cos 20^{circ} cos 40^{circ} cos 60^{circ} cos | vedclass

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

Question

(D) આપણી પાસે છે,$\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ} = \cos 20^{\circ} \cos 40^{\circ} \left(\frac{1}{2}\right) \cos 80^{

Vedclass · Accepted Answer

$\frac{1}{16}$

Vedclass · Answer

(D) આપણી પાસે છે,$\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ} = \cos 20^{\circ} \cos 40^{\circ} \left(\frac{1}{2}ight) \cos 80^{\circ}$ $= \frac{1}{2} (\cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ})$ સૂત્ર $\cos A \cos 2A \cos 4A \dots \cos 2^{n-1}A = \frac{\sin(2^n A)}{2^n \sin A}$ નો ઉપયોગ કરતા,જ્યાં $A = 20^{\circ}$ અને $n = 3$: $= \frac{1}{2} \left[ \frac{\sin(2^3 	imes 20^{\circ})}{2^3 \sin 20^{\circ}} ight]$ $= \frac{1}{2} \cdot \frac{\sin 160^{\circ}}{8 \sin 20^{\circ}}$ $= \frac{1}{16} \cdot \frac{\sin(180^{\circ} - 20^{\circ})}{\sin 20^{\circ}}$ $\sin(180^{\circ} - 	heta) = \sin 	heta$ હોવાથી,$\sin 160^{\circ} = \sin 20^{\circ}$ થાય. $= \frac{1}{16} \cdot \frac{\sin 20^{\circ}}{\sin 20^{\circ}} = \frac{1}{16}$

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

Similar Questions

જો $a \sin^2 \theta + b \cos^2 \theta = c$ હોય,તો $\tan^2 \theta$ ની કિંમત શું થાય?

$\frac{e^{4x} + e^{-4x} + 14}{4(e^x - e^{-x})^2} = \dots$

$\tan \frac{\pi}{5}+2 \tan \frac{2 \pi}{5}+4 \cot \frac{4 \pi}{5}$ ની કિંમત શોધો.

પદાવલિ $\frac{\cos 6x + 6\cos 4x + 15\cos 2x + 10}{\cos 5x + 5\cos 3x + 10\cos x}$ ની કિંમત શોધો.

પદાવલિ $\cos^2 \phi + \cos^2(\theta + \phi) - 2 \cos \theta \cos \phi \cos(\theta + \phi)$ એ

Vedclass Test Series

Exam Paper Generator

Online Exam Module