$\sin ^4 \frac{\pi}{8} + \cos ^4 \frac{3 \pi}{8} - \sin ^4 \frac{3 \pi}{8} + \sin ^4 \frac{5 \pi}{8} + \cos ^4 \frac{7 \pi}{8} - \sin ^4 \frac{7 \pi}{8} = ?$

  • A
    $\frac{1}{4}$
  • B
    $\frac{1}{2}$
  • C
    $0$
  • D
    $\frac{3}{4}$

Explore More

Similar Questions

If $y = \log_e \tan \left(\frac{\pi}{4} + \frac{x}{2}\right)$,then $\tanh \left(\frac{y}{2}\right) = $

The value of $\cos 1200^{\circ} + \tan 1485^{\circ}$ is

The expression $\frac{\tan \left( \frac{3\pi}{2} - \alpha \right) \cos \left( \frac{3\pi}{2} - \alpha \right)}{\cos (2\pi - \alpha )} + \cos \left( \alpha - \frac{\pi}{2} \right) \sin (\pi - \alpha ) + \cos (\pi + \alpha ) \sin \left( \alpha - \frac{\pi}{2} \right)$ when simplified reduces to:

Difficult
View Solution

$\frac{\sin ^2(-160^{\circ})}{\sin ^2 70^{\circ}} + \frac{\sin (180^{\circ}-\theta)}{\sin \theta} = $

If $\left[1-\cos \left(\frac{\pi}{2}+\alpha\right)+\sin \left(\frac{3 \pi}{2}+\alpha\right)\right]^2+\left[1-\sin \left(\frac{3 \pi}{2}-\alpha\right)-\cos \left(\frac{3 \pi}{2}+\alpha\right)\right]^2=a+b \sin ^2\left(\frac{\pi}{4}+\alpha\right)$,then $a^2+b^2=$

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