{sin ^4}frac{pi }{8} + {sin ^4}frac{{3pi }}{8 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) આપેલ પદાવલિ: $S = {\sin ^4}\frac{\pi }{8} + {\sin ^4}\frac{{3\pi }}{8} + {\sin ^4}\frac{{5\pi }}{8} + {\sin ^4}\frac{{7\pi }}{8}$$\sin(\pi - 	het

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(C) આપેલ પદાવલિ: $S = {\sin ^4}\frac{\pi }{8} + {\sin ^4}\frac{{3\pi }}{8} + {\sin ^4}\frac{{5\pi }}{8} + {\sin ^4}\frac{{7\pi }}{8}$$\sin(\pi - 	heta) = \sin 	heta$ હોવાથી,$\sin \frac{{7\pi }}{8} = \sin \frac{\pi }{8}$ અને $\sin \frac{{5\pi }}{8} = \sin \frac{{3\pi }}{8}$ મળે.તેથી,$S = 2\left( {\sin ^4}\frac{\pi }{8} + {\sin ^4}\frac{{3\pi }}{8} ight)$.$2\sin^2 	heta = 1 - \cos 2	heta$ નો ઉપયોગ કરતા,$S = 2 \left[ \frac{1}{4} (1 - \cos \frac{\pi }{4})^2 + \frac{1}{4} (1 - \cos \frac{{3\pi }}{4})^2 ight]$.$S = \frac{1}{2} \left[ (1 - \frac{1}{\sqrt{2}})^2 + (1 - (- \frac{1}{\sqrt{2}}))^2 ight]$.$S = \frac{1}{2} \left[ (1 - \frac{1}{\sqrt{2}})^2 + (1 + \frac{1}{\sqrt{2}})^2 ight]$.$(a-b)^2 + (a+b)^2 = 2(a^2 + b^2)$ નો ઉપયોગ કરતા,$S = \frac{1}{2} \cdot 2 \left[ 1^2 + (\frac{1}{\sqrt{2}})^2 ight]$.$S = 1 + \frac{1}{2} = \frac{3}{2}$.

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

Similar Questions

કિંમત શોધો: $(\cos 252^{\circ} - \sin 126^{\circ})(\cos 252^{\circ} + \sin 126^{\circ})(\sin^2 126^{\circ} + \sin^2 186^{\circ} + \sin^2 66^{\circ})$

સૌથી નાનો પૂર્ણાંક $n$ શોધો જેથી $\frac{1}{\sin 45^{\circ} \sin 46^{\circ}}+\frac{1}{\sin 47^{\circ} \sin 48^{\circ}}+\ldots+\frac{1}{\sin 133^{\circ} \sin 134^{\circ}}=\frac{1}{\sin \left(n^{\circ}\right)}$ થાય.

$\frac{\cos 13^{\circ}-\sin 13^{\circ}}{\cos 13^{\circ}+\sin 13^{\circ}}+\frac{1}{\cot 148^{\circ}}$ ની કિંમત શોધો.

જો $\cos x + \cos y - \cos (x + y) = \frac{3}{2}$ હોય,તો

પદાવલિ $\cos^2 \alpha + \cos^2(\alpha + 120^\circ) + \cos^2(\alpha - 120^\circ)$ ની કિંમત શોધો.

Vedclass Test Series

Exam Paper Generator

Online Exam Module