$\cos^3 110^{\circ} + \cos^3 10^{\circ} + \cos^3 130^{\circ} = $

  • A
    $\frac{3}{4}$
  • B
    $\frac{3}{8}$
  • C
    $\frac{3\sqrt{3}}{8}$
  • D
    $\frac{3\sqrt{3}}{4}$

Explore More

Similar Questions

$\tan \frac{2 \pi}{7} \cdot \tan \frac{4 \pi}{7} + \tan \frac{4 \pi}{7} \cdot \tan \frac{\pi}{7} + \tan \frac{\pi}{7} \cdot \tan \frac{2 \pi}{7} = $

$\left(4 \cos ^2 \frac{\pi}{20}-1\right)\left(4 \cos ^2 \frac{3 \pi}{20}-1\right)\left(4 \cos ^2 \frac{5 \pi}{20}+1\right)\left(4 \cos ^2 \frac{7 \pi}{20}-1\right)\left(4 \cos ^2 \frac{9 \pi}{20}-1\right)=$

If $x + y = 3 - \cos 4\theta$ and $x - y = 4 \sin 2\theta$,then:

Given that $\pi < \alpha < \frac{3\pi}{2}$,then the expression $\sqrt{4\sin^4 \alpha + \sin^2 2\alpha} + 4\cos^2 \left(\frac{\pi}{4} - \frac{\alpha}{2}\right)$ is equal to

Difficult
View Solution

If $\pi < \alpha < \frac{3\pi}{2}$,then $\sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}} + \sqrt{\frac{1 + \cos \alpha}{1 - \cos \alpha}} = $

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