \geq (\cos \theta)^{5/6}$
  • D
    $\cos \frac{7\theta}{8} \leq (\cos \theta)^{7/8}$

Explore More

Similar Questions

If $\cos 2B = \frac{\cos(A+C)}{\cos(A-C)}$,then $\tan A, \tan B, \tan C$ are in

$3\left[ \sin^4\left( \frac{3\pi}{2} - \alpha \right) + \sin^4(3\pi + \alpha) \right] - 2\left[ \sin^6\left( \frac{\pi}{2} + \alpha \right) + \sin^6(5\pi - \alpha) \right] = $

$\tan 81^{\circ}-\tan 63^{\circ}-\tan 27^{\circ}+\tan 9^{\circ}$ equals to

The equation $y = \sin \,x \sin \,(x + 2) - \sin^2 \,(x + 1)$ represents a straight line lying in

For $0 < \theta < \frac{\pi}{2}$,the solution$(s)$ of $\sum_{m=1}^6 \operatorname{cosec}\left(\theta+\frac{(m-1) \pi}{4}\right) \operatorname{cosec}\left(\theta+\frac{m \pi}{4}\right) = 4 \sqrt{2}$ is(are):

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo