2 sin(frac{pi}{8}) sin(frac{2pi}{8}) sin(frac | vedclass

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

Question

(C) ધારો કે પદાવલિ $E = 2 \sin(\frac{\pi}{8}) \sin(\frac{2\pi}{8}) \sin(\frac{3\pi}{8}) \sin(\frac{5\pi}{8}) \sin(\frac{6\pi}{8}) \sin(\frac{7\pi}{8})

Vedclass · Accepted Answer

$\frac{1}{8}$

Vedclass · Answer

(C) ધારો કે પદાવલિ $E = 2 \sin(\frac{\pi}{8}) \sin(\frac{2\pi}{8}) \sin(\frac{3\pi}{8}) \sin(\frac{5\pi}{8}) \sin(\frac{6\pi}{8}) \sin(\frac{7\pi}{8})$ છે.ગુણધર્મ $\sin(\pi - 	heta) = \sin 	heta$ નો ઉપયોગ કરતા,$\sin(\frac{7\pi}{8}) = \sin(\frac{\pi}{8})$,$\sin(\frac{6\pi}{8}) = \sin(\frac{2\pi}{8})$,અને $\sin(\frac{5\pi}{8}) = \sin(\frac{3\pi}{8})$ મળે.આ કિંમતો મૂકતા,$E = 2 \sin^2(\frac{\pi}{8}) \sin^2(\frac{2\pi}{8}) \sin^2(\frac{3\pi}{8})$.$\sin(\frac{2\pi}{8}) = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$ હોવાથી,$\sin^2(\frac{2\pi}{8}) = \frac{1}{2}$ થાય.તેથી,$E = 2 \cdot \frac{1}{2} \cdot \sin^2(\frac{\pi}{8}) \sin^2(\frac{3\pi}{8}) = \sin^2(\frac{\pi}{8}) \cos^2(\frac{\pi}{8})$ (કારણ કે $\sin(\frac{3\pi}{8}) = \cos(\frac{\pi}{8})$).$E = (\sin(\frac{\pi}{8}) \cos(\frac{\pi}{8}))^2 = (\frac{1}{2} \sin(\frac{2\pi}{8}))^2 = (\frac{1}{2} \sin(\frac{\pi}{4}))^2 = (\frac{1}{2} \cdot \frac{1}{\sqrt{2}})^2 = \frac{1}{4 \cdot 2} = \frac{1}{8}$.

$2 \sin(\frac{\pi}{8}) \sin(\frac{2\pi}{8}) \sin(\frac{3\pi}{8}) \sin(\frac{5\pi}{8}) \sin(\frac{6\pi}{8}) \sin(\frac{7\pi}{8})$ ની કિંમત શોધો:

Similar Questions

જો $(1 + \sin A)(1 + \sin B)(1 + \sin C) = (1 - \sin A)(1 - \sin B)(1 - \sin C)$ હોય,તો દરેક બાજુ કોના બરાબર થાય?

જો $\sin \theta + \sin \phi = a$ અને $\cos \theta + \cos \phi = b$ હોય,તો $\tan \frac{\theta - \phi}{2}$ ની કિંમત શોધો.

$\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ} = $

જો $\tan x = \frac{b}{a}$ હોય,તો $\sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}} = $

જો $2y \cos \theta = x \sin \theta$ અને $2x \sec \theta - y \csc \theta = 3$ હોય,તો $x^2 + 4y^2 = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module