${\sin ^4}\frac{\pi }{8} + {\sin ^4}\frac{{3\pi }}{8} + {\sin ^4}\frac{{5\pi }}{8} + {\sin ^4}\frac{{7\pi }}{8} = $
${\sin ^4}\frac{\pi }{8} + {\sin ^4}\frac{{3\pi }}{8} + {\sin ^4}\frac{{5\pi }}{8} + {\sin ^4}\frac{{7\pi }}{8} = $
- A
$\frac{1}{2}$
- B
$\frac{1}{4}$
- C
$\frac{3}{2}$
- D
$\frac{3}{4}$
Similar Questions
જો $\frac{{5\pi }}{2} < x < 3\pi $,હોય તો $\frac{{\sqrt {1 - \sin x} + \sqrt {1 + \sin x} }}{{\sqrt {1 - \sin x} - \sqrt {1 + \sin x} }}$ =
જો $\frac{{5\pi }}{2} < x < 3\pi $,હોય તો $\frac{{\sqrt {1 - \sin x} + \sqrt {1 + \sin x} }}{{\sqrt {1 - \sin x} - \sqrt {1 + \sin x} }}$ =
$\left( {1 + \cos \frac{\pi }{8}} \right)\,\left( {1 + \cos \frac{{3\pi }}{8}} \right)\,\left( {1 + \cos \frac{{5\pi }}{8}} \right)\,\left( {1 + \cos \frac{{7\pi }}{8}} \right) = $
$\left( {1 + \cos \frac{\pi }{8}} \right)\,\left( {1 + \cos \frac{{3\pi }}{8}} \right)\,\left( {1 + \cos \frac{{5\pi }}{8}} \right)\,\left( {1 + \cos \frac{{7\pi }}{8}} \right) = $
- [IIT 1984]
$\frac{{\sin 3\theta + \sin 5\theta + \sin 7\theta + \sin 9\theta }}{{\cos 3\theta + \cos 5\theta + \cos 7\theta + \cos 9\theta }} = $
$\frac{{\sin 3\theta + \sin 5\theta + \sin 7\theta + \sin 9\theta }}{{\cos 3\theta + \cos 5\theta + \cos 7\theta + \cos 9\theta }} = $
$\sqrt 2 + \sqrt 3 + \sqrt 4 + \sqrt 6 = . . ..$
$\sqrt 2 + \sqrt 3 + \sqrt 4 + \sqrt 6 = . . ..$
- [IIT 1975]
જો $\sin \left( {x + \frac{{4\pi }}{9}} \right) = a;\,$ $\frac{\pi }{9}\, < \,x\, < \,\frac{\pi }{3},$ થાય તો $\cos \left( {x + \frac{{7\pi }}{9}} \right)$ =
જો $\sin \left( {x + \frac{{4\pi }}{9}} \right) = a;\,$ $\frac{\pi }{9}\, < \,x\, < \,\frac{\pi }{3},$ થાય તો $\cos \left( {x + \frac{{7\pi }}{9}} \right)$ =